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i Solutions Manual forNucle ar and particle physics

W. S. C. WilliamsDepartment_ gf physics, [Jniversity of Oxford

and St Edmund Hall, Oifora

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In general, no answer is given that may be found in the parent text. Inaddition, referrals to chapers, sections,'tablef lna so on impty that text.Most numerical answers are given'to ttr"" significant figures even,,l.o.lqh

r probtem may eithergivJth;;"i;";;;re initiat quantity with aIesser precision or ask only foi an estimate.The varues of constants used have ur"oir,otr found by the directionsgiven in Appendix B of the p.t nt t"rt. eufgiir"t number is missins: rhcinternationa'y agreed vatue-to- four stc"ifi;;;ft;r"rl; ;-d;;;n,ri#i:The 'mol' is the sram more. Normafit;.-il'ilft system of units is usedwhich-mearS trralthis n*u".r,.r'io6;;;; ffi:or, x 1d6; this has bcendone frequi|.ndy and witnout comment.I ne value of e is the magnitude of the electronic charge. In some places

::l:I: tsed.energy ili _.is r"i"*i""e""'iiy, used mass nC andrnomentum pc instead of m.and p, or given i"ro.i1ig, io unit :omi*Xil"^. .o,,,."

^_1_1":f.UJems require the atomic ,fi"rr., "ip"rticular

isotopes. Theseare not given in the orincipal text so in each ;;;; have taken the mass tobe z{ unified atomic massunits.The definitive questions are those. of the reprinting of the fint edition.Four misprints have been corrected in tt

" ueriions given here. one omis-sion is.also-corrected: the wavefunctt"";""#bt probrem 3.g is now thenormalized wavefu nction

OxfordJanuary 1994 w,s.c.w.

Solutions to exercises in Chapter 1

Exercises

Try the following problems, keeping an eye on Sections ,|.2 and ,|.3, in order to rake rheunfamiliarity out of tho Rurherford scattering forrnrl.. -

1 '1 Csrcurate tho cross-section for the scattering of a 1 0 MeV c.particro by a gord nucreus(Z-79, A-197) rhrough an angta greater thania) io:, iif zo., (c) 30.. Negte* nucrear

1.2 For the same circumstance as in probrem t ,i carcurate the differentiar scanering cross-section da/do fmt sr-r for scattering at I 0.,

1,'3, .lr Rutherford scattering formura has ths properry thar as the angre of scatter, d-0..dold0-q' Tho practicar circumstances are that the incident parricro wiil be scanered byths whol6 .a^rom. Suggesr what happens to ddldO a; i_dln ,n,, ..rr.

1'4 show that the distance of crosest approach d, in Rutherford scattering leading to anangle of deflection 4 is given by

d =t0 +cosec 012).

whoro p is dofined in Fig. 1.6.IUse the consorvation of energy and angula, momentum.l

Solutions1.1 Use the formula in Table 1.2 for a(d > O):

oQ > a) =tz!,22 r+)', (*\ ' "oe ep .' 4 \atr€o/tc/ \ f i

Now Z - 79, z = 2, ?,= 10 MeV. For O = l0o:

o(d> lo ' ) - 1792 '22 1- | \2 /197.3 \2-l-- (tm,i/ (fr/. cot,s" f.,,= 5.31 x l0{ fm2 .

sinilarly for g = 20o, o = r.3r x l0r frn2, and for o = 30", o = b.66 x r03 fmz .tl3

latue of.4 for gold (u9) is not required but iudicates that the target nucreus is heavyand the recoil efects, which have beeo oeglect.d, Jiil. ,*"rr.

1.2 Use the formula for (da/d0) from Table 1,2:

rFh

do(O) _2r", I lgz.3 r2d=T(#ii#)-.**.f .Z = 79,2 = 2,7 = l0 MeV, d = 10. gives

do;5

(10") = 5.60 x 105 fm2 sr-r

d 2 - p d - b 2 = 0 ,

a= )e r *1@+76 ,1 .

But d can only be positive, therefore

2 Chapter I

. ' $ : ,. lt:,r.3 As d approaches r.i?^r:jll]iryt pa.rimeter, varies as d_r. This impliesl,ihatfor collisions expected to give decreasiog,.r^dr

"ogl; ol scatter, the distance of crolest

:i,::::f;Tffijil,:':,::9 the nucreaicharge;;;* rel gnecqive as it rs in*easirigrycross_secti on is m uch, "" ii:i'ji:i,lT[ I*, l"Jjnl, *il"lll;i# lfr.Xeppears electrically neutral ro e passing;;rr" ;;; etrd rhere is !o scatcerlog.1.4 In Fig, L5, Cho distnnco ofclocort nppronclr lo d. oBoforo tho colllslon tbo lnclclont d,il.j;],*

't dccurtlng nt polnt D. wo hovolklnotlc oncrgymechaaical iotentiatenergr

-!- tr*o',

angular oroa.otunr about O - 0,

At D it has: = mub.

kinetic energymechanical iotentiat energy

= *-*y'.,

, augutar momentum about o = Zze2.f 4nesd', I

Coaservation of "ogul".

-or.ot-u. giuo mab =X!,

:

Conservation of totlErimiaating r and usins rr. ;I1;;:"

I maz = ! iu' + (zze2 / nesd) .

' l / - ,o = r P ( l + y ' I + b r l p r ) .

I

o= rP

Using Eq. 1.4 gives

(' *..*. f)

Chapter t 3

' Exercler

*i;"1.5 DEfine the differentiar cross-section dalde in a scanering fuess, oerive Rutherford.sformuta. for the scanering cross-section ot c-parrictes iiliJitreavy riucteus wil;ilJ;

'nt whh kinetic energv r on a

where e is the erectronic charge and.d is the scatterrng angre..iindar what conditions dodeviarions from lhis formuta occur? lilusrrato y"ul;ffi;;y.discu;in; iil;.r"..rrg ;t15MeV c'partrcies bv rhin targs* of atuminrrr f.io.i. number z-13, atomrc weight

ffilf;.0 n.to (z-79, A-197).Assume the nucrr;;rbJ;; B is siven approximatory by the

R - l . 2 l r l r x l 0 - r r m .

nifiit fiom thc 1 988 cxomlnltlon In rhc Honourr schoot oiiNrrurrt sctrno, phyrlc|, unlvcrlty o,

1,0 show thot thc Ruthcdord dlttofontral ,olttorrng crorr,rootron tormurr oen bc wrlttcn rntorms of the aquared momontum transfer f as

o !a.-4xZ'z'a'(hc)z

t oQ. ft.tt

whore 4 is the fine'structur€ constant and u is the verocity of the dofrocted particre.

1.7..1f rhe,rarge-ans.re scatters obssrved by Rutherford had not been the effect of singrescattering but or murripre scanering lsee p. 's1,

tro* *oufiine yiord at fixed anlre *" *i,r,ths. thicknoss of ths target materiai? In the formei iise, negrecting attenuarion of rheincident beam, tho yierd wiil increase rinearry *if, rntrir*. what wourd be the depen.donce for multiple scattering?

fconsider a random wsrk probrem: each smal scattor is rike a hop on the surface of a spher,.starting at a pore, n randomry directed smarr rr"p, *iri iri, you a disrance from the pore.That distance wi[ vary with each trial of n hops but it tras an avorsge ovo, msny such trials.How is the msgnitude of that average distanco expecred io depend on n? The connectsdideas are those ot diffusion and Brointian ̂ oiiii.i--'--

Solutions1'5 see page 35 for a definition of the differential angurar scatteriug cross-section. Flompages 6 to 8 aud Thble 1.2 we have

do(0) | zze2 \2 , 0-an- = \iGA/

cose,'i '

For a-particles, z = 2, eo that

#F= ('n;Hi'eo';'

4 Chapter 1

#=(#*"'f)'Deviatioas wiU occur.if ,I..d:*Tt of_closest approach is less thau the nuctear

,:1,: !X:f i f.Ti l*:t:*k* r,.,. pJiir,J .rosest possibre approach occursf,k-,T-t, ,:L.l g":l ttiq. l.q. ir*llp . n;.;1;'rJ;::"Hil:T.TT:Rutherford fououla for tle scatteriui.rorr r..,ioo, ;.;r;;"fi;;:'fi'::.lt:T':ffwill lead to deriatiqps ffon the inveie souare t"*'f- C^,,,^-h rnra^. rrr^-.will lead to deriatiqps ffon the iou.i..qu*. t"*'f* il;#fi:'-:;

: i -., p

.= Zze2/4reoT = Zzahc/T

v- l 197.3 \ L- L . i - t - f h\ t32.04l ?MeV l - .

AIso, ueglecting the radius of tbe o-particle, the efiective R is given by

'slncs p < ,? for alumlnrum lut not for 6ord, dovratoas from Rutherford,e formutawill be expected from the former but not tho iatter. - -t

l|131,.r.i,* llu:t*T,?f j:pnrrtctor ln tbo iiocrrocrottc 0old of rho nucloua or tra:l*:,.y_qlf lljy r!!p.

"rJ.ry ii il.v;:ilffi;;'';T il::ili::: :l lTaluainium nucleus but not that ofa jold ou.irurl

:P3:*:,.": -"-:,rl.ij iyg,...."rlC1. conect Jalculation musr be done in the:fi1,:iij-ilT^:j,r,Tocouisiou

(see section ?.3). F;;;:;il;-;;;ilr;ffi;:'.:,".:;;in tbet frame ia

27frAts MeV = 12'14 Mev '

conclusion remains that the o-particles caa reach the aucrear surface.

Flom Table 1.2

from Fig. l.Z#=(ffi)'.o.,.'f,'

q=2Psinf * ,U.ru p=f f i , T=f ,mvr .

{2=8m?ein2f , aua cosec, i=ry

dq2 = 8m?sin d dd ,

d Q = 2 1 s i n 0 d d .

.R=1 .2 / l fm . I

RPotentlal energy of a-perticleat uuclear eurface

Chapter 1 5

Now

do da dO dd- = _ _ _dq' dn d0 dq2

= (#)'""",orznrior r#ffi

1.7^ Tbe motion of a single ga.s molecule is an example of a random walk. The rength oleach step is determiaed in a random way by the morecurar speed and by the time io thenext collision, Neglecting persistance of velocities, tbe direction of each step is raudomlyoriented with respect to the direction of the previous step. In these circumstances, overmany tri{s of a walk lastiug a total fixed time r, the mean square displacement of tbemolecule DT is proportional to the time l, multiplied by the ,o.uo ,qu"r. of the distancebetween collision F. If there is a mean velocity then the number ofcollisions is proportionalto t and hence the root mean square (r.m.s.) displacement is

(D')tt' x @P)l .Thus the r.m.s. displacemeDt over many trials is proportionar to the square root of thenumbcrofcol l isionsinthcwotk.(Joos,G.(l99l). Thiotvt icu! l ,hysict(2it<Jodn), l l luckic,Londoo,)

. Multlplo rcottorlng bohovcs ltr tho somc wny rlncc succcr.rcivc collisions hovc uncor-rcloted anglce of ecotter. Thc r.m.s, uglc of scaitc, aftc* collisious is proportional to1np-2)l where d-2 is thc mean 'quare of the scattering augle in a single collision. sincen is proportional to the thickness of materiar traversJd by the particle, the r.m.s. angreincreases as the square root of the thickness.

Rutherford was able to ehow cxperimentally that thc ,urnbcr oflurgc anglc sc&ttersfor a-particles increascd as thc lhickncss and could not thcrcforc bc due to the ellect ofroultiple scattering.

.4I

Solutions to exercises in Chap ter 2

Solutions2,I The (mechanical) potential-energy of two pointJike positive chtges Zp a:ad..Z2e,separated by a distance d is U, given by:

,_Zf izez _ZrZzahc

This applies equallv to t*o s113.icailJ1i,0*netric clarge disrributions, centres separatedby a distance d, provided the charge distriuutio* aoTot overlap. The latter is the case(unrealistic) in this question. In oui nuclear units:

U(MeV) = z,z, 197'3 |u 7 . 0 4 d f m '

2.1

(a)

(b)

(c)

2 .2

Exerclses

calculat€ the potontial energy in MeV due to the couromb repursion oftwo protons separarld by a distance of 1 fm;a gofd nucleus (Z=79) and an a_panicle (Z=2) wirhtheir cenlres lOfm apart;two nuclei Z=46, A=115, radius g=,|.2x4,/rfm, just not touching.

An investigation of the mean rife of the K'-moson yrerded 20 evenrs of the docay inode

K ' < p ' * v " ,

in which the time intervar between the arrivar of the K.-meson in a detector and the decaywas determined, Estimato lhe mean rife of tho K.-meson iroi ttrese m.asuraments (timo in

2 '3 Inasampreof ons l i r reo f carbond iox ider i t . s rpanavorageof 6d is in tegra t ions ,

r lC - ' f N+g-+9.

,lirlXr.:lr.rr,g#ure. Catcutate rhe atomic fracrion of rlc presont lf the mean tifa of,r

i

,,#-0"' ..,t ., :, l , 1. ^:. -6. ? - I s ro31.r 2.1 2,1 03 3C.4 i.t i.t i.t i.s ,f,..o

1 2 1 3 t 416 ,2 19 ,0 1 1 ,2

1 5 1 6 1 7 1 8 1 95,3 15,7 6,1 13,0 3,0

Then

--:' Chaptet 2 7

(a,) 21= /, = 1, d = I fm: U,= 1.44 MeV,( b \ 4 = 2 , 2 2 = l S , d = 1 0 f m : U = 2 2 . 7 M e Y ,

' (c) 21- 22 = lg, il = 2 x 1.2 (115)l fn: U = 261 MeV.

2.2 Thebest estimate of the mean life is to take the average of the 20 observed lives:this gives 9,9 ns. In practice the equipment may, for example, be uaable to measure veryshort lives and may miss very long lives; these effects bias the result and would have to betaken into accouut in a precision measurement.

2.3 One litre of COz at S.T.P. contains a number (N) of carbon atoms equal to Avo-gadro's nuobe! (6.022 x1023 mol-t) divided by 22.4 litres mol-r. The rate of dp,cay ofthe uC is given by

dNrrT

= -''"t'i :

whcre Nrr is the number of lrO atoms present end o ls tbe l'O decay transltlon rate. But(dNrr/d4 = -5 per minute. Thus Nu is given by \fu, where ar-r = 8267 years' Usingminutes for the time unit this gives

5 x 8 2 6 7 x 3 6 5 . 2 5 x 2 4 x 6 0-Ir;t-x22.48.09 x 10-13.

i

Solution i2.4 The mean life of the K+ meson is 1.237 x l&E s = r. The total decay transitionrate is o * r-L = 8,084 x 10? s-1. Multiply by a branching fractiou to obt&in the partial

Nrr- =N

Tha deQay of the K' meson

Th6 msan lile of the K'-meson is 1 .237 x 10-'s.

K ' - l l ' + y , '

K | a r l * lC , lK r i 1 1 . + r i + l t o

K ' r t ! ' + t t o+no '

,( ' * 1 ' + p ' * v , rK ' + Z + f c ' * Y -

8 Chapter 2 j:

transition rate for that mode. Thus for K+ + p++ z (aalled Kr2)

u(Kr2) = 0.635 x 8.084 x 107 s_r

Aad so on for the other modes.

= 5,13 x 107 s-l

Exercise

sample of gold is axposed to : '

ns por socond ara absorbed ,ll ,tr:::::lllttm of constant intensitv such rhar 10,0

t l lAu+n ' l r4uar.

Ih:lilli'.il1il'#:,".n{,i",?'o:;.'31',:#:H.ff Tl,!"3sedavsHowmanypresont at that time. assuminoequiribrium number or,,,o, ;,T;:;n''d;tlJff;:il ;ffitrJJ',HT:iiriif.Xlii:

ffii5t from thc 1974 erami*ion in rhe Honou. schoor of N'urar science, physics, University of

Solution t .

?'5 Ft /v(r) : nunbei*r:rrl nucrei preseur ",

,,-. J,'10* the rate of ross due tooecav is rulv(r) where ' "rgu a.."r-tt"or't;fi #J ;r the dpcav of this nuclide. If p israte of production (assung.o*t*t; ;;;.-""..

' i lt alyrrr *

This bas the aolutlon i ' ,

DN(t) = : (t - ocp(-r..rt)).

l.Ve have t = 6 days, r.r-r = 3,g9 days and p = 1gl0 r_r. SuUstituting with consistent units&ves

N(6 days) = 2.64 x l0r5 .This is' in fact' the number produced less the aumber decayiug in 6 days. t[g arrmls'produced is Pt = 5.1g x rotr. tt*.. tt. ;;;;;ff;". decayed = 2.s4 x lors ani this,tD turn, is equat to the amount

$ ,".It.ilil;;:'end

of 6 days. The equillibriumnuober /V(oo) of DrAu occurs wten arv[r=_i, il"'r'i *o.o I ie very large and

heace P=lf(oo)r.r,

,

N(o)=#=t.ru ,*".,.,1

r t

$t .

Chapter 2 I

Exercise

2 .6 ,138 i (mean l i f eT .2days )decaysby / -pa r t i c l eem iss i on to ' l l l Po (mean l i f e200davs ) ,

which in lurn decays by c-particle emission to'l3tPb. lf a source initiallY contains only pute,l3Bi, atter how long will the rate of a-particle emission reach a maximum?

(Adapled from ths 1960 sxaminstion in the Honours School of Natural Science. Physics, Univetsity ofOxford.)

Solutiou2.6 The decay scheme is:

! l

rro6r Number at time t = N,(t)6or Transition tate = o.

r!9pa Number at time t = Nr(t)x' v Transltion rate = @r

tPb Number at time t = Nr(t)

,l M. /r\

?=-<r r Nr ( t ) ( l )

tit

Now

dNrl;(t) = +orNr(t) -u)2N2(t)o,

If n (t) is the rate of c-particle emission at time t we have

&(t) = u2 N2(t) .

Since N2(f) is always greater than or equal to zero, & is a maximum when

dNr(r)--a;- =' 'The solution to Eq. 2 is

Nr(r) = ry [exp(-ar2t) - exp(-ar1r)] ., , t r J l - O 2 ' "

This give the maximum in rt"(t) when

u2exp(-u2t) = ur exP(!r;t) ,

which gives

t.o = --l- 1o 3! .u L - U 2 U 2

Now arfl = ?,2 days and tr2-r = 200 days hence

! - r r = -* r r3o= 6;766i'"A

d&vs

= 24.8 doys ,

(2)

Exercise

2.7 Natural potassium has an 616mis \ rsight of 39.Ogg and contains O.Ol 1g atomic percenlof the isorope f!K, which has two decay modes:

ifK-fiCa+/ +9. (fdecay), i r

+ ffAr'* v. (clectron capturet scc Section 5.3) i| . $'--i!Ar+7.

where f!Ar' means an excited state of f!Ar. In this case this excited state decays to thegrounct state by eminino & single 7-tiy. rn. totai-iniensitv of /.panicles emitted is2'7 x 101 kg-r s-r ef natuiar potassiur rna onir,a ru.r.na,nrr" are 12 Trays emitted toevery 100 p-partictes emitted. Esrimate th€ mea;life;;;{K.

l0 Chapter 2

i f K + c -

Solution2.7 The decay scheme for !!K is

l3ca

There are 2.T xr'a 0- cmitted per second per k'ogram of naturar potassium. There are12,y-rays emitted per 100 p- so that the tot"t r"-t-r'oii.."y, i,

t . l2 x 2.7 x 1gr s_r 1*_r

= 3.024 x 104 s-r kg-r .

1l:_., *"tr-"re from llK which has ar atomic fraction in natural potassium of t.l8x l0-.. Therefore the number, .rr, of !$K "tonr

p; ;l;;am is given by

N = l . t8x l0-rx 5ff i "6.022x t023kg-r

= 1.818 x l02r kg-r .tl: *',r.o1.0*ays is -(drf/dt) = uN where ar is the transition rate. Tbe mean life isra-' and is therefore giveu by

, - r - 1 . 8 1 8 x 1 0 2 13.024 x tOa

= 6.012 x l0l6s = l.g0 x lOe years

' Exercises

2.8 Natural uranium found in the Earth's crust corftains the isotop€s 23!U and 23!U in tneatomic ratio 7.3 x 1 0-! to 1 . Assuming that at the timo of formation these two isolopes wereproduced. equally, estimate the lime since formation given that th6 mean lives are L03 x 10tand 6.49 i I 0t years respecti\/ely.

2,9 Chemical analysis of a meteorite shows that it contains 1 g of potassium tnd 1g-a g ofargon formed from tho decay of r'|K. Using the results of Problem 2.7 and assuming no argonhas escaped. find a value for the age of the meteorita.

(Adaptod,from ths 1 966 oxamination in tho Honours School of Natural Science, Physics, University ofOxford.)

2.10 Write down the law of radioactivity dacay. Define the half-life and mean life of aradioactive nucleus and obtain the rslation betweEn them.The nucleus llRb decsys into th6 ground stato of !!Sr, with a half.-life of 4.7 x 10!o years and

a maximum f-energy ot 2i2kav, Discuss briefly tho difficuhies you might sncounter inanampting to mcasuie thls half.life.Five different samples of chondritic ryteoritas are found to havo ths following proportions

of !fRb, $Sr and !lSr.

Meteorlter llRb/llSr llSt/llSr

Msdss olao o,?!7Homonoad 0,t 0,76,|Erudorhclm 0,72 '0,747Kyuehu 0.0 0.739Bath Furnaco 0.09 0.706

Given that tho nucteus !!Sr is not a daughter product of any long-lived radioactive nucleug,show that rhoss data are consistant with a common primordial tatio llSr/$Sr and a commonaga for all thEse moteorites and find that agp,

(!,dspt€d from ihe 1979 sxrminstion in the Honours School in Nstural Science. Physict, University ofOxford.)

2,1 1 G iven tnat the carbon dioxide of iiroblem 2.3 is from a sample of carbon which wasfixsd in a biological specimen when the tlC/tlC ratio was 1O-ir, calculate lhe age of thespgcimon. .

Chaptar 2 ll

Solutions2.8 We ass.ume that at some.time, t = 0, the two isotopes were produced in equal

, fDUmOeHt l

I N2rr(t = 0) = N236(t = 0) ,

I

cl..?i

12 Chapter 2 ',

with obvious lotation. By the radioactive decay law

, Nns(t) = &r(o) exp(<u23sr) ,

. '/Vrsr(| = &rc(o) exp(:la3gr) .

At the present time, t,

.- lVz35(t) =?x l0-s/Vro (r)

hence :Z x l0-3 = exp(_u2j5t)/expt_a.,46t) .

Given r.,'fi = 6.49 x lOe years eod rr* = 1.03 x lOe years we solve for l. The result is

t = 6.07x10eyears

= the estimated age of uranium,

2'9 To do this problem we have to assume that tbe potassium in the meteorite hasthe sa.e age as the potassiu,q roura * ti.-zuil. "frh

ailows us to use the terrestrialvalue of the {K fraction to.determine tb; ;; li .oK so.uiuiog in the meteorite.19_10".:* of t'apped,oA (plus iufenea;frjaei.r.ioo when the ,ootassiuo wastacorporated into the meteorite.

The l g of potassiuo.coirtains. (6 .022 xro', /gg-og9) atoms of potassium of which afraction 1.18 xl0-4 a^re.oK. fnus tle nunU.r.f'r"*i"f"S .rI( ";;rr;'-s

vr wu

iVr = 1.82 x l0lE .

The 10-5 g of '0Ar contains.(6.022 xro23,x 10{/40) atons (assuming an atomic weightof 40 for thislsotope). But tho decay of .r p.o'uiao io' atoms of .'ce for 12 of {oAr,Tbua tbo total numbor ofdocay producte ls

'

n2t r O = E

= l ,4l x l0rt ,By tbo radloactlvo docqy lrw

Nx = (Ne + Nx) exp(-ufi ,where I is the time the meteorite was formed and r.r is tbe decay transition rate found inProblem 2.?. Then

' = jr (No *Nr/Vx ) {

= 1.09 x lOe years.

Chapter 2 13

z,t0 The radioactive decay law is N(r) = N(0) exp(-rat). The half life, tr, occurs when

N(i i) = ( l /2) N(0), hence fr =(l /w) ln 2: mean l i fe x ln 2.' The rubidium decay scleme is

---\- SinO r= 4.7 x 1010 years

p- \\ szc,

3 8 e '

A measurement of a balf-life of tbis magnitude requires a determination of thenumber of decays per second from a sample of 87Rb of a known weight. (Table 2.4, (3)).This sample must be suficient to give a statistically significaot number of detectabledecays (see Section 2.11). It is understanding the detection efficiency which causes themajor problems. Relevant points are:

l. The kinetic energr spectrum of the elections emitted is continuous from zero to272 keV. (See Section 5.3).

2. The eficiency for the detection of such electroas depends critically on the behaviourof electrons of this energr range in the material of the source and detector.

3. The efficiency also depends on the geometry of the source and detector system.

4, Some of tbese efiects are very dificult to model so that an accurate calculatioa ofdetection efrciency may be impossible; subsidiary and calibratiug measulements Daybe necessary.

Age of chondritic meteorites. A constant amount of 665r is present in each meteorite.For a given meteorite, let a be the ratio of the number of atoms of 87Rb to tbat of 865r

and p be the ratio of atoms of 675r to 865r. These ratios are changing. Let o(0) aud p(0)be tbeir values wben this meteorite was formed, aud a(t) and d(t) their values at the timeof measurement. Applying the law of radioactive decay to the amount of t?Rb gives:

dg(t) = -ro(r)

a(l) = a(0) sxp(-q,r) ,

where t.r is tle decay transition rate for 8?Rb. lt followi that the accumulation of 675r isgiven by

p(t) = p(0) + o(0) (1 - exp(-cat))

= f(0) + o(t) (exp(u.rt) - 1) .

The data are o(t) and p(t) for five meteorites. If these five cases have the same p(0) andhave a common age, i, then a plot of p(t) agaiust o(t) will be a straight iioe with slope(exp(ot) - l), where u-t = tt/h2.

14 Chapter 2 , 1tj

0.76.'...--

0;75

0.74

nr., 0'73Pt(,,

0.72

ModocHomesteaduruo€meimKyushu

Bath furnace0.71

0.70

0.69t.00.80.60.4

a(t)

The points are sufrciently close to a stnight line to conclude that they h"u*,u .oo.ooage aud a common primordial ratio of "s./;rsr. fi;;ratio is p(0) and is tbe interc.eptwhea o(t) = 0; irs value is 0.2. The ,f""j h-i.oii srving I = 4.8 xlOe years for thecommo'. age. - ,

2'11 The atomic fraction:l l,a::

therample was found to be 8.09 xl0;r3; this is theresult of ageing from an initiar ratio of r.o "i-o-".

ih'. o'."o rife of .c is g267 years.The radioactive decay law then gives ;; -"g.rJ"

t = 8267 r (#) = lzso years

i t

Chapter 2 lE

Solutions i ' I

2,12 Tbe total cross-section (o) of zrsli 1o, treutrons of 0.29 eV is the sum of the givencross-sections, that is a = 2.7 x 10-26 m2. We give tbe attenuation as the fraction of the

r$', ,li

Exercises

2.12 A beam.ol neutrons of kinetic energy 0.29 eV, inlensity 't 0r s-i traverses normally a {oilof l3fU. thickness 1 0-t kg m-2, Any neutron-nuclous collision can have one of three possiblerosults:

(1) elastic scdtterlng of neutrons: r 12x10'30mt.(2) capture of the heutron followed by rhe emission ol a y-rcy by the nuclaus:

d.r7 x lo-tt m!,(31 capture ofthe neuton followed by splitting of nucleus into two almost equsl parts

. ( f iss ion):6r , r fx lQ'r i6z.

(a) the attonuation of.the neutrdn baam by tho foil;(b). the number of fisslon reactions occurring per second in the foil, caused by tha incident

beam; | -(c) the flux of elastically scattered noutrons at a point 10m frorh.the foil and oul of the

incident beam. assuming isotropic distribution of the scattored nEutrons.

(Adaptod trom lha 1970 ersmlnation of thr Honours School of Naturst Scioncs, Phydcs, Univ.rslty olOxford.) j

t2.13 Define what is maant by tho terrh 'cross.section'applied to.a nuclear reaction.

A tiquid hydrolen target of volums l0:' mt and density 60 kg m:! is immersed in a broad,uniform, ihonoenergetic beam of negativo pions, of intensity 10t particles m'2 s{. At a beammomentum of 300 MeV/cthe onlyreaction, apartlrom elastic scattering, which takes placeis

t!- +Piz'+n

whh a cross-section of 45 mb (4,6x 1O-s mt). (Tho nourral pion decays in the modo tto-21,with negllglbly Short lifetlme,)

Calculate tho n-umber of 7-rays emitted per second from the tafget.l .

(Adapr6d trom tns h7l bramination of lhe *onourc School of Nstursl Science, Physics. University ofOxford.) i2.14 Atargetofnatf ir{boron (aromiccomposit ion20%'g8,80%'lB) inthelormof athinfoilwhich has a mass pbr unit area of 1 kg m-, is traversed normally by a beam of neutrons ofkinetic energy 1 keV, The only significant noutron absorption is by the A-10 isotope forwhich the crol$-section has the value of 19.3 barns at I kev. This absofption laads to th€emission of arfc-panicld. The residual nucleus is lefi in its ground state in 30% of theroactions 6nd foi the remainder in an excited state at abo&t 500 keV from which it decays byrlngte l-ray emlialon to ihc gfound rtrto, Strto thc Zrnd A of th. r..ldurl nuclcur rndoaloullto thc ylolp of gemmo royr'whon tho bcom hil rn lntonrlty of l0r por rcoond, Th!offoot of clortlo roltterlng ol noutronr In the torgrt mry br n$l|oi.d,

(Adoptod ftom rhcrtgSS ox!mlnltlon ot rhe Ho^our. Sohool ol Nlturrl 8olcno., Phy.lor, Unlvo.rltV ofOxford,)

16 Chapter 2

iocideot bea,rn transmitted and that is exp(-naa), where .n is the number density of nucleilu the target and c is the thic&ness. This follows from Eq. 2.16 which is an integratedversion of tbe operational Case 2 in Table 2.2. In this case as is the super0cial targetdensity nultipligd by Avogadro's number and divided by the target atomic weight: thatis -.

, * * , -

This gives

10-rx6.022x1026n t = -

235I

exp(-noa)'= 0.9931 .The fraction, 0.69%, of the incident bea,m interactiug gives meau collision outcomes inthe retio of the cross-sections; thus the fraction that lead to fission is (2.0/2.2) so thatthe mean aumber of fissions in the ta.rget is

2.0 0.69, ' , r * . x 1 0 5 = S l l s - r .

The elastically scattered fractior is 2 xlo-to /2.7 x 10-26 of the number interacting whichgives 5.11 x l0-2 s-1. These neutrons are scattered uniformly in all direction, ia th.mean intensity at a distance of 10 m, through a surface facing the target is that dividedby 4n&, d = 10 m. The result is

4.02 x l0-5 s-r E-2 .

2,1g Cross-section is a meas,ue of collision probability between two particles. Classicallythe coucept is etraightforward. Tlansform to the frame in which one particle is at rest; themoving particle will suffer a collision with the particle at rest if the former's trajeooryintercepts some disc representing the area presented by the stationery particle: the cross-section is the srea of the disc. In statisticrtl situationg the concept is taken over to calculate,for example, the collision rate between particles in a gas in terms of a cross-section,In atomic aud subatomic physics the same ideas can be applied, and the operationaldefnitions ia Table 2.7 con be used. Applied to a nucrear reaction, a colision cross.section will hsve to be subdivided into partiel ooss-sections each defined by a specificoutcone of the collision.

- Th. numerical part of the problem is a csndidate for the case I operational defini-tion ofcross-section given in 1bble 2.?. The ta.rget coatains

atoms of hydrogen. Each proton, by case 1, has a probability of suffering a collision by anllldent r--mesou leading to zr0-meson production that is 4.8 ;16-ro [r * 19r r:zl-r.Thus the mea,l number of reactioas is 161 e-t. Each reaction produces oae r0 which decayswith a mean life of about l0-r6.s into 21+ays ltauie s.q. Thus the number of ,y-rayspro{-uced in the target is 322 s-t. since the chanie of a 7-ray interacting in hyd,rogen issmdl, this number can be taken as the number emitted per second fromlhe target. The'elculation has to neglect the beam atteuuation as it travemes the target since no targershape has beea given.

z'fa w9 u9e equations 2.r7 with )t = no,Now n is the number of roB nuclei per unitwl1n9 of the ta.rget. This translates into nc being the superficial mass density of l0Bnultiplied by Avogadro's number and divided by t-be etomic weight of this isotope. we

Chaptet 2 L7

teke the atomic weight of the isotopes involved to be l0 and 11. A superficial mass deusity

of I kg m-2 becomes (20/108) kg m-2 of r0B as the only effcctive target' Then

20 6.022 x 1026n"= i6T t - - I 0 - .

Tbe neutron absorption cross-section is 19'3 barns (= 1'93 x 10-27 m2) so the number of

absorptions is (Eq. 2.17)

105 (l - exp(-nor)) = 2129 s-r .

A fraction oI 70To of all other absorptions lead to a single 7-ray so the yield of 'y-rays is

1491 s-t '

n.+tfB + o +!Li 1or ]Li' +lli + ?) ,

so that the residual nucleus has Z = 3, A = 7.

The reaction is

Exerciges

2,15 A particle counler registers 8n avorage of 0.453 counts per second. What is the

probability that it registers 2 in any one second?' A largei counter rigisters 1 296 in tan minutas so that the best estimate of its rate is 2'1 6

per second: sslimate the error on this measured rate and the ptobability that this measurs'

mont is low by moro than the enor. Assume that the Poisson distribution P(n'm) lor large m

is approximated by a Gaussian distribution with mean m and standard deviation m!/2.

2 .16 Ana.par t i c leo f mass4MandSpeed vpassesnear thenuc leusof anatomof a tomic

n u m b e r Z a n d a t o m i c w e i g h t A ( > 4 ) . T h e c . p a r t i c | e h a s a n i m p a c t p a r a m e t e r b a n d i sscattered through an angle d' Derive the approximate relation, for small d'

Q= (Zd)l(4xeoMo'b).

A beam of a-part icles with speed r.2x10tms'r is incident normally on a foi l , thickness

l0-rm, of gold (z=79, A=197, density 1.9x', l0.kgm-r). Estimate the proport ion ot d-

particles suffering doubls scattering, where eact scalter is of at least 10'. in Vaversing the

foil, ',

(Ad6pted lrom rhe 1 970 Examinslion in lho Honouts school o.f Nalu,al science, Physics. univqrsity of

Oxford.) \

Solutions2.15 The context of the question implies that the counter is detecting Particles which

are firing the counter randomly at a mean rate of 0.453 s-1. The number of counts in

oou ,..oid will be governed by the Poisson statistics. ftom Table 2.8, the probability of

detecting n (integer) in one second when tbe mean is rn is

P(n,m) =t#

18 Chapter 2

with n = 2 and m = 0.453 tbis gives the probability to be 0.06s2. (Note that a realcounter may not be able to give a Poissoo distribution. It may be incapable of resolvingtwo particles closer in time tban some minimum; thus the probability of detecting n,particularly when n is large,-will be less than-thot given by the poissonformula.)

The best estimate of the rate is 2.16 s-r. The best estimate of the statistical error.on 1896 is y'T806 = 36. Tbis represents a fractional error oD rgso oi (ff96);. tleestimated fractional ergor\on the rate is the same giving an error on 2.16 s-r of b.oo s-t,Such a result is often expressed as 2.16 * 0,06 s-r

Assuming an approximately Gaussian distribution about the correct value for mauytrials, then the probability one measurement (trial) is low by at least one error is about16%. (This is an approximate value for the amount of a Ggussian below one standarddeviation from the mean.)

2 .16

Target nucleus, charge Ze

In the figure, the force on the o-particle when at p is

Zzez _ _.4neo(b2 +*) ' b - ' r

In a time 41 = (dx/a), where o is the a-particle velocity, this gives au impurse, transverseto o, of

Zzez dc b

@n; (0,+,,jTThis causes a small chauge in momentum trsrsverge to the path. The total change intransversc momentum AP1 is therefore given by

a P r = g [ r * d '

- = 4 2 e 2 ,Atreso l-* $2+*1t -

Anesab'

Thus the o-particle, which has a mbmentum 4Mu, suffers a small deflection d given by :

,=APr = , ' ! l=- 4Mu 4TcoMu2b

This calculation allows only small d to be contemplated since the impulse is calculatedassuming a straight trajectory.

The cross-section for scattering through an angle > d is

o(> o) = n[b(o)], = " [#*],

The probability that an a-particle scatters through an angle greater than l0o is (Eq. 2.lz)

P-X+dx,

P(, > l0') = I - exp [-na(d > 10")tl

. Chapter 2 L9

wheib I = 10-! m, tbe thic.kness of the foil, end n is the number density of gold atoms

ry xl.exro{m-3.

Tbe result is

is P(l,n) = In €-T = P(0 > 10') so that

rn = 4.65 x 10-2 .

f(' > 10') = 4'44x 10-2 ' ' '*" '

The scattering is a rendoia proceos 80 that tbe number ofeve-nla, e8.h with 0 > 10o, in asingle traversal of the foil, is given by the Poisson formula. The.probability of finding one

The probability of obtaining two scatters in a single traversal is.

P(2'm\=ry =1'03x10-3:

Note that this result is close to m2 l2.The reason is that the probability of one scatteringoccuring is rn and tbat, oD the average,. half the foil remains to be traversed. Thus givenone scatter tbe average probability tbat a second occurs is ml2 afi thus the probability

that two occur is rn2/2.

I. i-,.

u

Solutions to exercises in Chapter 3

Solutions3.1 The geaeral formula for the form factor is

F(c2) = * I I I alexp(iq.r/,r)dv,

where the total uuclegr charge ie

z"= t I ld, lav.For a spherically symmetric cherge distribution we have p(r) = p(r), and

ze = ur ll ,1,1,2 o, ,

F(a2) = ! r" t+r r@ ' 'z, lo !_, Jo o(r) exp(iq.r/,1) 12 dr dcosdd/ ,

where the polar axis is oriented {oog ql so that q.1 = grcosd. The integral over theazimuthal angle / gives a factor of Zr. Tie integrjover cos 0 gives

. F(q') = .fi lr- Albb(qr/h)/(qr/h)l4nr2 dr .

Exerclses

3.1 Show that. for a sphericalty symmetric cfurge distribution,

#1.,-.1-#,'where (d) rs tire mean 'quar' 0f the erectric charge distribution. [Hrnr: expand si n (cilh) inthe formuld in Table 3.l l

3.2 Show:h:j^lho torm tacror to, rh6 char06 distrtbution of modet I ts

, : ^or-3lsin(qalh):(q?!.!,)cos(qolh)l. . , : .

\ z ' G 4 w t '

3.3 Find ,lr.l:.T lacror for a charge disrribuildn . ,. , , , . { a * , \ , : ' : . :

| "Ii" P(r)'Poe'h/r.' . . .

3.4 An electron of momentum-33oMeV/c is sb.attered at an angle of 10. by a calciumnucleus. Assuming no recoir, find the momentum transfer and its ,rauceo'ae aiogriewavelength. Also carcurate the Mott differentiar cross.ssction (point.rike nucreus), and bywhat tactor it is reduced if the carcium nucreus (A=40) can be assumed to be reprosonted bymodef Iwith a-12Anlm.

Chapter 3 2l

Expanding sin (qr/h):

F(c') = ,L" I,* ̂n['- f *l (f)'.. .] an,'a".Including all factors, the first term gives 1. The second term is

/ n 2 \ r t t a

tH e lo* o{iao'^a') 'The secoad factor is the mean square radius of tbe charge distribution which we representby (r2). Thus

F(c\=r- #(r2)+o(q{)and

dF(c') | (r')

ll-lr,=o= -;p

3.2 The model I charge distribution has (Fig. 3.1)

Now

Putt ing s-qrfhg:v?s

p ( r ) = p , r c a

P ( r ) = o , r > a .

It lsin (qr I h) / (qr / h)lar r'z drtf hrz dr

F(c') =

F(c\ = !###_ 3[sin(qoltr) - (qaln) cos (qal,i)l

(1a/n1t

3.3 This charge distribution is spherically symmetric so tbat we canfrom Problem 3.1:

.. "

F(c') = [ p(r) lstu(qr /h) /(qr /h)l 4rr2 d,r/o- p(") 4rr2 dr '

woerep(r) = psexp(-r/a)/r .

If we make the substitution x = r/o, the formula becomes

use the formula

F(o2\ - (h/qa) i- exp(-c) sin(qar/[) du- \ : /

/p- cexp(-c) dr

ll, ,{r if

22 Chapter J t' hll' i

Tbe integral iu the denominator is l. The result for the other integral gives i; I1,. ;

F(c')=ffi,

3.4 The momentum *uorr., q is given by

A9 = 2P sin

i = UUO sin 5o = 57.5 MeY lc .

The reduced de Broglie wavelength,\ is

^ _ h t _ 1 9 7 . 3 M e V f r _ o= A = 5 7 5 M t v

= 3 ' 4 3 f m '

._.The form factor for calcium at this momentum transfer is, for Model I,

tr ii

a = 4 .10 fm,

T = ''"radians '

F(q = 57.5 MeV/c) = 0.864 .

^ . . \ 3 ( s i n z - o c o s r )F(q ' )= - - - ; i . - - : , r=qafh .

Now a = 1.2.4i fm, z{ = 40, so

and

The Mott scattering formula is given in Table 3.2. since u = c, with srrftcientprecision for electrons of 330 MeV/c, tbe cross-section becomes

do(0) _ Z2a2h2C , le\ f. _,_, /d\ldn

- 4E cosec ' [ . t i

[ t -s i " ' i t / ] .

Z = 2 0 , 4 = 3 3 0 M e V , d = 1 0 " g i v e s

l n l lno \

? = 32'74 fm sr-r '

The expected cross-section is F2 times this value. that is 24.4 fm2 gr-1.

'' Exerclrar 'll.* i. '

3.6 What ls rhe mechanlcal potentlal energy of on electrori'ii'ihd cenrre of r goldnucleue ,(Z-79, A-1971 glvgn thot thc nuclour ir a unifomlybhtrged spheto ol radlucR a 1 . 2 A t t t l m .

; . .

3.0 Estimate tho rsdius of t1i'first Bohr orblt of a p-.meson about

(8) a proton, . :' ,(b) arlC nucleus.

What is the principal quantum number of the Bohr orbit tor I lJ' which lies iust outsido I l68dnucleus (Z= 82. ,4 = 208). if the nuclear radius is 1 ,ZAtt lm?

Using Bohr theory, calculato ths transition energy from the next higher orbit into this orbit.

3.7 Calcutate two things for a p--meson in a 2s state sbout a r:C nucleus:

(a) the binding energy. assuming a point-like nuclear charge distribution;(b) the first-order perturbation of this energy using model land assuming a=1.2Aatm.

3.8 Suggest how the calculation of Problem 3.7 can be improved.

Some information for these problems:

(1) 'Estimate' means get it right to one significant figure.(2) The mass of the p--meson is 105.7 MeV/cr.(3) The wavefunction for the 2sstate ls Li"'

*o = (*!)* 1 *)''' \r- z i) ":"*\

4tteohzro=-FT

Solutions3.6 Outside the nucleus, the electric field is 8s if aI tbe cha,rge were ct tbe centre.Therefore the electrostatic potential at radius .R > o, the nuclea,r radius, is given by

7eY(R) =

ff i., R> o.

i

7. c tr=R

V(n) = f f i - l ,=. E(r) dt . . .

F o r R < o :

:il

a

:l!

i:|;

24 Chapter 3

The electric field,.'E, at r is, by Gauss's theorem, gven by the charge inside tbe sphere ofra{uull t:

l /-!\n(r)= fra \A) t,.Substituting a,nd integrating gives

v(R)= 3- - F:!\ 1" , R<a.' 4Tesa \ 2ot 'J hrcsHerce I

vrc\= sze8aeoo

'

Thug the mecha,nical potential energy of an electron (charge -e) at the centre of a gordaucleus is

, _ _32e2. 8rcsa '

Fot Z =79,A= 19Z,rr. = 1.2(dl fn this gives

8.6 The formulg for the radiur of the fmt Bohr orbit for a negatively cbarged particleof mass m moving about an infiaitely heavy centre, eharye +Ze,is

' . , . '4ieshz I lgz.o4 x l9z.B _

: , . 70=m=V- r - r ^ .To allow for the nuclear mass re-place rn by the reduced mass for the two-body systee.:l

Proton: M, = 938.3 MeY/C, m, = lbs.?MeV/C, Z = 1;then r = 284.6 fm.b) Carbon: nuclea^r mass = 12 x 9it.5 Mey /C, i = a; r = 48.0 fm.

Learl: nuclear mass 20? x 9Bl.S Meyf C,Z =g2; then r =3.12fm. But this radius is lesstban the nuclear radius, a = L.2Al = Z.ll fm. ih. Boh. radius for an orbit oi priocipAqYPiuE 'umber a goes as a2. Therefore n must be 2 for tbe muon to have au orbitwhich, in the sense of Bohr's original model, is outside tbe nucleus. (The auswers givenhave a greater precision then sxt..,ed frtlm ihe problem's ,estinrate,.)

- The bi!'ling energy of a particle of pass m in a Bohr orbit of principal quantum

number a arouad a nucleus Z is

meaz2 nLC o

@=f f iaz22 .For the p- mesou around a lead aucreus making the traDsitio' from a = 3 to n = 2, thetransitiou energy is

;m,ca2z2 (i - *)

= 2.63 Mev.

'3;1 ,Th bi-"$* enerry for the electron iu a ls state of the hydrogen atom is 13.606 eV.

} ?:^T.ti{"') yd Z: for.au orbir of priucipal guantum number n and a nuclear charge."-: 1'.t-o-t

o*lron ia replaced by o negative mu-me8on then the binding energy increaseeDy t hct.r oI milma. Thue, assuming a poinblike nucleus and including rJuced mass

Chapter 3 25

effects, the bindingenergr ofp- meson in the n = 2 orbit about a 2 =6,A = 12 nucleusis

13.606 x f rso"f f i "v .= 25.1 keV.

If the nucleus is finite, the p- inside the nucleus a distance r from the centre has amechanical potenti euerry whith is not

_zo2

fo^, but ffi 1to'-r'1

where o is the nuclear radius (question 3.5). Therefore the perturbing poteutial v(r) dueto this model I nuclear cba^rge distributiou is

r < a :V(r ) = 9- - =4-(3a2 -12) ,Atesr 8reoa3

r ) o : Z ( r ) = 0 ,

where o = L.2Al = 2.75 fm .

The first order shift in enerry of the 2s level from tbat for a pointJike nucleus is given byperturbation theory:

tp = Io*

g'(r)v(r)rt(r)4nr2 dr

where

,l6=he)' Q-';) *,(-#) ,,.. =!' '#' P

e 2 mand m = reduced mass of mu-meson and carbon nucleus. Then

r p = 2 5 8 f m .

It is convenient to put

which gives AE

o, =: (*) #1" ('- L)' "*e1-2,/a,i (i- #) 4,,r2 dr .

Since o, = 86 fm, we have o 11 ap and we can expand the first two factors of theintegrand as a power series in rfar. Tbe integrand becomes

(r-n ' *7" * \ / t - 3o2 - 12) 4nr2 dr .\ o ; '4 - I \ ; - u ' 1

The first term of the power series gives a contribution AE1 which, from a straightforwardintegration, is

z l2rr a,,

AE1= t(*) (#) e-1 = no,.u

26 Chapter 3

The second term gives

AE2 = -gBr 19 3 = -2.2 eY .z4 a,,The next term is expected to be reduced from Ag2 by a factor of or der af orand thereforecan be neglected. Thus the total perturbation is

' "F --- ---

AE = 38.2 eV

This is the change in the real energy; since tbe real energy is the negative of the bindingenergy, this positive perturbatio'r- de*eases the binding energy. (An iicomplete o,u*fuo.-tion was given and used to calculate ao incorrect answer in the'book).

' -'-

,i3'8

. The calculations of probrem 3.7 can^be improved in several ways. we grve uo uor*i,for the general case, not one specific to 12C or io a 2s level.1)

9:.. Tgdd II (Fig' 3.1) for the charge disrribution, or any other believed to be a morefaithful representation.

2) The p-meson is a heavier version of the electron and has tle same. spin. Dirac,srelativistic theory of the electron (section 9.3) can be applied to tri p-meson wiihan appropriate mass change. The equatious can be solved an"tyti."uy roi lylroguii-like atoms with point charge nuctei. For non-point like nuclei, ilu .qo*iooi i;* 1.obe solved numerically. There are other imporiaut efiects that must ue io.tuaia io,o

. complete calculation; the relative importance of each depends on z, Aaod the orbii3) In cases where the nucreus has a magnetic dipole or electric qua.drupor. ,o;;;;;

there are level splittings due to the equivalent of hyperfile inieractions io "tb.iiphysics.

4) Quantum electrodynamics gives some corrections to the Dirac results: for example,the equivalent of the Lamb-Retherford shift in hydrogen (Section 9.6).

5) In the case ofsome mu-meson orbits, there is some siguificant screening ofthe nuclearcharge by tbe electrons.

- ;

6) In heavier nuclei the calculation of wavefunctions and energies can be complicatedby the perturbing effect on the nucleus of a mu-meson wavefunctiou that has asignificant value inside the nucleus. In addition, mu-meson transitions can lead tonuclear excitation, reduced muonic X-ray energies, and the emissiou of uuclea*y-rays.

Solutions to exercises in Chapter 4

Exerclses

4.1 A slmple problem in elec{rostatlcsl Show that rhe potential ,rnergy due to olec{rostaticforces of a uniformly charged sphere of total charge o and.radius g is 3cl2l(20rr6oB).

4.2 The Coulomb term in tho eemi-empirlcal mass formule ls i.

' l acttAlts,

Using the result of Problem 4.1, calculate the value of a. lnnuclear radius ls glven by I = 1.24 x Atr! fm. "lv.lev/C. Assume that the

,Y.ti:n jhg y_rlu.:8.of av, a€, and ,^ glveir In Table 4.1 and the fsct that the blndlng energyof i"jra is J4&l}lev, check your value of as. comment on any,dlscrepancy you riay find.

Ine nucrsus'iiu c8n undorgo spontaneous fisslon (see chEp,1e1 6l: one ol the manyflssion channels is, .

'

' : { ' t " i '

,-l!jU, jler + 'gLa + 3n. , ."i.1.,

Estimato the energy released in thls channel.

lAdoptod from th. 1979.x.m|n!tlon In rhc Flnrl Honourr School In Nltur!l scl.ncc. phyrlc., unlv.rrtyof Oxford.l

i

4 .3 The_Fermi-gasmodel of thenucleussssumesthat thonucleuslsasphereof volumeV=4rR3l3 (take 8=Fo x A1t3 tmwith Bo:1.2 fm) and solves thii SchrOdinger elu*ioi tofind the number of states avallable up to a momentum p; that number is7,y',wglzaf .show that for a nucleus with equal numbers of protons and neutrons lhe energy Fpof theFermi level is given by

Er = ll2t2M87l (9trtllzts,

where M is the mass of a nucleon. Estimate the total kinetic energy of the nucleons in anucleus.of r!0.

For a hucfeus eAl tho asymmotry term ls aalV-Azl'/., Assuming lN-Zl <A, use themodel to ostimate dA.

{Adaptod troni thg .|978 cxamlnltlon In th. Flnll Honoun School In Nlturll Scbnc., phvrlcr, Unlvc[ttyof Oxford,l l

Solutions4,1 A uniformly chargedConsider coustructing suchexisting sphere of radius r

(Q) sphere of radius .R has a clarge density of. Je/4rR3.e.sphere by eddiug a uniform layer of thickness dr to an< .R. The electrostatic potential V.(r) at the surface of tbe

28 Chapter 4

ephere of radius r is due to a charge Qr' / ff , effectively at the centre: therefore/),n2vb)=;fu.

There is work done h bringing up the cha^rge dq coutained in the layer thickness dr. Thatcha,rge rs dq =3Qr2dr/ft3 andrhe:'-ffiff) dq.'Iherefore

Rt , AnesRl

7.lr\2= #Bfdr .

Integrating from r = 0 to r = R gtves

w= 3Q220resB-

Tbis work dooe is, of course, tbe mechanical potential enerry of the assembled chargedistribution.

4.2 For epherically symmetric nuclei Q = Ze,wherc e is the proton charge, eld I =1.214, fr.. Therefore

w=&.2Mes$.2aAt)

Tbrs is equivalent u o,Z2/Al . T6 obtain c. in MeV/c2 put

e 2 1 9 ? . 3 . , . , ,G=;;;5;nMev

rm

and equate: then -

a. = 0.692 Mev.

The biading energy of a uucleus (2, /), 4 odd is

. ; B (z ,A ) :ovA-u f i - q * -o^ (A*z ) ' .

Using the nalues from Table 4.2 we have lor (Z = ?3, / 1 181)

rnii:ruev = 2816.4 Mev - sbr.32 Mev - r5?.s9 u"v - qq,t .

Ai

wbid givesoc = 6S3.4b AI lZ2 = 0.694 MeV .

Thrs is in very good egleement witb the previous value for o.. Of course, the semi-empiricelmass formula is a physically motivated parametrizatioo of the observed values of nuclearmasses throughout the periodic table, so agreement is expected.

Io the spoata,neous fission

'zlNU -$nr +'!f La + 3n ,the euergy releaped is the inoease in nucleer biudirig energies. Now

aE(35,8?) = a(8?)-c(841-"ffi -o^9#,

Chaptet 4 29

BE(57,145) =, a"(145) - a"(145)l - " ffi

- o^ ttntiljtn'' ,

BE(s2,23|,) = a,(23s)-0,(zss)i -" ffi

-o^ (23s----taa)'.

Then the energy released is

E = -3o" - 9.153 a" + 476.7a * Lll9or

= 154 MeV .

It is interesting to note that by far the greatest part of the energy released is from thechange in Coulomb energl.

I

4.3 The number of states available to a particle confiaed to the volume V of a nucleus,up to a momentum p is

4trd V'=t@.Each state can be occupied two ideutical fermions (spin up and spin down). Neutroo aodploton are not identical 80 we car have ar occupaDcy for each state of two protons andtwo neutrons. Thus the number of nucleons having momentum between 0 and p is

Ltuf v3 i2?r,')T'

For nucleus (Z,A) rhis number is.4. So the greatest momentum, the Fermi momentumpp, is given by

(1)16rp$ A(2rh)l

3 = v 'Now the Permi enerry Ee = p$l2M where M is the nucleon mass (assumed tbe same forneutrons and protous). Putting V = (arft) Rf;,4 where the nuclear radius is R6,41, gives

^ rr' /9ur\ rlor= m4 \T )The number of states in the range p to p + dp is-

4np2Vo n = @ o ' ' " ' '

The 4dn nucleons in this ra.nge each has kinetic euerry @'lzM) so the total kinetic enerryis

,=t#on=|,'"n#W4 | hc\z I /9r\ 3

in \6/ m (t/ A'

For r0O, r{ = 16, MC =$8 MeV, .Ra = 1.2 fm, givlng

E = 3 2 0 M e V .

30 Chapter 4

Near the Fermi level, tbe level spacing is grven by

Now

so that

Hence

^2

m=od p Md E p

Yot'

o" = (#1,,)= (at rp2V dp1 \ - t

\(2rri)3 dElar/

aE = -Qrhf-AnpFMV-

I f i n a n u c l e u s Z , N = A : Z , N > Z , o n e p r o t o u i s c h a n g e d t o a n e u t r o n , t h e n i thas to be moved up the ladder of energy levels to find an unoccupied etate. since protonor Deutron occupancy is two, the number of steps bas to be (N - z)/2 and the energyrequired to change froa Z, N to Z - l, N + I i8

But since AZ = -1, this is equal to

_ dEldZ le .

lt z, a N; = Al2 the asymmetry energy is zero so that tbe enerry required to changefrom Z; = A/2 to Zr

,^ = l:::;,+ffi1^az = - l:;,ff o, o, = - l:;ff aaaz= [+(A-_22) 'z or] t , '=(A-2zt)2 ̂ n

t s - - J e t z 8 - " '

we can drop the subscript f to deaote the z of the nucleus obtained by successive chnngesof a proton to neutron. Thus the coeficient oa of the ssyrq'etry term:id the mass formulais predicted to be

. A Ea e = A l _ :

Using AE from above and substituting for p1 from Eq. I above we find I

o^=* ( * ) 'En = l lMev . .The observed value is 23.3 MeV. The discrepaucy is due to the assumption that the eolecontribution to the asymmetry e'ergy comes from the kiaetic uo.rgrof th. nucteone. t[isis not so. For nuclei that are lot symmetric, that is z * /v, th; is a,n extra pot.iiiJ.energy that also increases I (z - N)2 inoeases from zero. This potential ererry occrirsbecause tbe nuclear force between two protons or two neutrons is difiereut fiom.tbat

Chapter 4 3l

bbt*,eeo a neutron aud proton, with the result that the effective total potential chaogesas the neutron to proton balaDce is,altered.

Solutions to exercises in Chapter b

Excrclrct

6.1Erp|ainthi ie, 'alnthceemi.empir icatao|icmass|ormu|al.*::_j

l(2,7): ZMx+ NM.- al * arAn + dczrl At, + i^ (A _ 2Z)2 | A * ar1 tqut.show for largc 4 rnd Zrhat the cncrgy roreascd whsn a nucreus (Z,A) emirs an d-particrc iggiven by .:.,

Q,- - 4a"*EaJ3An +4arZ(l _ Zl3AllAtr _ 4a^(N _ 4rlA, + B(2,4).

"T-: 111,4, is rhe bindins enefsy of the a-particte, 28.30 MeV.

",';l,iJ::rf:y-"::llg_,::,:f: :f sirver and gotd.are ,sAs and TtAu. Discuss rhestsbrtty or theso nucrerin ,f Ga;il;';;;#:ffi;,El,iJ."illJ#:'iff li:

. coefficlents r glven ln Table r0.1.

lAdlptodtrom the rggo sxamrnrtron In the Honour schoor In Nsturar science. physrcs, University of

6'2 How crn thc rpin of thc nucrcus rfNltrogcn bc mearured by morecurrr rpectroocopy?5.3. Dlscugr thc .xpcrlmenral..€vidonco for thc cristence o, tho ncutrino.The nuclidc flAa decava ro ?lNo wittr ilt..ri*iin oii-pli,ron. The radius of the nucteusli*:^r

masg number 21 is 3.6 tm. e.rin'r,r ir,.iffil-rinrri. enorsv rhe posirrons can

sfJ:i ttom lhc t076 0rlmlnrtlon In th. Honour schoot In Nrrurlt sctcnc!, phy.tcr, univc,.tty o,

6'4 ln an und'rurbcd ore c^o-ntarnrng 0,rg6 by werght of z!!u there w* be como rf;Ra.calculate the woight of this irotope or radium ri, u, rirnJ ,n one motrio ton o, oro, wh.l ItllililtJ.!:t"rtlon of hcllum sct tn ts pc' viirlitii, Jrorn,or oro? uro rho dl.o srvon5'5 Carcurate values for a- and a^of.probrem 5,1 making ucc of tho folowing focr* rlAromltr portrronr wrrh o mcximum kincrro .ncd;il;iiiii ono ,flBr rr rhr rrrbJc rrcbriolmlrr numbl t3t, Erprcr your enrwon In MoV/or.

ffil# trom rho i'o? cromrnoiron In rho Honour schoor or Nerurrr Scronccr. phvricr, unrvc.rty of

5'0 Find some nucrear rabres and trsco th€ mo$ rikely route fiomz4. - 245, 2-g6 to $gl by:.:if :T:,,j:: :1 ff;;;H'*'yr,

cach .ith.; ;: ;; ;:;;cay. which i, r,, 6ig,,i ri"dln principle lhere are four decay series (A(modulo 4).0, t, 2, 3). tf rhc progcnlron wcro

il#:.,T,n:*[irj:n:::,*::1T.?:il;-il;,h-.";urarrbundrnccconErrrhorrhc[:?tr,jfi:".;Hllicridc oicach reri,,. D;-;; ;;;;tffiffi:iffiil1_.:ll.:lll;

Chapter 5 33

Solutions6.1 For an a,uswer to the first part, attempt a precis of Section 4,3.

Tbe enerry, Q., released when a nucleus (Z,z{) emits an a-particle is given by

q" = IM(Z, A) - M(Z - 2, A - 4) - M(2,a)lc2 ,where M(Z,A) is the mass of the nucleus Z,A. Tbe binding energy of the a-particleB (2; 4) = lzMt + 2Me - M (2, 4)lC.Therefore

Qo = M(Z,A) - M(: Z - 2, A - 4) + B(2,4) - z(Me + M^),

where we have dropped the C since all coefrcienk in the mass formula are given in MeV.Substituting for M(Z,A) nd M(Z - 2,A - 4)

Qo= - o "Aqa , l l +o "z2 l l l +a1 r ( l - 242 /A

* a , (A -4 -a , (A -4 ) l - o " ( z - z ) ' ? / (e - t1 l

- ot(A - 4 - 2(Z - 2))' l@ - 4) + B(2,4) .

Since .4 is large we have ueglected the pairing term, Iu addition, we expand terms like(A- 4)" using the binomial expansion. For example,

@-n1t = , r3( r - i ) t= / t ( r - * ) ,

(l - ay-l = ,4-l (t . #) , and so on .

In tbis approximation and neglecting small terms we find

.Qo= -4a"+Ba"lgAI + aa"z(r- zlgA)lAl - 4o^(A -22)2/A2 + Be,4) .

For tlfAg: Qo = -1.93 MeV.For tffAu:

Qo = *3.10 MeV.If tho formulo ia corroct, tho negativo Qo moans thot tor?Ag c&tr novcr spont&ncouslyemlt en a-partlcle, The posltlve Q" for fr?Au permits spontaneous decay by a-emission.In chapter 6 we give the simple theory of a-dccay: it will be cleor thrt the tronsitionprobablllty for auclr a dportrooou! docoy wltlr o Q" - J,t MoV wlll br: vcry ururll(- 1g-zr s-l).and the element is effectively staile.

U,2 Moloculat tpoctroacopy 0ncomp&860i tho vlbratlonrrl nnd rotntlonnl atntoa of molo.cules. Naturel nltrogen ls 99.6% one ieotope tlN qpd therefore the great majority ofits dirtomic molcculcs are homopolor. The nature oi this molccule's rotational states isdetermined by the nuclear spin (j), as the following argunrent shows.

The possible states of total spil angular momeDtum due to the nuclear spins are 2j,2 j - \ 2 i - 2 , . , . . , l , 0 . w i t b s t a t i s t i c a l w e i g h t s 4 j + t , A j - t , , . 4 j - 3 , 4 j - 5 , . . . , 3 , Irespectively. under exc.hange of the uuclei, the corresponding wave functioo is symmetric,antisymmetric, ... alternately trom 2j down. Tbe symmetric states are labelled the orthomodification aud the antisymmetric stetes are labelled the para modification. (See Sectioo9.6 for this labelling applied to positronium.) counting, the ratio of the number of orthoto pa^ra states is found to be (j + t) I j.

The rotational otste€ csn be specified by tbe integer quaDtum number {. Even (odd){ states are aymmetric (antisymmetric). If the nuclear spin is half odd-integer then the

34 Chapter 5

total wavefunction must be antisymmetric and even(odd) / rotatioual states have paia(ortho) total nuclear spin states. If the uuclear spin is integer tbe association is even(oad)/ goes with ortho(para) states. .

Tbansitions between rotational levels must preserve tbe symmetry, The nuclei areso weakly coupled and isoleted that tbere is little chance of a cbange in ihe ortho or paiespin states. Thus there are two sets of transitions, one between ortho rotational statesaiiJthe otber between para ldtational states. They have relative intensities uquJto ii. ,"tiooJ tbe statistical weights (J' + l) / j . The lines are easy to identify since. they alternatej'6stbe wavelength changes. Thus the rstio ofthe intensity ofalternate lines gives (i +L)/i.For rfN this figure is 2 and thOrefore j = t.

A more complete discussiou of molecular spectroscopy and of tho role of nucleican be found in several textbooks. see, for example, Leightou, R.B. (lg5g). principtesl"o!Modern P/rysdcs, McGraw Hill, New York.

5.3 The evidencb for the existence of the neutrino is discussed in chapter rz. rue flrvedecay reaction is

!f tta -!!Ne + e+ + v, .since the nuclei are relatively heavy, the decay energy is shared betweeu the positron andtbe neutrino, since the neutriuo has zero or a very small mass (< 20 ev), the positroucan bave any kinetic enerry up to a ma:<imum equal to ep, where

Qp = M (21, lL) - M(21,10) - rn

and masses are expressed in MeV and rn. is tbe positron mass. using the usuel massformula for the two atomic masses:

Qp = Ms - Mn 1:5 (rr, - to2) - rn. .2li '

The value of o" is not given and the implication is that the change in coulomb enerrymust be calculated from the given nuclear radius R. Ftom problem 4.2 we know that, theCoulomb eaergy is

* _ 322e220reoR'

Tbus the change in Coulomb energy id

aw =2 ;11'z- 1o') / ' t" ) 't.t 'S \__ - - ,

\4 reohc) R.

Using the usual numbers and R: 3.6 fm, as given, we have

Al7 = 5.639 Y.Y .

Now, since the mass of tbe proton Mp: Mx - m.

Q p = M p - M , + A W ,

= -1.293 * 5.039 MeV- 3.75 MeV ,

and this will be the maximum kinetic energy the decay positrou can bave.

fs,ry' i O.'i,

I'

.i

{

,tu,'4.ri

' : :T

1ist".t:':tl-.8

i * Cbapta 5 36

5.4 One metric ton is 1000 kg and that quautity of o.. .bot"io, I kg of ?|tU. Thisuuclide slowly decays aud populates a secular equilibrium of reliti'rely short lived daughterproducts: iu tb.\.9qu{i!$lm .aV is tbe sane f9q 28iU as fqq;,q*t!g rrnstable members ofits decav scheme. Tbus

' r.rrr3N2fg = U226N226' t '

(The assunption of 1rulV the sa^roel is not true where branchinli'bccurs. However there areno significant branches around,,f!!Ra.) Now

_ ,i,

n,--- _ Ne l.tog ,

rY238 = _r5E_ |

where iV1 ls Avogadro'e number, in mol-!. Hence

^, . N,t X 101 ra43,,2r,6_{

G

,.rFF-*,

The mass of 2!!Ra is:226 226 u2ssNrr ffiTF

= *ffikc.."-.Now r,.r2r, = (6.45 x 10e)-t yea,r -t

, ind u226 = (2,3L " 19:)t year -r . )llence

nass of 2!!Ra = 0.34 mg . ' 'tl

The decay of uranium feedg the secular equilibdu. ".oog

it lproducts. The decey chainto 2!!Pb iuvolves in all the emission of eight o-particles so that, uuder the equilibriumconditious, each urauium decay effectively adds eight helium atous to tbe ore. The uumberof 2i!U decays per year is

rrrgeN2sa = #

- o*:16

ye&r-r.

Hence the weight of helium gas generated per year is8 x 4 1= ff dfurksyear-r

= 2.09 x 10-ll kg year:! . ,

5.5 The argon decay is

ilAr ,rllCt + e+ + v. + 4.95 Mev-.,;--

The mar<imum kinetic energy ofthe poditrons (4.95 MeV) is given by

?L* = c[M(35,18) - M(35, L7) -2m,1

(from Table 5.1) where M(Z,A) is the etomic msss of nuctide (Z,z{). Use the massformula, as in question 5.1, for example, Absorbing the c2 in the defiuitiouri of mass ardin the coefrcients we bave: ..." ,

T.o =i Ms- M,+ fr tra' -fi\;31y.

Therefore ",::., ; 4 . t . - - . i :

4.9S = -o.782*0"(35)3-2r0., :. i ' ,a n d i

l : ,

36 Chapter 5

ac = 0.631 Mey/cz

The most etable isobar of A = 23S will be close to the point at w[ichaM@,A)T = U .

NowAMO.A\ Dn Z- Mn- M,+2q3 - l

(A-2z l,. az

Thrs ts zero whqn z = 56 and ;; "'o'IF - *a^

--!'

.itt at'=2r"4ucv1"i '

5'6 The longest lived meober of eactr of ihe four decay series, with its mean life, are:z{ (modulo 4) = Q;d (nodulo 4) = t1z{ (nodulo 4) = 2:/ (oodulo 4) = 3:

we assume that the nuciides more massive tlan these decayed rapidry to these long livedtTl.o-T that-very soou after t = o there Jsied equar numbers of each of thesenuclides. Now, 1010 yearc later, the uumbere,;r;iri;; will be in the ratios given by theradioactive decat' law: 's vse rolreu t'rver

/Vo : lVr : JVI : /Vr = exp(-r.rst).: exp(_r.r1t) : exp(_r.r2t) : exp(_r.r3t) ,where t = 1gl0 years aud r.r is il. t .ipto.a or iu.'..uo life ia years. This givesIVo : /Vr : Nz Nr= l : 0 :0 .34 i :g .E t ; i0* : - 'All four radioactive series have alternative routes to the final etable lead or bismuthnuclide' Most of the alteraative routes iu.odd e *J.i are udmportant since they are

ld by ptp-decay branchiog tractions whire-*J'" deviation from the main roureis pcssible' These deviatiors come about when ,u..G". a-decays leave a nucleus aeu_tron ricb and far enougb from the Z for stability egainst /_aeca' tUat t;;r;;"_*competitive witb the uext a-decry In some ...* pla..", wins and a,ll the series includeeeveral such decays (see for ocample, Table 5.2). I; the case of the odd A nuclei there isouly ooe 'oass curve aad as each su."ori". "-'i.."vli.r.r.o

the neutron richaess, thenucleus smoothly deperts frll-r_Ur. ooio, .f ,t"UiUti "grnst

p decay. (See Fig. s.6). Ontbe other ha^ud evea A nucrei have tw9 mass curvEr irii.s.o;tilil;;i". "-iu"tu,opportunity for deviatious'fro-

" raolioiu-;ffi.o.! over several stages.

2!!Tb, r =2.034 x l0ro y.'8 lNp, t=3.08?x1oov.'8f U, r = 6.446 x fOe v.?8 iU , r= t . 0 t4x rOrv .

Solutions to exercises in Chapter 6

Exorcirer

6.1 Describebrief lythephysicalprocessesoccurringinc-decay.withoutdetai ledcalcula-tion, givo a qualitative explanation of the dependence of the transition rate on the Zof thedaughtor nucleus and on tho onergy released in the transition, O.

5.902 MeV'iict

0.294 MeV0.142 MeV0.043 MeV

0.0

6'

2'

o'

Tha figure shows lhe c-decay scheme of 2$Curium to rllPlutonium. ThE transitions aremarked with tho branching fractions in per cent. According to a simple formula the transition.ate, ,1. for the ground stato to ground state transition is given by ln 1e C- DZlAtn, whereC-132.8 and D-3.97 (MoV)'n when I is in s'r . Calculare rhs mean l i fe of 2l lCm.

f f the same C and D arc assumed to appty to rhe rransition from the ground state of ,llCm tothe 6r level of 2flPu, show that this formula for c-decay transition rat6 overostimates tho ratefor that transition and suggest a reason for the discrepancy.

(AdePtod trom rho 1988 exsmination in the Honour School of N8tural Science, Physics, Univsrsiry olOxford.)

6.2 Explain why naturally occurring nuclei of mass number A:60 aro not a-unstable.

with neglect of nuclear spins, and for coulomb potential barrisrs largo comparod lvith the d-particle enorgy. tho dopondenca of the mean life r of an c-unstable nucleus on tho chafge zand c.pirticle velocity v is rspres€ntsd by

| 4nZd\t .c cxp \a;#;/

. i

Discuss the basic assumptions made in the derivation of this exprassion.

ln the decay 2l!U + t$Th + a, the omission of c-particles of onergy 4,1 95, 4..t47, 4.039 MeVis observed. Use the above formula to estimato the relative strengths of these three lines.what ara the most probable spin-parity assignments for the states of 2srh produced in thisroaction? stato briefly why consideration of the spins of the statas of ri.Th would changeyouf,ostimatos of the relative line strengths.

(Adapt.d fiom the 1975 cnminrtion in the Hbnour School ol Natural Science, Physicr, Univcreiry ofOxford.)

T

38 Chapter 6

Solutions6.1 If As is the transition rate from $cm to the ground state of 2$!pu then its value isgiven by

h)6=@ - DZ/Qrlz, Z = atomicnumberof daughter.

Hence

and . .Jolo

= 132.8 - 3.97 x 9altffi, ,

' lo = 9.1? x 19-10 r-r

If ) is the total transition rate. then

)o = 0.767.\ '

ThereforeI = 1.20 x 10-e s-r .

The mean life is )-t = 8.36 x 108 s = 26.i years.If the formula were correct for tbe transition rate ()6) to the 6+ excited state of.2f!Pu then

Inl5 - 132.8-3.9?x 94/\f f i

Tberefore the predicted rate is )6 ='1.72 xlg-ll s-I.In fact the observed decay rate is 3.6 x l0-5) = 4,32 x 10-1{ s-r, about 400 times

slower. The formula given applies only to transitions in whi& there is uo nuclear augu-lar momentum chauge. In the transitioa coasidered, conserrration of angular momentumrequires tbat the final state relative orbital angular momeltum be 6t; the angular mo-mentum barrier reduces the transition rate from a value expected wheu there is no orbital.angular momentum in the final state.

6.2 Near u4 = 60 the binding etrersr per nucleon is about g.76 MeV (Fig. 4.6). Thus :the removal of two protons and two neutrons requires, therefore, about 95 tvtev anathis cannot be compensated by the binding energy, 2g.3 MeV, of tbese nucleons in an,a-particle. Thus a-decay is energetically forbidden in this region of the periodic table. ::

- In the decay of $U - ,33Th * a, the a-particle kinetic energies are 4.195, 4.14? li

and 4'038 MeV. The mass, Mo, in MeV of an a-particle is 3z2z MeV. The velocity of an .ia-particle of kinetic energy ?-is given by o = ct/T/zMo. This gives lf c= 4.745 x 10-2, il4.717 x l0-2, aad 4.655 x l0-2 respectively. (This is a non-relativistic calculation and we ,are neglecting recoil). The quantity 4nzelfLnesha can be rewrittet as' ;zal0, o,here l0 = u/c and a is the fine structure coostant, The trausition rates are proportionnal, .therefore, to exp(-AraZ/p). That is to

exp(-I23.94) , exp(-t?4.92) , and exp(-12?.80).

This means the relative strengths are expected to be

I , exp(-1.03) and exp(-3.36) .

That is I : 3.6 x l0-l : 3.5 x 10-2.The ground and lowJying excited states of ,!$Th are expected to belong to a

0*, 2*, 4+,... rotational band. The three involved will therefore bave jp = 0+, 2+ and4+ and the transitions with kinetic energy 4.14?(+ 2+) and 4.038(- 4+) MeV will beinhibited by the angular monentum barrier, as in problem 6.1.

:

Exerclses

7,1 Prove the formulae lor.U, P", and I. given in Fi9.7.4.

7,2 Prcvethe formulae given In thil caption to Fig.7.5.

7.3 Computo the O-values for tho reactions

d+! r n+ !He,- d+ t {n+d.

given: mass defect ol the neutron - 0.008 665 u.mass defect of deuterium atorn = 0.01 4 1 02 u,maisdefectoft. i t lumatom -0.016050u, -* '

mass delect of helium 3 atom -0.016 030u.mass dsfect of helium 4 atom -0.002 603u.

What are the maximum energies (in the laboratory) of neutrons that can be produced using4MoV deuterons incident on stationary targets of deutorium.and of tritium? Use non-rslativistic kinematics and assumo nuclear masses are Axg31,5MeV/C in the translorma-.tions, but usa th6 O-values already calculated.

7.4 Show lhat, if the proton recoils folward with a kinetic energy of 2 MeV from photonscattefing by an initially stationary proton, the incident photon energy is 32 MeV. (Thisproblem .equirss i6lativistic treatment.)

7.5 Show that if I particl€ of mass m and velocity v elastically scatters irom a nucleus of.mass M, tho grsatost.velocity.the rscoiling nuclous can haveis'2frv/(n+M1.'", Hence the ratlo of.the great,:sq,recoil velocities of protons (vr)-and*ot nitrogon ntrclei (v*)bombarded by neutrons of the same energy is given by

where M., Mn and /Iy'* are thb'masses of the neutron. prgtciii,'and nitrogen nucleusrospectivslv, Chadwick's values were vr-3.3x10?ms'r and vil-4.7x10tms'r. Find hisvalue lor MnlMr,

Solutions to exercises in Chap-ter 7

Solutions7,! Usirrg tbe symbols of Fig. ?.4, the velocity of the centre-of-Fass must give pa^rticles

I and 2 equal and opposite momentum in that frame' Therefore

M{V;U)= MzU ,

u = Mrvl(Mt+ Mi .

tiitllilJ

fifl

t

t'!

hence

40 Chapter 7

Now P. equal to the Domertun of M2 is the centreof-mass. Therefore

P, MtMz " PrMz: i712v =

@rvr= 6g;.

Tbe total hnetic eoergr in the centre-of-na$ (4) is the sum of the two kinetic energies.Therefore, since the two particles have the sane mtmentum' r"=#.#:m7.2 After a reactiou has occured, the total enerry in the centre-of-mass is { = T"* e,This ie dividl{ b.,.$ry.e.o the two fnal state particles so as tq give each equal, but opposite,roomeotun 4: T!.e19fore -:

' p n D a

r _ i ' i d ' r

This means ":;:"4=,fffii,

where the redgp*:rygss M is given by i _..' t 1 . 1 .

_ = _ + _M M t ' M t '

The uext step is to fud the momenta of the pa*icles es would be observed in the labo,ratory. Tlis means adding velocity u (Fig. ?.4), in the directiou of the incident, particle,to. the velocities of the particles in the centre.of-6eqq. ta this addition, the transverse (tou) component of aay. velocity is unchanged. Thug fcrr pa^rticle 3, for example, in Fig. z.E

P , H

frsin0" = frsin0r,

7 D -

. U+ftcosl" = fr*tr.

Multip$ng by M3 grves tbe formulae in Fig. 2.5 for d = B.

7.3_ The Q-value is given by the decease in the mass defect from the initial collidingnuclei to the product uuclei. For d + d+ n + lHe + e we have

a = (2x14.102-8.665- 16.030) x 10-3u' = 3 .5 l x l 0 -3u .

Now ly = 931.5 MeY / C, therefore Q = J.27 MeV. Similarly for d + t + n * a * e, e =17.6 MeV.

For the hnematics of the two reactions we use the system of units described inTable ?.3. The velocity of a 4 MeV (T1) deuteron 's

Jffi *nere M1 =2 x 931.b MeV.Jle.resutt is 0.06553 (= 7,). (Rcnember, the actuai velocity ig this times the velocity oflieht).

Chapter 7 4l

T r = 4 M e V d + d

In + l H e

d + tI

n + caK. E.'r\ centre.of-mass:T.=\Mzl(Mt+ Mz)

Velocity of centre-of-mass:U =VM/(M* Mz)

Centre-of-mass kineticenergy alter the collision: {

Momentum of neutronafter the collision:Pl=lM 85.81 MeV

3.27 MeY

2 MeV

0.03267

5.27 MeY

172.61 MeV

17.59 MeV

2.4 MeV

0.02621

19.99 MeV

Referring to Fig. 7.5, the final neutro! will have the greatest energy in the laboratoryif it is produced &t d. = !o il the centre-of-mass. Therefore the neutron in the laboratorvhasmomentum P=f+M^U.

7.4 The proton recoils forward and the scattered photon recoils back. Therefore theproblem is only one dimensional in space aud all momenta are along or against one axis.

Total EuerryMomentum

lnitial state

" t po

4 lMPcEr/" I o

+Final state

? pH

g , lE r=T+Mrd-q/."1 /S 4c1eThis table defines quantities for individual particles before and after the collision. Con-servation of enerry requires

E, = E, +T .

Consenation of momeutum requires

Et = -Er+ rf4'f Mic ,= -4+lT\zrM,c.

P

The kiuetic energy is P2 /2M,

116.2 MeV

7.25 MeV

24.4MeY

197.0 MeV

20.8 MeV

Eliminating Ef grves

q=;g+, /n+nun.

42 Chapter 7

Putting T = 2 MeY, Mrc2 = 9J8.3 MeV givo for the incident photon etrergl

Er = 31.7 MeV .

The table defines quantities for the particles before and after the collision, Conservationof energy requires

I*o'=!*r" +!uv'z z 2and conservation of momeutum requires

n1! = -fn!' + MV .

Eliminating u' gives

v= 2num * M '

If m = Mo, the mass of the neutrotr, and Vx is I/ when M is the mass of a nitrogenuucleus and V, is V when M is the mass of a proton, we have

7.5 The recoiling nucleus of mass M has the $eatest kinetic enerry in the labora-tory frame if the incident p?rticle mass rn, is scattered back along its i;ident direction. ,The problem is thereforeonl dimensional. We can use non-relativistic kinematics as thi,velocities involved are small compared with the speed of light. , :

Velocity

Momentum

Kinetic energy

initial statem e u l o M

, l oIrnu I t

i^o' I o

==+' Final gtateI t ' a l l n l M * v

-1)' I v

-m!'

|

*'

tr*ro I tuv,

Taking Mx/Mp = 14 grves

Therefore Chadwick's value is

Ve Mn+ Mx

G= M"+ I h '

M"/M" + 14 3.3 x 107+M " / M e + l 4 . 7 x 1 0 6

t r

# = 1 ' 1 6 't r r9

ChepterT" 43

' ." . Exerclses

i '' , 6 | . : ;

7.6 Define the l6rms atomlc mass unlt, mass defect. and averagc bindlng ehergyper nucleon as applied to tho neutral atom. Draw a diagram showing how the averagebinding energy per nucleon varies, with 8tomic mass number. and discuss briefly itss |gn i f i cancefor thes tab i | i t yo ! .nuc |e io f |a rgemassnumber . : . . ' , ' ,

Whsn the rlC(a,7) cross.soction b measured, a peak is found at an c.particle energy of7045 keV. Explain this observation 8nd find the energy of the 7-ray emitted. Discuss whetharit would be possible to confirm your explanation by measurement of the rtN(p,d) cross-soction as a tunction of proton energy.(The mass defEcts of the proton. d-particlo, [N, ltO are 7289, 2425, 1 00, and -4737 keVrespoctively,)

(Adsptod ffom the 1 977 examination ot iho Finrl Honours School of Nstural Sciencs. PhFica, Univclsityof Oxford.)

7.7 The reaction rlC(d,p)tlC has a O-value of 5,952 MeV and I rosonancs when theincident deutoron has a laboratory energy of 2.45 MeV. Do you expect the reactionrlB(c,n)tfN, O-0.158MeV, to have a resonance? lf so, estimate the laboratory kineticenergy of tho d.particto at which it would occur. The following pieces of information arenaedod to answsr:

1. The /-decay rlC+rfN+9-+9. has a O-value of 0.156 MeV.

2, The neutron.lrydrogen atom mass d.ifference is 0.782 MeV/ct,

7.8 Define the O-value 8nd tho threshold of a nuclear rc8ction..Outline the compound nucleus hypothssis in nuclear roactlons and give one piece ofexperlmental evldence for th6 hypothosls. Describe conditions under which it wllt fail.

The tabls shows tho energies of the gamma rays and alpha particles smittsd whenresonancos in the r!F+p roaction 8re studled.

Proton rosonshcg Gamma ray energies Apna particte energiolenergy (laboratory) (centr'i of mass)

keV MeV llMeV

668 :

843

874

6.1 3 .1.306.92 1.47

' ' 7 .12 2.10I

. no gamma rays .7.14! 6.13 i .46

6.92 1.627,12 2.25

Draw a diagram to show th€ levels in fiNe and rlO involved in thbse reactions. Suggestwhy there is no gamma radiation accompanying the (p,d) reaction at 843kev. Whatexperiments mighl be canied out to verify your interprebtion of the.levels in r!O?

(Adsptod from tho I 980 cxrminltion ofihe Final Honourr School of NrturufScienc., Phfics, Univcrsityof Oxford.)

44 Chapter 7

7.9 The radius of a nucleui of mass number ,4 determined by the measurement of thoelastic scanering of high-bnergy electrons, can bo approximated by the expressionB'1.2x Atn lm.The cross.soctions fof th6 radiative capturo of 0,025 eV noutrons by manynuclei are of tho ordaf 10'2'to 10-8mr. Explain why these.two ststements ar€ notinconsistent.

The radiative captufo cross-section, as a function of neutron €nergy. has'tho followinggeneral form for many nuclei: for energios <1 oV rhe cross-seclion is proportional to v-t,whore v is ths neutron velocity; for 'onergies > 1 eV, ths cross-soction varies rapidly withenergy, having narrow poaks superimposed on a relativqly small background. Give aqualitative explanation bf this energy dependence,

Describe an experimental arrangement suitable fo( the m€asur€ment of a radiativo capturocross.section.(Adaprad from thc I 977 oxaminrtion ot ths Final Honou.s School o( Nsturrt Scienco, PhyEica, Universityot Odord.)

7.10 Describe briafly and explain the principle of operation of detectors for (a) thormalneutrons and (b) noutrons of energy 1 00 MeV.

How may a suitable neutron beam and a target whh atomic number and mass number(ZAl bo used t<i investigate .

(1) rho excited states of the nuctido (eA+1),(2) rhe oxched states ol the nuclide (Z-1,A)?

An anomaly is observed in the rcattaring of 10MeV c.particles by"hydrogen, At whatenergy does this anomaly appear in the scattering of protons by helium?(Adaprcd fiom thc I 975 crumlnstion o, thc Fin!l Hgnour3 School of Nltural Science. Physice, Universilyol Oxford.)

7.1 1 Can you find an oxplanation for tho fact that radiative de"excitation is more likely thanre-emission of the neutron atter slow noutron captur€, oth€r thsn that givgn in the text,Section 7.8?

Solutions7.6 The ca.rbon target reaction is

o + r A C * 7 + r f O

The Q-value for this reaction (Qs) is the decrease of the mass defect'

Qr = A(o)+aftC)-affO)= 2425*0- (-4737) keV= ?162 keV.

We interpret the peak in the cross-sectiou to conespoad to a compourd state that is etroccited state of t!o:

. 12^ r6nr - , 16o+r8c* fo ' - . y+ r3o .

This excited state has a mass (in energr units) tbat is greater. tban the masn of an a-particle plus llC nucleus by the ceutre-of-mass kinetic energy at the resona,lce; fromFig.7.4:

7 ,= tM2 l (M ta M) .

Where ? = ?045 LrV, Mz = !2 o,Dd Ml = 4. Hence T, = 5284 keV. Then this er(citedstate of tlO is 5.284+?.162 = 12.446 MeV (= Eo) ebove the grcutrd state. If the resctiongoes to the $ound state tben this erergy goes into the yny (Er) and the nuclea^r recoil

i

' Chapter 7 45

kinetic energy. Using Eq. 11.19 gives

4= Eo(r_ #1,where M is the mass of the recoiling oxygen nucleus: now Md = 16 x g3l.b MeV, theconectiou due to recoil is small and we fi.nd

Et = 12.440 MeY .

The reactionp + t f N - a + l l O ,

clearly has the same neutron and proton numbers as does the compound state of interest.That state will mauifest itself as a peak in this reaction cross-section as the incidentproton enerry is iucreased, provided the mass ofa proton plus a rfN nucleus, both at rest,is below the mass of the compound state. See the figure below showing tbe energr levelsin this case. The test for this is tbat the Q-value (Q2) for tfN(p, a)tfC is less than [, rhecentre-of-mass energf at which the peak appears in the tfO(a,,y)lfO reaction.

Qu = A(p)+a(TN)-a(a)-a( '3c)= 7289 * 100 - 2425 - 0= 4964 keV

This is 320 keV below ?". Thus a peak due to the compound state would be expected tooccur in the p * lfN cross-section when the collision centre-of-mass energy is 320 keV.This corresponds to & protoo laboratory energy (stationary rfN target) of 320 x (16/lS) =341.3 keV.

Thus our interpretation ofthe peak in the coss-section for tfC(a,7)1fO as due toa compound state could be confirmed by the observation of a peak il tlre cross-sectionfor tfN(p,a)tlC at an incident energy (laboratory) of 341 keV.

Centre-of-massEn€€y(mass)

Ir"LII

5284 keV

12440 ksv

II

46 Chapter 7

7.7 Todeal.with this questiou we need to find the mass (ooergy) difference between iheproton plus rlO uucleus (Me + M(6,1a)) and the neurron ptur'illltaf" iluit, ,r)), Ii;;we are told

and that

M(6,r4) - (M(7,14) * nq) = 0.156 MeV ,

. \

M" - (Mp * rn ) = 0.782 MeV .

(Mp + M(6, 14)) - (M, + M(Ttr4)) = 0,1b0 - 0,7E2 i;= -0.626 MeV . . .l l

Wc cnrl now rlrnw lhn otrorgy, lovol dlngrnrn approprlnto to tbotlO(<t,py,lO ruueUi,passlng through a rcsonan.,e (compound etatc) at a ccntre-of.mass i.ioeirc unergy of2.a5 x $3/15) = 2't23 MeV. wo place on it rho lever for n + rf N, orila rs oJzi"Mevabove that for p + rfC, and that flt o + rlB, which is 0.158 MeV above the n + rf!.The figure below sbows the result. Tbe ma.ss of a + rlB is below that of the compoundstate which is therefore expected to sbow as a resoo"rce in rlB(o, n)rfN. A centre-oi-uasskinetic erergy of.T.29L MeV is required to reach the compound staiel tnis co*esponds toa laboratory o-particle kinetic energy (t!B at rest) of.7.29t x (18/li) = 9.942 frioV. :,

Centre-of-massEnergy(mass)

7.291 M€V

o *'1t o.)r.

""u ).,r.

""u

- t

Chapter 7 nz

?.8. 'The

firet step ie to rewrite the table go as to give enugie!,ail they are in the cetrtre-of-mass. Co]umn one nirmbers hbve to be nultiplied by 19/20'tb"giw the centreof-masskiuetic energies. Gomma ray energies sbould be corrected for.iruclea^r recoil but this isa aegligible effect at these energies. The last column also requires correction for uuclearrecoil to gve tho finsl state centre-of-mass euergies and now t19.9$ect ls not uegligible.The reaction has t9 bre; .: ,;r ;.

. .,. .

so that the o-pa,rticle centre-of-mass energiee must be raised by iffrbtor of.20/t6 to obtainthe total centre-of-mass kinetic enerry in the final state. Tbe table now sppe$lt as

Now we note tbat in box 1 (in MeV)

6.13+2.625 = 8.?55

;.6.92+1.838 = 8.758

7.12+1,625 = 8.?45

Tbese are the same total within the precision given and we.conclude that a 7-ray emis-sion and a-particle emission occur one after the other with a total energy release of8.?5(3) MeV. The sa^me kind of result ie obtained in box 3 with a totat energy release of8.9a(5) MeV. We uow compare the uumbers from the first column with thdse we havefound and'the no 7-rays'eoergy', :

' o.63i 8.?s30.801 8.9250.830 8.945

Within the limited precision the energy change from row to row is the same in both

columns. Thus the increase in centre-of-:nass energl in the proton plus rlF system is

appearing directly in the'y-ray plus o.pbrticle plus l$O systemj"-The resonances in p + rlF system must be compound states that are excited states

of !$Ne. The two sets of three 1-rays repeating the same energies must be from the de-

e*citation of llO produced in the emission of o-rays from two of the compound states(the o-particle has no bound excited states). This gives the enetgy level diagram shown

in the figure. : i

48 Chapter 7

. -?r :*prrrrwhy ihere are ,.o ?-rays in the case of the second reson&Dce we shouldconsider the spin and parity of the i'itiar snd final states in the reaction

where fgo i, io'iu. ground state. ;t1;;1il, n * an even-even nucreus so borhhave-epin-parity 0+. If the l$Ne. srate has siiu-parity ir tl* iU. t; l* ,"]pp.", i"tbe final state as rerative orbital -gur* .o.uotun z = i, with iarity coiir-iuutioo(:!'. T9 cou;erve parity aud a,ngubrtomenthm, therefore, ii ,"r, iAoie;;;r.1160+,1-,2+,3-,... Thus tbe secood corpouoa stal.ust have jp U.f""g-i"'ini, ,*i*.conversely the first aad third must have each a jp not belongini to tuis ;eries, otber*isec-particle ehission producing a ground state t!(i would occur. These two states thereforeeach-emit an o-particle producing excited st-&tes oJ {o, presumably tr.*uy Jioo,i'gaogula,r-momeotum aad parity conservetion by a different route,

The conclusion that the Lr"vs r". eniited by excited etates of l$o is consistentwitl lhe data given. To verify this conclusion the uest course is to produce excited statesof r$o in a nuclear reactiortbar cannot involve tjF ", igN" ou!."*ii""

"r r-iu'u.-.o<cited stat€ would o<clude the latter nucrei as being the souce of the 7-.uy, our.ruua.

Contre-of-massEnergy

, o-particle emitting transi.tions are labelled by the

. laboratory o kin€licenergy ln MeV

7.12 Mev )9.9? l,!.Y f Excired srates of r,!o6.13 MeV j

a+ rlo

Chapter T 49

7.9 The determiuation of the charge distribution in nuclei by measurement of electronscattering requires the use electrons with a reduced de Broglie wavelengtb much less thanthe nuclear radius (see Section 3.4). Neutrons cj kinetic energy 0.025 eV have a reducedde Broglie wavelength (X) of 2.9 x lOa frr, wL-ch is very much greater than the nuclearradius, aud it is this length that sets tbe magnitude of the absorption ooss-section, notthe uuclear radius (see Section 7.8).

Section 7.8 provides the material trecessary to aDswer most of the second part ofthe question and only the o-l dependence of the cross-section below 1 eV remains to beclarified. This cross-section includes two factors. On is d2 where I = ft/ P, afi P" is thecentre-of-mass Eomentum; for low energr neutrons P" can be taken as the laboratorymomentum and we have r2r2 s. u-2. The second factor comes from the probability theoeutron enterg the nucleus. By a time reversal invariance argument (see Section l3.l),this probebility is the sa,me as the probability the neutron is re-emitted. Fermi's GoldenRule number 2, iutroduced in Table 12,6, shows that this probability is proportional tothe density of final states. In this case, tbat quantity is proportional to u. Thus the totalcross-section varies as o-1, near u = 0 and away from resotrances. Ofcourse, the fa4t thatthe neutron can be re-emitted means tbat not all the cross-section is due to radiativecapture. However; in medium and-heavy nuclei the largest fraction is radiative, more soas o become,b very small.

The measurement of the radiative capture cross-section in medium and heavy nucleican be done by measuring the transmission of a ueutron beam through a thin target ofthe material. The 'target in'.and 'target out' response of tbe monitor to a steady beamas iu Fig. 7.2 would do tbis. A correction for neutrons scattered (expected to be small)could be determined by measuring the inteusity of scattered neutrons. That is a statementof the principle of the method: for the details concerning Deutro[ sources, beam energyspectrq, shielding and neutroa detection, and other relevant matters, interested readersare rdferred to one or more of the following:l. Curtiss, t.F. (1959). Inlroduction to Neutron Phgsi,cs. Van Nostrand, Princeton.2. Melhods of Erperimental Physics (1961). Voluroe 5A, pp. 461-581 (eds. Yuan, L.C.L.

and Wu, GS.). Acadenic Press, New York.3. Methods o! Experimental Piysiu (t986). Volume 23A, (eds. Skiild, K. aud Price,

D.LJ Academic Press, Orlando.4. Knoll, G.F. (1989). Radiction Detution and Mmsurement (2nd edn.) pp. 48r-SS4.

John Wilen New York.

?.10 Almost all particle detectors depend finally on.the detection of ionization charges.In the case of moving cha^rged particles this is normally straightforward as they lose energyin a material by ionization. For neutral particles sucb'as photoas or neutrons, other stepsmust occur. To be detected neutrons must first cause a nuclear reaction that gives moving,detectable ctra.rged particles, or a product that has a radioactivity that can be measuredduring or after exposure. For example:

//

50 Chapter 7.

Thermal leutrols. A typicel reaction is

n + rjB * ltte + lt i', e =2.J|MeV (see problem 2.14) .The borou is in the form of boron trifluoride gas in a proportioual counter whichcan detect the ionization produced by the a-particle and ihe recoiliog t i oo.r.orlr.uSection 11.2).

b) 100 MeV neutroag'lUse !€utron-proton scattering: a suitable material is plastic scin-tiuator which contains a large fraction of hydrogen. ?he recoil.proton is ieteited bythe light it causes to be emitted as it slows down and stops (see section rr.ij.There are many otberreactions and methods. In addition theru

"r. r*y i*por-

ta't aspects to neutro' detection; for example: efrciency as a function of euergy, eoerrymeasureneBt and resolution, nature and arrangement of the neutron absorber ioa of tlaionization detector, Interested readers are referred to the references given iu problem 2.9,.

1) Excited states of (z,A+L) using target of, (z,A). Expose tie t"tg.ito o.o*oo,rn an energy renge up to &bout 100 kev with a technique for varyiag in a known waythe energy of the neutron bea-- Resouarces in the absorption .ro.r-ruaiioo iudicate com-pound states and these are excited states of. (2, A+l), This technique only explores levelsia asmall rauge above the neutrou separatiou energy of the ground xriuoit 1i,i+t1.Below that level, the states must.be explored by a measurement of the euergy spectrumof 7-rays emitted after absorption of thermal neutrons (see Fig. 2.12),

2) Excited states of (Z-L,A) using target of (Z,l). i*por" target to & mono_energetic beam of neutrons of say 20 MeV. furange matters so that the eiergy spect.umof protons from the (n, p) reaction can be measured. ihis spectrum reflects ti-e spectrumof ground aud excited states of (Z-1, A), as in Fig. 2.7.

The scattering of r0 Mev o-particles by hydrogen mears a centr+of-mass kineticenergy of l0 x (1/5) = 2 Mev. The anomaly occurs at this energy. To reacb the samecentre-of-mass energy iu proton-helium scattering requires 2.5 Mei-incident protons. 1

7,ll section 7.8 stresses the high excitatiou above ground state of the compound stateformed by thermal neutron capture. There are ootrn"lly several routes ofde-excitatioi sothe system will favour the fastest and is very likely to have an El or an Ml first step. A2 MeV El can approach a transition rate of l0r? s-r (Sectioas 11.6 to ll.g). r. competing with this is the emission of a neutron. Fermi's seconi golden rule(Table 12.6) shows that a transitioo rate depeuds on two factorg. The firsi is, in thiscase' the square of a matrix element between the compound state and the groundstate'of(2, A) p|* a neutron. These states are very unlike because of the bigh u*litutioo

"o"rryof the first and tbe matrix element is small. The second factor is the deosity of final states(dN./dE in Table 12.6). For this two-body decay, it is proportional to p, ihe momentumof the emitted neutron. For thermal reutrons p is very snall conpared with equivalenimomenta iD most strong interaction processes. The consequence is that the transition ratebecomes small compared with that for photon emission. , [.

i:;

Chaptet 7 5l

Excrclrcr il' :

7.12 Neutrons of kinetic energy below 1 MeV are scattered isotropicalty in the centre ofmass by nuclei. mass numbei A. Show that averaged over many.collislons. the enorgy lostper collision is a fraction 2A/(1+Af of tho Incident neutron kinetic energy.

7,13 The r6son8nc6 at 1,46eV lor neutron absorptlon by rllln (Fig. 7.l0) is almosrcomplotely lsolated trom eny othor rgsonancs and the cros3-s6ction may be descdbed by theBreit-Wigner formula. In this case it describes the varlation witl\reutron klnstic energy I ofthe neutron total crbss-sectlon o(li) by the formula .;;; .

o(r")-dti1:ffiy14t tl*!:

'

where |-h/Pe P. is rhe neuron contr€-of.mass ror.nt,ffio is the value of ?i atrosonance (1.46 eV). The quantities f.//t and Vll are tho tran3ition rat6s for tho rosonsnceto decsy into tho s661p6l n+rllln and into all channels respectiVely (1. is called the neutronpartial width and F ii the total width of the resonance.) Th6 sratistical factor g is given by

, - (u+ l )" (2r+ l)(2j+ l) '

where s- * is the neutron spin. / is thp sgin of $f n. .l i, tho spin of the compound nucleusformed.

Inv€stigate tho meaning of this formula (start at Section 2.9) and calculate tho valuo itgives for th6 cross-section maximum and compare that whh the value in Fig. 7.10.

11 - 7, t = 5, I"=3.0 ncv. | - 75 mcV.l

Solutions7.12 We use the notation of Figs. 7.4 and 7.5. The label l theu refers to the incidentneutron and 2 to the target nucleus (,4). We take Mr = Mo and, M2 =.AMo. The centre-of-mass momentum is given by,

A' P"=P'fr'

and the centre-of-mass velocity is given by

P r lt t = - --

M " A + L

i i

52 Chapter 7 .

Couider & neutro! ecattered at an angle d" is the centre-of-mass. Using the formula,eat the end of tbe captiou to Fig. ?.5, it eppea,rs ot ar &rgle dr in the laboretory withnomentum P3 (3 = scattered.neutron) given by

P3aind3 = P.sinds,

P3cosd3 = P.cosp. + ilf"U .

It followe tbat the laboratory kiaetic enerry, ?9, of the scattered.neutron is given by

-":.' rr=#=ffi*r'i"*s"a'ry-. ' 'The average, (4), is found by averaging over equal intenrals ofcos0" from cosO. = I to@sd" = -1, foi which (cosdJ = 0. Hence

The average enetily lo$r is "r

- (?s) which !s a ftaction

2A

izTlFof the incidert eoerry ?1.

7.13 At resonance ?" = ?o and tbe peak total ooss-section becomesrr

o(Ts) = ar,rrof .

The reduced de Broglie Wavelength A = ft/Pu Iu thie case P" = P(115/116),P = \/rIM, is the laboratory momentum of the neutron. Hence

,2 _ (hc), .. /116.\2x = riffi*\rrs/

Also .r = Lf2t i = 9/2 nd "I = 5 which gives

2J + te= (@il;lt11

Subotituting these,values, la = 3.0 mV and | = ?5 mV

o(Ts) = 3.99 x 106 fmr= 3.99 x 10{ barDs.

{-s.,**

where

Chapter 7 53

The value obtained from Fig. 7.10 is 2.8 x l0{ barns. The measurements shown in this

figure bave been made with uatural indium, which is only 95.?% rlsln' and at normal

temperature. Therefore o(?6) must be multiplied by 0.957 and corrected for the Doppler

efiect on the sbape ofthe resonance in order to predict tbe observed value. At 293 oK the

Doppler effect reduces the peak by a factor of approximately 0.77. Tbe final result is

2,94 x 10{ barns.

A measurement has given 2.96 x 10{ barns. This value was extracted from a pub

lished curve (Landou, H.H. and sailor, V.L. (1955). Plrysiccl Reuieu,98, 1267-1271).

These authors obtaiDs values for I and o(?q) froIo a Deasurement of many points over

the resonance end a complete error for one point is not available. In addition their analysis

avoided the need for o Doppler correction and they did not report a teEpelature.

A more advanced trestment of the Breit-wigner formula than is posible in this

volume may be found in Krane, K.S. (198?). Introductory Nuclear Physics. John wiley,

New York.

/r,{

,nts

) t i

- Exerclses

8.1 AsinglenucleonEtatoisanoigonstatoof theoporstorsJr. l .r , SrandJ,wlthquantumnumbers 7i, /, 9 and ,1, respectivoly. show that lt is an eigonstato of tho operator l..s with theeigenvalue

*t i( i+l ' t - t( t+1) - s(s+ 1 ) JD,.

8.2 Prove tho formula of oquation (8.4).

8.3 Nowputthoshell model toworkforyoursalf.Belowisatablooftheodd-Anuclei uptorlF' Ootormine tho type, noutron or proton. of tho odd nucleon and use tho lov6l ordoring forthe 'nuclear potential with spin-orbit interaction' of Fig. g.6 to decide its configuration.Pencil your .esults lightly into column 2. Then put the sxpocted spin-parity into column 3and work out your prediction for the nuclear magnetic momont and pencil that into column4. column 5 contains tho measured values of tho magnotic momont. lf other readers havebeen here before you, 6r8se their pencll marks or check theml

Solutions to exercises in Chapter 8

Odd nuclaon Nucleartype and spin-parityconf iguration

Magnetic dipolo momentnuclear magnstonsCalculated Measured

iH

!He

lLi

!Bet t n

i l a

' ic' iN

' ;N

'!o'lo' :F

' !F

2.9788-2.1276

3,2564-1,1776

2.6885-1.0300

0.7024-0.3221

-0.2831

0.7189-r.8937

4,7224

2.6288

Solutione8.1 Let the single nucleon etak: be represented by the ket lX,8,o) wbere

= iu +L)h2l i , t ,E l , ,^ . := t(l+ 1)42U,{,s) ,E r(r + l)aru,r,r) .

J = I r * S ,

".ff;;.-,-,,, Cnaiier e ss

Now

therefore

In addition we have

and

tJ2l!,'t, clL2li,2,sls2li, r' t)

J 2 = L 2 + 2 L . s + s 2

and

L . s = ; ( J r - L r - s r ) .

Therefore l

L. Sl j , / ,s) = ;A( i +\ - qe +t) - s(s+ t) lh2U,I ,s) .

8.2 A particle is in an eigenstate of J2, L2 and 52. Its total magnetic momeut consistgof contributions from its intrinsic spin magnetic moment aud from its orbitat motion,Tleating all these quantities as vectorg:and writing them as operators we have I, I " = 9 'S '

, P t = g tL t

p = g i J .

J = L * S , . : ' +

,. i.i

t t = g i J - 9 1 L * 9 , S .

Taking the scalar product of both sides of the last equation witf, J we have

9 i J , J = 9 1 L . J * 9 , S . J= scl,. (t + S) + 9,S .(S + t)= s t (L2+t . S) +9 , (S2 +L.S) .

i .

9i.0. .tro!.,r o,ii!ti.;{v.; i9affif rhe nucrous in which ,rrr nrffi rr" consrdered ro bemdVlng in a ceniral poiential preiicts'shell closure' effocts in nddii'ir'properties for neurron

:.or proton.Qumbors.2q 8.20,'34;.40. 58,92, and l3g. The.otii5iiied neuron o. protonnumbers bt which these closure e{focts scluaily occw.arc 2,g,2eilgiFO, g2, and | 26 (126for neutrons only). Exptain howt modificafion to the singte.particls potontial accounts for

66 Chaptet 8".

Apply this operitor equation to the ket lj,(a) and use the result ofproblem 8.1 to find

as required.

8.3 Use equations 8.3 and 8.4.

8.4 The a.nswer to the question's 6rst part cg^o be obtained from Sections g.B and g.4The nucleus !!Ne is eveu+ven and so must hove spin-pa^rity jp = 0+. The nuclide

lfAl has an odd proton in the 1d572 level so has jn = 5/2+ ondllSc has an odd proton iulhe lfTp level and so has jn = 7 /2- .

See Sections 8.3 and 8.6 for the assumptions that have been made.

Exercise

8.5 Find out how nuclear magnetic dipote and electric quadrupole moments may bemeasured.

Solution8.5 Many teclr"iques exist for the measuremeat of nuclear Eoments. We therefore referreaders to gome Sources. Two classic books are:

Kopferno", H. (1958). Nuclur Momenls (trans., Sc.heider, E.E.). Academic press,New York.

Ramsan N.F. (1953). Nuclur Moments,John Wiley, New york.However, there have been many technical developmenk since theee bookr appeared, De-scriptions of some typical modern measurements Eay be found in

Krane, K.S. (feaZ). Introductory Nuclear Pliycics, John Wiley, New york.

type and epia-parity uuclear magnetou

pnPD

PD

D

ppnnp

8Ll28r12P3l2Ps12p312p612PL12Prl'.PrlzPu2dstzdqz

u2+ -r.913 -2.t276312- 3.793 3.2564312- :1.913 -t,1776312- 3.?93 2.6885312' -1.913 -1.0300Ll2- 0.638 0.7024rl2- -0.264 -0.322rLl2- -0.264 -0.283rl:/2- 0.638 0.71895/2+ -1.913 -1.893?5/2+ 4.793 4.7224

Chapter 8 57

Exercise

8.6 Thespin-par i tyT?andexci tat ionanergy€ofthegroundstateandasequenceofexci tedslates of tho nuclous rflHf are given in the table.

i ' o * 2 * 4 * 6 ' 8 'f,(kev) 0 100 321 641 1041

Account for this series of states, and calculate the moment of ine(ia of the nucleus in sach of

tho OxCitod states, Commenl On YOur tesults and comparo thom with lhe momsnt of inenia of

tho nuclous considered es a rigidly rotating sphorg. Thg moment of inertia of a sphere is

I M R2. Take 8= 1 .3 x 1 0'rrAut m.

(Adaptod irom the I 983 examinstion of rh€ Final Honours School ol Natutsl Science, Physics. Universityof Oxford.)

Solution8.6 The expected rotational energy is giveu by equation 8.7:

Ei=iu+\h2lx

We calculate 7c2 fot earh of the exited states in units of MeV fm2

td = j(i + t)h2c2 f zgi .

jP EiMeV

!c'MeV fm2

{,7

2+4+

o '

8+

00.1000.3210.6411.041

1.17 x 1061.21 x 1061.28 x 1061.35 x 106

The classicat moment of inertia of a rigid sphere about a diameter is 2M R2 /5. Tbea

xd =2M&R2/5. Now for rflHf:

Mc2 = l?0 x 931.5 Mer/,

R = 1.3 (t?o;trr 6,

therefore

xt = 3.29 x 106 Mev fm2.

This is roughly twice the value derived from the enelgies of the rotational levels: for

coDEents on this result, see Section 8,10.

/

Solutions for exercises in Chapter 9

Solutions9.1 We represent the gauge transformed felds by a prime. Then

A ' = A+g tadA ,

6 ' = Q- * ,otwhere A is a scalar function of position and time. Now, by definition (&. g.1)

B' = curl A, = curl A +curl grad A = curl A = B,therefore

B '= E | .Also (Eq. 9.2)

E' = -grad d -# ,-$ad o+$s,aa r -# -$s,aa,r ,

_ _$ad 6_T,therefore

E ' = E .Thus the fields B and E are unchanged by the gauge transformation.

9'2. consider e- + e- *'y. suppose this process can occur: transform to the rest frameof the initial electron; the total energy iu this frame ia m"*. No* .;il;; th. i*i ,r"r.,

., axorclrec

9'1. s.how that the gaLge rransformation.of equations (g.3) reave unchangod the magnericand elecrric f ields of equarions (9..| ) and (9,2),

r 'rv !!!ev"er

9'2 show that the processeirepresented by Fig. 9.2(a)-(d) cannor satisfy rhe rure lhar'nergy and momentum ara conserved at a vertex ir ail three particras in each case ara rir".(This means rhat a[ of theso processes are physicaily impossibro una* $,ur coniiti-.j-'

9.3 Find the two leading-order diagrams for erectron-erectron orastic scatteiin!. (The twoto bo found are both two-verrex diagrams.) Find one and raber arr ftre in-goiig;il;;t,going partictes with typa and four-momentum. Then find a second which is different notothat there can be no annihi lat ion diagram l ika rhat of r ig. S3ibl.- - - i ! ' Yr ' | !vrv"v "v'Y

9.4 Find the two reading-order diagrams for tho production of an erectron-positron pair bya real photon in tha electric field ol a nucleus,

lf both electron ond photon ore real, and siuce they muet pa.rt company with equal aadopposite momentum in this fra,me, they must have a total enbrgy greater than the resrmass of the electron. Thus energy condenation cannot be setigfied and the auppositioo lelncorrect; the process is forbidden for real partlcles.

The same argument, nbmely energy consenration in a zeromomentum frame foreach process or lte reverse, ehows that all these processes are forbidden if all particlesinvolved are real. .

9.3 The figure shows the two leadinglorder Feyuman diagrams for electron-electron scattering. when the free electron linei are labelled with energy and momentum, these twodiagrams are clearly different diagrams,

Pr, Er Pr, Er

Cl"pt"r e fS

' Pz'4

9.4 The two diagra^ms are:

tt: t'

Exerclses

9.5 Considor a system conslstirig of two non-identical spin-| fdrririons: list alt the atiowedstates that are simuhansously eigenstates of the operators L?, 52, J2 up to ,-3 by theirspectfoscopic notation. For each find the value of 1t, whero,P is the parity. and thesymmetry..Check that rQ and tS, are the only states with 1r=1..

All tho states thst:you have listed are availablo to the neutron-proton system. but only haltare availabla to the noutron-neutron and proton-proton systems: list these.

9.6 Use etsctrostarics to prove tho formula given in Fig. 9.1 2 for the mutual potentiat energyot two eloctric dipoles. Ghoose a non.trivial set of values for d,, dr. and 9, and show that theforce acting on one dipole does not act slong the line of centres.

Solutions9.5 We use the notation 2'+rtrj, and denote the spingarity by je and the total symmetryby s or a, for symmetric or antisymmetric under exchange. For { < 4:

./

/

jP

1Fr, ,Fr, 3F, 2-,s-,4-

lgl l+ IlPl l- s

3Ps,3P1rsP2 0-r l - r2- era,8' D r 1 2 + a

3Dy3D2rsD3 l+,2+,3+ I ,s ,gttr'3 3- 8

a

- Thl only states with jr = l+ are 3,9r and tDt. In calculating the parity wc have madethe usual assumption that two non-identical fermions have even relative parity. (Beware!A fermion and its antifermion have odd rerative parity.) For two identical t..ioor, to.exanple, proton-protou or leutron-neutron, the pauli exclusion prhciple allows only theaatieymmetric.(a) states. These are

60 Chapter9

9.6

/ r E

E1

The electrcistatic potentiar I/, at a poiat X, coordinate r, d1 from an eleciric dipoleP1 at a point O.is given by l

'-Il' ' v =ry?hea the raaiallana transvenie compone!*';'; electric feld at X are

I

E- = -Y-2qqosiei' 0r 4ne6t ,

F^ = _!0V _ prsindr '' r 0h 4tres73

'

The remainhg orthogonal coroponent is zero. The mechanical potential enerry of a diporep lviag in aa electric field E is -p; E. Applv this to the dipie p2 pr*.J"?i; "ri"t.awith anglee defi"ed in Fig. 9.12. The nechanical potentii eneigy'of tl. ,yrt.i- u, i,therefore grven by

| (J = -p2oos02E,*p2sind2cos/.&1 ,= -ffi{zros0rcos,z - sin0rsing2cos/}.

Tlivial ralues of the angres occur, for ocample, if all are zero in whid case theforce does adt along the line oi cerrtree. i" g**J"'.'lave to show that the force ou p2tla'Dsverse tc r car be nou-zero. we use a virtuar work method ir tbe case wben 6. = o and{fowing a dieplacement rddl ofp2 wlthout rotatiou ofp2 so that 02 changes to 0z*dLt,Then the work done by tbe force f, actirg in ths direciion of tbo diaplacetent, ie Frddl

Clnpter 9

which must be provided by the decrease in energy of the system. Therefore

Frdbr -

Hencep = -@,sr in (d1+d2) .

4Tesr'

In general this is non-zero, tbere is a transverse force, and the total the force is not alongthe line of centres.

u t

-(#.H)*,

I

/

' iF

r ' ; r i '

Solutions to exercises in Chapter 10

Exerclses

10.1 Use tha informarion in _Fig, i0,7 and rhe densiry of liquld hydrogen (7i kg m-r) tocalculate the mean free path for 5 GeV/c K. mesons in tiquid hydrogeni

10.2. Look up tho nacessary masses and carcurate the o-varues for each of the foilowingr0actrons:

, t ' + P i K o + n ,K - + P + z o + X o ,

K- *P+ l (o4go,

P + P - K t * E ' * n ,E - + p - n 1 n ,1 t - +Pr Ko+Ko+n,K- +P-16 ' +Ko+O- .

write down the valence quark content for each of the diff'rent particr€s and check that thoconservation laws of olectric charge, flavour, strsngeness and baryon number are ssrisfiedthroughout. Draw a quark flow diagram for the lasi reoctlon.

10'3 Two of the foilowing rsactions arnno, oaa* under any circumstances and a thirdcannot occur by strong interactions. Find th6so and indicate tho reasons in each case:

(a ) K- +P+(01n,(b ) z '+p+ + K+ + ! . ,(c ) z - *p+f (+42o"un- ,( d ) r - + p + 1 1 - 4 1 * ,( e ) K o * p + 6 - 1 p 1 n r ,(f ) p+p + 1.+6. *x- +v- av.,(9) z' +p - 11o1Xo* zr * K+ +Ko,( h ) K - + p + D t * n * z - .( i ) z - + P + f , ' + I - + K o + p + t , + n ,

0 ) z - + p 1 E - + E o + p .

The notaion r means the antiparticle to the x and tho sign convontion is that E. has charge+1 and is theroforo tho antiparticlo to the !-.

10'1 Jhe tatms hadrcn, repton, boson, fermion arc used in rhe crassification of particres,Explain their meaning, giving exampres of theh uso, which can be appried to quarks?

Clapter I0 63

Thef'lowest mass (non-strange) baryon states are:

Electrlc charge- 1 0 + 1 + 2

nuclgons: n pA-baryoris: A- Ao A+ ' A+ r

(a) What are tho quark constituo-nts of each of these states?(b) Assuming the quarks are in states of zero relative angular momontum. what tundamental

difficulty bppoarc to be assoiiated with the A statos. which hivb i -1. and how is it. resolved?(c) How do you explain the occurronco of excited ststos of tho

"u.fioi and A wlth hlgher

values of l?(d) What approxlmate 6gqu6nco In the parlty of these higher atatis wortld your slmple

model predict?The A0 and tho n (strangeness -1 ) both decay to pfoton and r'.meson, E;rplaln why tho

A0 moan-l l fo h - 10'rt ! whl lo that of tho A l ! 2,6 x l O't0 r,

(Adopted from I 984 exsminotion of the Finsl Honours School of Nstural Sciencr, Phyricr, Unlvc'tity ofOxford.) |

10.6 Descrlbo tho exporlmsntal ovldence whlch iupports the quark model of hadronstrucluro. " :

ln o eimple quark model the llghtesr hadrona cre congldered al Odrjnd ctatsl of u, d rnd rquarks or antiquarks. Glve the compositlon of the mesons n' , f , K!, K' (S- + 1 ) and (o(S-1) and tho baryons n. p and A according to this model.

State, whh reasons. which of the following reactions can proceed via the stronginteractlon:

l(-+P+(o+1Ko+n + A +no

K-+P-A+floKo+p+K' *n .

t .

ln the additive quark modil tho total inreraction cross.sections ol high.onergy hadronsare assumed to bo duo.to the spmg of.the interactlon cross-soctions of tho constituontquarks, Taking account of the diffeiing cross.sections for different, quark pairs (qq) andassuming that d(qq)-o(qq), use thii model to prove the relation . ,

r(Ap) -o(pp)+oKop) -o(z.p). :

At 8 laboratory beam momentum of 100GeV/c tho total interaction cross-sections areslowly varying functions of momentum. Given that

oKop) r20x10-t bit ' (r 'p)-24x10-r br (pp) r39x10- ' b

mtf[' i l l t. t t l iit

i

llili ,ril 'rll:l i

i l it;l :III

I

/

64 Chapter 10

lll.^.:.::,:if":t the interacrion cross saction oi the E- hyperon (quark composition dss)with pfotoffi. "-" ''"

(rurptca trom iliiffilrmination of thc rinat xonours siioot of Natural sclenco,'physics, universlty olo x r o r d ' ) . . ; . ; : : ' * : ' - ' - " , ,

'

10.6 outtin6'tli6Guarr moueroerlrriprioni* rrrefrioii,'*ii;n n.ri.-o-. ciu" ii:. quarrrn^odel d_escriptloh and strangeness of other mesoii..*ith ir-O- wtricfr .* U. *;irJiiarrom s, q ano s quarrc. How do.mosons with /r-!;.grise. in this modol? What 1? would youoxpoct tho tirst 6iiited statos of tho 0- and 1 1 meJjns ro hauel ..Thep mesonfiasstrangeness s=o.ir=l- and a mass of zio uev. what wourd yoi expectto be the principal decay mode gf .. o1" why can the / meson not decay to zozo? wharwould you expoct to be the principar decay moder and approximats rifetime of tho ,o?

:fflll'Jj:lt;;; *: y' misht be nroducei 311 itgj:ii;;;;-il ffi;-.;;i1,.(Adapted from r9g5 oxsmin.tion of thc Finar Honoure schoor of Naturlr science, physics, university ofOxfad,)

10.7 stare what is meant by (r)..charge .ymmotry (D) charge independence of nucroarforccr, lllurtratc ahirrgc rymmctry by rcierrrng to ttre .ncrgv revers of ilght nucrer.Explaln qualltatlvely the followlng phenomeno:

(i) At a csrtain momentum tho.cross-sections for z.-neutron o(n.n) and ,!--protono(t-p) interactions are equal whereas o(K.n)-22mb and a(K-p).6i66.(ll) The p0 and Ko mesons borh {ecay pradominantly to (2. +a-). Tho mean life of rhe po is1 0-a whereas that of the Ko is O.tig x i g-ro,.(iii) The mean life of tho E'-m'son is 2.6xr0-rs, whereas that ot rhe zo-meson is0 .8x10- t .s ,A-beam consisting of a' and r' mesons strikes an iron absorber. Given that th6 crosseoction for n. interactlons with hon is 600mb/nucleus calculate the fhickness of ironnac.$ary to attonuate the z' beam by a factor of r000. Exprain why tho .nron, *orti Luattonuated by a different factor.[Tha density of hon is 7g00 kg m-r. Tho rerstive atomlc mass of iron is b5.gs.j

(Adrptcd trom | 98r orrminarion o, the Finsr Hon6un schoor ol Nrturat science, physics. University olOxford.)

Clnpter 10 65

Solutions10.1 The total cross-section, a, read from Fig. 10.7 is about 19 mb = 1.9 x 10-30m2.The meaa free path (Eq. 2.15) is (na)-t wbere n is the number of protons per cubic netreof liquid hydrogen.

n = 7L x 6.022 x 1026/1.008 .

Hence( n o ) - t - 1 2 . 4 e .

t0.z The concept of Q-value is described in Section 7.3. It is the sum of the rest massesof the colliding particles less the sum of the rest masses of tbe collision products. Thus,for example, the reactiou

r - + p + K o + A

has Q given by

Q = 1 3 9 . 6 + 9 3 8 . 3 - 4 9 7 . 7 - u 1 5 . 6= -536 MeV .

The calculation of the Q-values is therefore straightforward and the remaining values are,In MeV,

104, -391, -746,29, -957, -1232.

The valence quark contents may be formed found in Tables 10,3, 10.4 and 10.S. Aquark flow diagram for the last reaction is: ,.

K - su

UpH

10.3 Reaction (f) does sot couserve electric charge and reactiou (h) does not conservebaryon number. Since neither conservation law is knowo to be broken, these reactions arenot expected to occur,

Reaction (d) has a two unit change iu strangeness. The weak interaction can cbangestra,rgeness but has a very small cross-section for lASl = 1; requiring lASl = 2 *oo16mean a nggligible cross-section.

10.4 For a defnition ofthese terms, follow tbe index, Quarks are strongly interactingand are strictly hadrons; however, usage is now restricrting hadrons to be tbe observablecombinations of quarks,(a) See Table 10.5.(b) The three identitical quarks in A- and in A++ appear to have s total wavefunction

that is symmetric. This violates the Pauli exclusion principle. This problem is resolvedby the introduction of the colour quantum number, as described in Section 10.?.

(c) Excited states will occur when there is orbital angular momentum between thequarks.

(d) Each additional unit of orbitol angular momentum brings anothcr factor of (-l) tothe reckoning of parity and increases the mass. Thus we expect the parity sequence* (ground state), -, *, -...

K+

lc

r,lrI

66 Chapter 10

The decay Ao * r- + p conserves ell quantum numbers and therefore is caused bythe strong interaction. The Q-value is large (- 150 MeV) and so it has a mean life oforder 10-23 s. The decay h + n- * p has a straugeness change of *l and this Beang &quark flavour change; such a change can only be brought about by the weak interaction. ,; ,It has an efrective strength 10-r{ that of the strong interaction so the mean life is of order i,"'l ,10-o g,

. .r

10.5 The experimental evidence for the existence of quarks and for the quark model "f

lii{hadrons is listed in Table 10.11. The quark coutent of these lightest hadro.s is given in. iX.:il$Tables 10.3, 10.4, and 10.5. . ,ti. 'i,!

Reactious K-*p - K0+n, K-+p - A+r0, aud Ko+p * K++n proceed Uf t le l [ , i istroag interaction siuce all quantum numberg are conserved, The reaction Ko+n * A+,nohasachange ins t rangeness(As=-2)audthere forecannotproceedbythes t ronginteraction.

To show o(Ap) = o(pp) +o6op) - o(tr*p). we use the quark content and theadditivc quark model:

o(Ap) = 2o(iu) + o(sd) + zo(uu) + o(ud) + 2o(du) + a(dd) .= 2o(su) + a(sd) + 2a(uu) + 3o(ud) + o(dd)= 4o(uu) +a(ud) +o(dd)= 2a(su) + a(sd) + 2o(au) + o(Ad)= 2a(su) + o(sd) + 2o(ud) + o(dd)= 2o(uu) + o(ud) + 2(du) + o(dd): 2a(uu) + 3o(ud) + a(dd)

where we have used o(du) : a(ud) = a(du) = o(ud). Then

o(pp) +o(Rop) - oQr+p) = 2a(uu)+3o(u<t)+o(dd)+2a(su)+o(sd)= o(Ap).

We now need o(E-p); this is found by the following steps:

o(E-p) = 4d(su) +2o(sd) +za(ua; +o(dd)

= zo(Rop) - (2o(ud) + o(dd))

= 2o6op) - (2oQr+p)- o(pp))

= zo(Rop) + a(pp) - 2o(zr+p)

Numerically

o(E-p) = 40 x 10-3+39 x 10-3 -48 x 10-3 =31 x 10-3 barns .

10.6 The 6rst part of tbis problem may be answered after reference to Sections 10.8and 10.4. The first excited qQ above 0- and 1+ are tPr, 3Po, 3Pr, md 3p2 having,P = 1+,0+,1+ and 2+ respectively.

The principal decay mode of the p+ is expected to be and is

P + - t + + r o .

This mode conserves all quantum numbers so is caused by the strong interaction, Allother euergetically allowed modes are suppressed:

o(pp)

o(Rop)

o \ f r 'p )

Cbaptul0 67

uecay modeo * -

Critical factor caueing suppreesiolrelative to r*ro node

' -vobserredbruchlng fraction

1t+1 Electromagnetic - 4 * 1 0 - r(J?rr- r (0n)-, ?r ' ? 1. Forbiddea by strong interactions

aud can only be caused byglectromagaelic interactiong. between 'Unobserved

,iif&,r h,.,iiiusditdit quar:ts'.:2: Sifilll:lifi{t' itate pbase iir'dii;

[ 4 r ) r ' SmaU [Dal state phase 8pace. .,-< 2 x l0-r

' These decays violate SI conservstion of G-pa,rity. See Perkins, D,If. (1932). Intruhrctioa toHigh Energy Plyricr (3rd edn). Addison-Wesley, Readiug, U.S.A.

The p0 has ie = I- so tbe decay into r0r0 can only corurerve total angula^r InoDeD-tum if tho two r0 are in an / = 1 atate of relative orbltal angular mdmentum. Thie stateis antisymmetric and tberefore forbidden for two ideutical bosons !y the Peuli exclusioqprinciple. This principle requires thst s state of two e more ideutical boson must have awavefunction tbat is symmetric under exchange of eny two. ' ' ..

For a production pethod see Fig. 2,7, and for infening the lifetime see Section 2.9.

10.7 Charge symmetry and charge independence'are described dnd illustrated in thecontext of nuclear foices iu Section,9.Z

(i) The states ur-n and n+p rure cbarge symmetric since uuder the opr:ration of cba,rgesymmetry n =r p and r* + r-. Therefore the collision cross.sectioDs should be identicalapart from the small effect of the n - p mase differeuce and of the electromagnetic inter-actiou. The states K+n and K-p ele not conoected by the operation ofcbarge symmetrysince they have opposite 8tralgenessr therefore there is no reason why the corespondingcross-sections should be equal. :

(ii) Tbe decays are

p o - r + * r - r r = 4 x 1 o - 2 r g ,

K 9 * r + + r - , r = 8 . 9 x 1 0 - u s .

I

As we bave see4 in quegtion 10.6. The p0 decay corselves all quautum numberg a,nd iscaused by the etrong interactiou. The K0 decay does not conserve stralgenesn rod ca[only be caused by the weak iuteractioo, bence bas a mean life $ester tbatr thet for tbepo by a factor of order 1013,

(iii) The decays are

rr - pt +vp, T = 2.6x 10-8s,

ro -^ l * ,1 , r = 8 ,0x 10-17s .

, l. l

i

68 Chapt* 10

These are the domiuaut decay modes in each case. The first occurs by the weakirteraction, the second is obviously electromagaetic. Tle former is ."J ".i1,

,i* ,0.latter aud leads to a much longer meen life toi tru oi decay thau for the ro decay.The a*enuarion of a beam of particles passbg througL;;;;;fi; ,i,J,u, Bo.2 . 1 6 : . .

/V = /VoexP(-c/l) 'where ,\ = (za)-I, tlg nean free path. For pioas in ;iron" i "

. ; -:",..o = 800 mb/audeus = b x l0_2e m2 .

For iron n = 2900 " 9:0F , |O26/ES.8S. Hence :

) = 0.235 m.Tberefore to obtain rn attenuation by e factor of 1000 require a thickness ofiron r, given

exp(r/l) = 1000,that is

t = 1 . 6 2 m .The pi-mesons interact strorgly with the nucleons of the nuclei of the iron absorber,eo that the interaction cross-section is appto*iratety geonetrical (- tr}z,when fi is nowtbe auclea^r radius). Mu'mesons do nof interact r;;.;gly with nucleone but do interactelectromagnetical; bowever the erectronagn.ii..r"*-rl.ti.ns ere ress than strong inter-action cross-section by a factor of order. toi. trus ne

"it.ou"tioo for mu-mesons is muchless than that for pi-mesons.

+€€4* l rL : . ; ' ! - i l t r : i

Chapter 10 69

Exercises

10.8 oraw diagrams to illustrate tha mochanisms by which leptons and hadrons are creatodin e'e- annihilation at high energy.

In an electron-positron collidor the particles circulate in shorl cylindrical bunches ofradius 1 mm (transverse to the direction of motion). The number of particles per bunch is5xlOtrandthebunchescoll ideatafrequencyofl MHz.Thecross'sectionforp'p-creationat 8 GeV total enorgy is 1.4x 10-!g cmr; how manyp'p- pairs are created per second? What istha rate of hadron production at this energyT

lndicato. with the help of a sketch, how g' and p- frorlr o'e- annihitation can be identifiodand suggest the principal sources of background.

(Mapted lrom the 1985 eraminadon ol the Finab Honourc School of Natural Science, Physics,Univerrity of Orford.)

10.9 Ouantum electrodynamics predicts the decay rate for 1rS, positronium into threo y-rays to be givon by

,,,_2d(n'-9) m,C' 9 n h

Assuming ths same formula applios to the decay 9(3097)-three gluons followedby fragnlentation to hadrons, ths tragmsntation occurring whh unit probability, calculatethe effective a, for tha ccg vertices in this docay, The full width ol the rl(3097) is 63keVand 82% of docays are to hadronic final states, Assume the mass of lhE charm quark is1.65 GeV/c,

Now consider the radiative decay p/ d t+g+g. This, followsd by fragmentation of the twogluons. woutd rspfesont tho mochanism for radiativa decay io hddronic statos not contain'ing charmod quarks. In it one ccg vortex of yr + 39 becomos a ccl vortox' Modify the formulato give your prediction of the dEcay rate and estimate tho branching fraction for this radiativedecay mode.

10.10 Thehyperf inespl iningof thelsstateof positroniumisaboutl50timesgreatarthanthe analogous splitting for the 1 S state of the hydrogen atom. By considering the origin o{tho splining, can you explain why this ratio is as great as this?

I 0,11 Show that the onergy of the lovol of positronium viith principal quantum number n isgivon by ,,

" ' E=-a 'Y ' ,c r l

where m. is the mass of the electron.

70 Chapter t0

Solutionsl0'8 Two of many posslbre quark flow dragrams for hadron production in high energyelectron-positron annihilation:

e'+ e- + zit+ +ztd + no e + + e - - ) A + K + + r o + r - + f r

These diagrams do not show the gluon exchange lines which create tbe secondary qQ pairsaad which arrange the final productiou of ciourless hadrons. rn aadition, we hlie ootshown productioa through an intermediate state of a Zo (see Fig. l2.Z), : -- '

To an electron in the electron bunch, the positron bunch -appeais

to have E x 10'positrous in a disc of radius 1 mml that is an appaient surface density of s x totr/(r* tg-u;e+ m-2' The p+p- production cross-section is r.4xr0-3? Jil;;;;ilr; ri,ll.r..r."has a collision producing a p+p,. pur in crossing the other bunch is

5'x 10rl,r x lo;

x 1.4 x l0-o' .

But there are 5 x r0rr electrons and the bunches cross 106 times per second, Thereforethe expected production rate of p+p- pairs is

(5 x tgtt ;z x 1.4 x lO-3? x 106- = . t . t l X l U - - S - - .r x 10-6

8 Gev it is possible to produce the first two generations ofquarks so we expecta (Fig' 10.10) to be r0/3. ?hereiore the hadron p.oa,i.ti,io rate is expected to be-'

1.11 x 10-2a = 3.Zl x l0-2s-r .

A simplified detector wourd have compo'ents shown in the next figurg, this showslongitudinal aad transverse sections.

6h"pt , to Tt

Electron aud positron bunches cross et o, the centre of the eolenoidal detector. A is e thtuwslled beem vapuum tube. A magnetlc ffeld puallel to tbe uttg tr produced W a cunentin the solenoid C. Tbe.returu peth of tbe megnetic field ts by.;dte€l cy[nder, p-S is a,low doueity dotector giviug apaco coordlnatoe aloag tho troJoctoit& of chargod pa,rttcleg.E is a detector which gives trai:k coordirates ofparticles penetiitiia the coil aod steel.

The collisioDs a,re betweeu e+ a^ud e- of e4ual and oppositripiiineDtum. Thus eveatsgiving a p+ ap.d lt' alone produce these particles witb equel.,iriid;opposite Eomeatumrtheir trejectories bending appropriately h the magnetic 6eld and.both easily penetratingtbe coil and steel (c+D) to be detected in E. The tbidoess of the steel is choseu so tbatin the energy region of interest, there is a minimum probability- that hadrons penetrstethe steel without absorption or scattering.

Major backgrounds are due to:l) Low multiplicity hadron production, in particular,

liII

2)

3)

e+ + e- - n+ + n- * neutrals ,

where the two charged particles are nearly back to beck, botb penetrate the steel aldmeasuremetrt e$ors on their momenta make these eveuts difrcult to classify correctly,Low multiplicity hadronic events where r - pu decay in flight allow easier peaetra-tion of the steel by the observed charged particle(s).

Tau pair production (see Table 12;1)

e + + e - ' t + + r - t

where the t&u-mesons both decay by a mode with a single cha.rged particle

r i p + v + v

t + t * u t

::::and both charged particles p-enetrate the steel.4) Cosmic ray mti-mesons passing near O can also imitate p+p- pairs." . All these backgiounds haid features that allow most evei6io:be properly classifed.

However, apparatus iroperfections do allow some event topologies to mimic p+p-.I

10.9 Reptace a by as and rn. by rn. to obtain the transitionlrate'oi l3,gr cbarmonium(r/(3097)) decay to badrons. Thus

'hu =,rg?4Jl,-n,".cz

The full widtb is 63 kev &!d tbe branching fraction to badrons.i! 82% so that the partialwidth fo! this mode is 0.82 x 63 keV; this is &.r. Thug

^ 9rturtd : = --J

2Qr2 - 9)m.C

as = 0.287.and

72 Chaptqr 10

For the radiative decay, one ccg vertex becomes cc? and one as b replaced by an a. Tbusthe.partial width (n r) for decay into one 7-ray plus hadrons is alrar/as = l.3l keVpredicting a transition rate of 2.00

" 16tE r-1 and a branching fraction of 2.og%,

10.10 The hyperfine splitting in positronium and hydrogen is due to the difierence be-tween the iuteractios energy of the two constitueuts' magnetic moments iD the s,sr erdr.9s states. The magnetic momeat of a particle of spiu s is (see Sectioa g,?)

inlr,= 0tfis,

For the positroi. g, = *2 and ry,C = g.EUlMeV.' For the protol, g, :- *5.b9 andMec = 938 Mev. Thus the positron naguetii!'inoment is about oso tin.s that of theprotou. Tbe reduced mass of positroaium is half that of tbe electron ia the hydrogen atomso that the magnettc moments ir positroaium arei on the average further epart bi a factorof 2- thon they qpe in hydrogen. since tlie magnetic iateraction is inveroeiy prolortionalto the third por* of the eeparatiol (Fig. 9.1t), thig makes for a factor of i. hon tlissi^mple argument we o(pect positronium to l"*

" uyp.tnne splitting about g2 times that

of the hydroger atom. In facr tbe ratio is about 14s, ile annilihtio-n diagram (l.ig. 9,3b)increases the positronium splitting above that expected from eimple mag;etic inte-ractionarSurDents.

lo.ll. The rqr!! for. tbe euerry of a bound state of an electron in a hydrogen atom,gven by solving Schr6dinger's eguation, is . . .i

^ mieao"=-W,

In positronlum tho effectlve mass lg tbe rcduced mass, m./2. Then for positronium

r _ rflrel azmrio"=-64nt ] f f i " - -F .

Clnptcr 10 73

Exercises

10.1 2 What is the isotopic spin of the O. meson ('So, i t=0-, cS) ? Why would lhe decay oftho D: - D.+ rf be forbidden under the effect of tha strong interaclions? (D': is the 3Sr statool c5.) ,

1 0.1 3 Exptain how the isospin and strangoness quantum numbers arise from tho existenceof multiplets of elemontary particles and from experimental observations on their ptoductionand decay. Stato tho isospin selection rulss observed in strong. electromagnetic. and weakinteractions, and with rsfetence to these saloction rules discuss each of the followingStatements:

(a) the z' and z' mesons aro of equal mass, but the X' and E- baryon masses ditfer by8 MeV/cr;

(b) the mean life of the Io 'baryon

is many otdots of magnitude smaller than thosa of the Aoand Eo baryons;

(c) the reaction l('+prf,++l(+ has not been obs€rved at any K'energy.

(Adoptod from 19?7 cxeminrtion ol tho Fin!l Honours School ol Notu?61 Scisncs, Physics, Univorsity ol

Oxlord.)

Solutions10.12 Both the C and s guarks have zero isotopic spin. Therefore the D" and Dj have

zero isospin, The zr0 hos isospin l. Therefore the dccay

Di- + Df +ro

involves a change in isospin and canuot be caused by the stroug interactious.

10.13 Tbe first put, of the problem may be answered by reference to Section 10.11. The

isotopic spin selectiol rules express the allowed changes in total isotopic spin quantum

numbeis (t, h). The dange ie ftom all tho hsdrons in tho initisl steto to ell tbe hedrons

in the final state.l) Strong lntoractions: Al = 0, Alo = 0,2) Eleciromagnetic iateractions: for A - B +'y wbere A and B are hadroulc states'

3) tffiifi*lttjl'=urtinarcn*a in tho commontr6n p, 230, tho chansoe in iootopicue not usefully summarlsed. Ae en example, the eemi'leptonic decaye (Section 12.1)

lnvolving u, d or e quarks (or antiquarks) only, hovo

A 5 = 0 , A t = 0 , * 1 1 A l 3 = j l '

orAS = AQ = *1, AJ = *l /2, 66 = 6912 .

(Q is the charge of the hadronic system and S its strangeuess.)a) The r+ and zr- belong to the same isotopic spin multiplet but tbey are also related

by charge coljugation (Section 10.3) and must, therefore, bave the same mass. E+

and E- are not related by charge conjugation and need not have the same mass'

r {74 Chapter l0

They belong to the same isotopic spin multipret and therefore will be expected ha*L\ a mass difference of a few MeV.

ess lqersrure wtu oe expectedD/ l'he Eajor decay modes are

E0 - A + ?: 1l

= -i'. At3 = 6, electromagnetic ino-o-+ l ; a t= q112, i i i= lvz ,A,9=at , pu"1.E0 + A0,r-no, at = q1'12', iii= _itr,As=a1, q,eak.,The electromagletic*teraction is a factor of about r0rr times stronger in its effect:111r*::1i

so tho mean 'fe or thi ii is ,i]r"".o., ot magnirude tess rhan thoser;) The reaction

K * + P * K + + D tbas A5 = -1' It can onry be caused by the weak interaction aud the cross_sectiou istherefore so small as to make tl. ,.*ii*

"ffir*ur..

lilI :i :

L

Soluiibn's to exbrcises in Cfradier 11-"

.. . Exsrclccr . .:...:l

11 .1 Show that the maximum energy thit can bc transierred to .i .i..tron, initiary at rosr.by a particla of rast mass tt and total energy E is glven bv

2n,C(8,- MrC) 2rt,Cgy"" M|c+nl/+2Emi tg.fEr\ M / , \ M I

where/c is the verocity of the particra and y-(1- f2 )-r, as usuar. This is thc rerativisticsilyexact formufa. Show that for the extreme relativistic casa E> M2czlrrr. .'"-.'--',r

t u- '-E'

and for a less extreme case when , n fic and E< M.c2lm,

v*,*2m,f)!c\ .,.

Non.refativistically (E-Mcz+7, T<Mcz) the maximum encrgy transfer is given by

u = 4m,MT^, (m.+Ml'

Calculate the value of v-. for p.mesons with E-5 GeV.

11'2 show that if energy v(<<ra.cr) and momentum g are tansferred to a freo.slationary erectron, then g.g =2rn v. rf v> z.c the probrem L r.r.mti.ii" ir,il1*, o iithe four'momontum transfer. Look at Fig. 9.6 to find out what thiIineans ana ,no* tiriq2n-2mrv .

11.3 A particle a, of a mass Mr(>n,),'chrarso ZJll.,and initiat kinctic energy I has arange R.' show thar rhe range in rhe same materili ior parricre b i, gr"ri il ii, JJi"slaw

, M . z lRr(M v Zv Tb)'

i, 4 R,lu,, 2,, T,- M,TJ M ).

llili:::,,1_i,1ltly,l-9^r]:-:.r',r1. or.positrons.) whal assumptioli are impticit in rhe useof this scaling law? ls it valid for relstivistic. magsive pa.ticfesf

11.4 UsingthergsurtsgivenrnTabrelr.l,showthatrhekineri. in.rgy(r)specrrumofd-rays knocked out of atoms by a fast heavy parricle is given approxlmately by

d N KdT T2

for l< I< v_,, where l( is I consrant.

. ,::;' ' . . :

76 Chsptet rr :':'

, '

Solutions11.1 In the centreof-mass, the collision that trausfers the greatbst momentum is oneia which the electron reverses its direction of motion. If u" is the velocity of the centre-of-mass ia the laboratory (electron initially stetionary) then in the centre-of-mass, as acotureque[ce of the collision, the electron revenres its velocity from -u. to *u.. Ttansfor-mation back to the laboratory after the collisiou nea,ns adding velocity u" to i to obtainthe poshcollisio4 vglocity of the electron in the laboratory uhb. using the relativisticadditiou of velocitiei,

us -l u6-rlrb = - ,r + 0 ;

, .where all velocit'ies are in units of c. The energy transferred to the electron is jur;t its finalkinetic energy 4 and it has its greatest ralue in the collision described. Therefore

Yu=Tc= tzh ( f t b -1 )

where

?lo=m;

a.nd m. is in uaits of enerry.Now the velocity of the centre-of-mass in the laboratory is the total momentum of

the initial system di',ided by its total energy. Therefore

. , . - . . . . . ' ! c = f f i '

where P2 = E2 - M2 end "

-t;:':#H'ituting we have

Hence

')leb = ,, (E + m')2 + P2

{ ((82 + m,)2 + P2)2 - 4P2(E + m.)2

. = (E +m,)2 + P2,w,

Therefore '

t/nu=7,%(T.u-l) = ffi

_ 2m,(E2 - M2)M2 +.m!,*2Em,'

which is the result given ia the problem, apart from factors ofC and c{. For the incidentparticle we put 1 = E/M, then p = PlE, E2 - Iu{2 = M2p2f end.

.. 2w02t2u * = @ .

Thrs formula for z.o can also be written as

Chapter 11 77

For the extreme relativistic case E >> M, E >> M2 f m. and E )) m" therefore

Umu= E '

For the less extreme case where M 2 m. and E < M2 f m" then the denominator terms(^,lM)' afi 1(m"lM) = (Em,lM2) are both much less than I and

v^u=2mo92,f2 =2rn,(82 * M') lMt .

For the case where the incident particle has a non-relativistic velocity tben E = ? + Mand?( M,g iv ing

r/mu=ffiffi1_ 4m.MT-

(M +rn" )2 '

Consider a 5-GeV muon. The mass of the muou is 0.1057 GeV. Therefore M f mo =206,9 and, M2 /m. = 21.86 GeV. This case qualifies for the less extreme relativistic formula:

/mu = 2m,(82-M') lM': 2.29 GeV .

11.2 Energy v transferred becomes electron kinetic energy and momentum q trans-fened becomes electron momentum. Since y (( rn c2, the velocity of tbe electron is non-relativistic and 2zrn. = q. q.

The relativistic case has Iour-momentum q = (q,v/c) and. q2c2 = u2 - q.q8(Fig.9,6). Ifpl is tbe electron four-momentum before the collision that transfers energyand momentum, and p2 is the final electron four-momentum, then conservation of energyand momentum means that

q = p 2 - p l

q 2 = q , q = p 2 2 + p l - 2 r r . . p z .

Now p = (P,Elc) atd p2c2 = E2 -P ,Pc2 = m2ca. h particular

Therefore

pr * (O, rnoc) , -

p t = (P2,m"c+v /c ) ,

Pt 'P2 = mucQnoc * vf c) ' . '

q2 = 2m3t -2m,c(m.c+vfc)

= -2mrv .

and

Thus

11.3 The range, R, in a given material, of a particle of mass M(>> m"), cbarye Ze andinitial kinetic energy ?s is given by (Equation 11.2)

to ATR(M,Z,Ti = Jr"fiG

D

'.t

78 Chapter lt

Now dfldo = -22!(T/M) wberc f (TlM) is essenrially a function of the velocity of theparticle. (T/M = 8f2 not-relutivistically and T/M = 7 - t, where t = UtllT/C,for a relativistic particle.) It follows that

' v

1 - . I 1 o d r g )

. rtnM,z,rol= -n |r",,ffiand therefore that

'-n1u,z,ro'1=- f -YM Jn I@)

whey V = T /M aad, yo = To/M . Hence all particles with the same y0 have the same value,of z' R(M , z,Ts) / M . That is, that two particles, a and b, with initial kinetic energies ?iand ?6 have

ff "r*",r^,^)= frR 1uo,zr,rr1,provided that T,/M, = TalMv. Therefore

1 .2.

R6(M6, z6,Ty) = p$nfu., 2",T. = TbM.l Mb) .tutr 26

The assumptions implicit ale thatThe loss of energy is by ionisation alone.The rate of loss of energy by ionisation is a function of the velocity and charge, at:all velocities, but not of the mass of the particle.

The formula canuot be used if one of the particles is an electron or a positron sinceassumption I is not valic in their cases. Both lose energr by bremsstrahlung at a rate..greater or comparable to ionisatiou loss at all but the very lowest energies (see section,,11.3). The next heavier particle is the muon: at all but very high.oetgies muon energyloss by bremsstrablung is negligible.

Relativistic particles may have ranges that become comparable to or greater than themean free path between nuclear coltisions, In tbis case nuclear collisions can frequentlycause large, single, energy loss events which decrease the range and which cause largefluctuations in the distance penetrated by a sample of such particles.

11 .4 Thecross-sec t ion fora fas t ,heavypar t i c le to loseenergy in therange u lov*d ,vin a single collision with an electron is (doldv)dv given in Table 11,1, siuce r, = 1, tr560nolgy 0{ tho knocked 0n eloetron, tho dlllarentlal erosr"r60tl0n for protluelng a,n €lootronof kinetic energy T is given by

da i la I [ . u l . l z r \ ]f r=&"7[, ; ; ' \ , -V)J

The frequency distribution in energy of thc ejected electrons is

#"#['-#('-5)]For a fast particlc V/c: I nnd

d N l

7Fof r

T{T:" t } ' . - ' i . . . , . t r , ' , ;

[l'Ciapter II 79

Solutioniir'ttii. radiation leii6h is aCiinea in Section 11.3. Electr;;ll;i dnirgy 8.2 GeV/cemerge from crossing a radiation length of leed with an avera!'e enerJ of 3.2/e =1.18 GeV. Thus this average enersf loes is 2.02 GeV. . ..;; .

";,.Tbe root mean,square multiple scattering angle is epprocldi-i*ely giwD by. , " , & , . .

for liangversar or ri'aisii:lii .c ,(E6'.:rr.n). For p I' 1000 MeV/&Ht L = Lal2b, rhisetgld,is 4.? mr, An.elechon.emitling a photon of enerry 87 iffiibii tlet at ao a^ugleapproximately m,C/E, to its dirdction of flight. [u th6 case of,liliis,problem this angleis 0.6 mr. Thus the augUlar distributioD of these pbotons is rodStly.determined by themultiple acatteriug of the radiating electron.

Eierclse

PeakABcDE

F

Energy (keVl1 368275417322243| 1532520

11 .0 Flnd a formula for the maxlmum onofgy thNt 0 photon of sncrgy E, csn transtor to. fr66electron.

Thc figure shows tho encfgy spoctrum obtained when a gmall rolld stltc detector bexposed to t.fsdlatlon ffom flMg. Account for thc structuroc at E and F. The peaks D snd Care 51 1 keV and 1022 keV lower in energy, rospectively, than 8. Explain how they arise.

(Ad!ptcd ,rom th! .| 984 crrmlnrtlon of thc Fln!l Honou,r School ol Nrturrt Sciincc, Phyrlct, Unlvrrrljyol 0xford.)

I

1'1 .5 Show that the average radiative energy loss of electrons oi i"liJbev/ccrossing oneradiation length of .lead'lc rbout 2.0 GeV. ' . - . r j .

A bcam of clectronr of cnrrgy 1 GcV trrvcner normally r loll of,lcad fith ofa rtdlatlonlength thlck. Show that th6 angular distribution of bremsstrahlulg,photons of energy800 MeV is determined mdre by multiple scanering of the elEctroriS rhan by the angulardistribution In ths basic radiation piocoss. .

/

80 Chapter 11

Solution11.6 The $eatest energy is traasfened to the electron when the incident gamma raygcatters back aud the electron recoils forward. This is then tbe game situatioa as iuproblem 7.4 where we forrud the gamma r&y ererof required to give a forward recoilingproton a bnetic energr, T, equal to 2 MeV. Solving the formula for E, as a function ofT, fot T as a function of .B", with MrC replaced by mrc2, gives

T = ZEl '=,-

2E., + tncC - 'mo'

This is then the ma.:cimum energr that car. be transferred to a stationary, free electronby a gamma ray of energy Er.

The excited llMg enits g&'oma rays of energr 136g kev and, of 4L22- 1368 =2754kev. Photo-electric absorption ofsuch,y-rays ofthese two energies in the detectorgenerat€ the pea&s A aud B respectively. compton scattering oI these,y-rays gives recoilelectrong with kiuetic energies from zero up to 2loo. This has the ralues 1153 ani 2b20 kevrespectively. Thus E and F represent the top end of the kiaetic energy spectrum ofcompton recoil electrons from the two primaryy-rays. The ldeal spectrui-is not flat butrises to a pea^k at ?.ol'the peak is smeared by the efiect of the electrons being boundand not free.

_ _ - Th. 275a k3Y'y-rays will, in some cases, produce an electron-positron pair in thefielil ofa nucleus'in the detector. fire positron loses its kinetic energy, .orn., to rest, andwiihiu the time resolution of the detector will annihilate witb an tlectron to give two7-rays each of 0.511 MeV

a ' 1 e - + ^ l + . 1 .

If one 7-ray from the aanihilation escapes from the detector theu the enerry deposited is27s4 - iLl keY = 2243 keV. If both ?-rays escape tbe energy deposited is ress uy another511 keV, that is 1?32 keV. These energies corespond to peaks D and C.

Qhapter 11 87

Exercises

11.7 Show that oquation (11.14) is dimensional ly conect.

11.8 Consider the docay of equation (11.12): given that tho spin-parity of tho r i l8a'

excited state is $- and that of th6 ground state is 1}', dacide on th6 multipolarity of thisphoton-emitting transition and calculato the expectod transition rate using a formula fromTablE 11.4.

Determine the multipolarity of the radiative decay f,orAy (equation (1 1.1 1)) andestimato tho transition rate using a formula fromTable 11.4. 0o you sEe any modificationsthar should be made to the formula that you wish to use?

11.9 Classify the following transitions by their multipolarity and calculate the oxpectodtransition rates:

'lllu'(i') + rffl-u({+) +y+ I l4 kcV,

!!Co'(2') + $Co(5' )+ 7+ 58.6 keV,

f iNi '(2') r $Ni(0')+y+ 1.33 McV,

!isc'(l -) - !fSc(2t )+7+68 keV,

:lsc'(+') + !fSc$-)+ y+ 767 keV,

@ = 5 x l 0 ' s - r ,

@ = 4 x l 0 - r s - l

ar= 9.5 x l0rr s- |

a)=4.5 x 106s- l

& r=2 .6x I 06s - l

a r : l . l x l 0 -2s - r ,

@ = 5 . 8 x l o - t s - l

Comment on any discrepancies betwesn expected and actual transition rates wherever you

can.

11 .10 Use the given transition rates lor the following decays to attempt a dslermination ottho multipolarity:

! !Rbi-! !Rb+y+556kcV,

llNbr -!fNb + y+ 236 kcV,

Look up the transhions (try Lederer ot al.,1978l. to find how noar you aro to tho cortoctclassification.

Solutions ,11.7 Equaiion 11.4 gives the transition rate of El photon emission:

" ' ' u(Er)=*(3) ' . f . " ro i , ' .

The fine structure consts,nt is dimensionless; the quantity hc has the dinensions of (energyx length) so that (Erlhc)3 has the dimensions of (length)-3. The velocity of light is(length/time) and the squared value of ( r ) is (length)2. Putting these dimensionstogether gives the right-hand side of the equation the dimension (time)-r as required fora transition rate.

11.8 The decay is

t jutBa' - t3jBa+?+662 keV

82 Chapter lI

. " l l - - .q+J '=T - j .= iThe smallest cbange in angurar momentum is 4 units and there is a change of psrity sothe tra'sition could be M4. rlowever, Es is likely to have a comparable ;te. d;g theformula froo Table ll.4 we fnd tbe Weisskopf single particle rates:

. - * M 4 i u = l . 5 0 x l 0 - 3 s - l ,

E 5 : w = 3 . 5 4 x 1 0 - 3 s - rAlthough the. latter is closer to the observed value (4.5 x 10-3 s-r), the approdmatenature of these formulae will not permit a 6rm cressidcation of the transition.

The decay

P o + A * ? + Z 6 . 9 M e Vis l/2+ + 1,/2+ so that Ml is the only allowed transition. using the formura for Mr gives

u = L.43 x 10Ie s-l ,The weisskopf rates assume that the radiating particre is I proton and that it has amagnetic moment of.2,79 nuclear magnetons. The io is not a nucreus and the only cuargeaparticles thet lt contains are quarks. Assuming that quarks are point-like fermions thenthe intrinsic megtetic Eonent of an up quarkifor exa'mple, is expected to be

/2\ eh&\5 / d ' ,where m' is the mass of,Lo

If quark and 2/3 is for its charge. Now quarks have spins =.1/2 so, by analogy wlth Dirac'e theory for the electron *hi.h giuo 9" = 2 for thatparticle (see Section 9.6), we assume that quarks also have g. = Z. tl'erulorl" *u ir"io. outfrom the weisskopf formula a proton magnetic moment (g"ehf 4Me,9. = s.bg6, Table g.4)

:gl"red,and replace it witb.an op qu"ri."go.ii. rir.tt (g.ehf6mu, & = 2) squared.we cau only take a reasonable value for Eu, e-eo uev7c2 (Table 10.6), so that the changefrom our previous result for the Eo + Af transition raie is a factor

/0.667 938.3\ 2

\z:zs3 $o i =o'462 ,

whicb gives a prediction for the transition rate

r..r = 6,60 "

1gl8 r-l .This quark model calculatiou cau be improved by a proper treatmeDt that tekes itrtoaccount the quark spin wave functions of the Do

"oi n ptur t iour"t" .e.* rv* n"y.",R. aad Weisskopf, V. (1967) Nuovo Cimento, Vol. AE0, 612_645. Errata, Vol. AEl, Sg3)to give

u = L.32 * 16le t-lThe measured value is 1.4 xl0le s-r, with a 10% error. :11.9

a) r { f lu ' ( t ) * t f f r ,u ( i * )+?+n4kev ,ar=bx l 'es - r t ,This decay is expected to be Ml or E2 and using the appropriate formura from Table 1r.4, , .rwe predict

Ml : u = 4.67 ri lgto r-t ,

ClapCer II 83

E2 t u = 1 .37x106s - l

Thus Ml is'the most likely classification.

b) ! fCo'12+; - !$Co(s+)+?+58.6keV, ar=4 x 10-5 s- t .

This decay is expected to be M3 or B{. We predict

M3 : r.r = 5.81 * t0-e t-t ,

E.l: ur = 4.80 x 10-12 8-t .

Thus M3 is the most likely classificatiou

c) ! !Ni ' (2+) + tgNi(o+)+7+i .33MeV, ta=9.5 x l0rr s- r . . 'Tbis decay has to be

12. We nretlict t:::t,

, .'.|d =7.t2 y 1gr0 r-r , " .'-.-7"#.,, . -1

E2 transitions fiequenity have transition rates enhanced Uifffii.t' to the'tireisstopfrates (eee Section 11.6). :..,:.d) l!Sc'(l-) * !!Sc(2+) +? +68 keV, ar = 4.5 x 106 s-r . ."

This decay is expected to be El. We predict

u = 4.02 x 10rl g-l .

e) !fSc'(**) - t ls.(l-) *,y*76?keV, u=2.6x 106 s-r .

This decay is expected to be M2 or E3. We predict

M2 I w = 7.73x107s- l

E 3 i w = 1 . 4 0 x 1 0 { s { .

Apart frorn d), the most likely predictions of the transitiotr rates are within a factor of 30of thr: obsewed values; this is good by the expected precision of the Weisskopf formulae.The llSc predicted decay rate is greater than the observed value by a factor of about l0t;there are no obvious reasons but the most likely is thet the uuclea^r matrix elemert ismuch smaller tban anticipated in the Weisskopf formula.

11.10 For the Rubidium decay: R = 1.2(86)r/3 = 5.296 fm, E" = 0.556 MeV andthe observbg vahrcufm - 5.027:x 10-2{ fm-r. We 6rst coDstruct a table for electrictransitions of this R and enerry (tnits MeV aud fn):

Lz (L+ r ) | 3 \

W\Fsi (#)'".'R2L a/ca

fm-lI

234

2.500,x 10-1 ....4.800r 10-1. :)

"6.047.x:10:1 .::,5.142,x 10i7' : '

2.238 x l0-5L,777 x L0-r31.411 x 10:rt1.121 x 10-23

2.805 x.101.7.867 i'i0112.206 x,1-0.L6.189 x 106'

't.0tt9 x tu-"6.710 x 10:131.882 x 10-1t3.567 x 10-2{

I t! !i :

i" tue jjiceeaing three'cohrmnsand is the expected rralue of ufca $ven the assumptiou for'.L"iu'the 6otr sotunn, uFsurning electric transitious. Evidentiy L = 4is a possible.fr.ridotioo. W.-*Oo ft4t

84 Chapter lt

might predict a sinilar value for wf ca; ix fact that assumption gives a prediction of2.9? x lo-m fn-t.

A sinilar trial and error for the Niobium decay shows that, by the order of magni-tude, the observed decay rate is cb$istent with eitber al E4 or an M3 transition.

These trial and error metbods need uot bo used. Graphical presentation of tbe resultsfor transitioo rates or lifetimes as predicted by the Weisskopf formulae are available.(See, for exo,Dple, Erge, H.A. (1966). Introduction to Naclear Physiu. Addison-Wesley,Readiug.)

Solutions11.11 A source b emitting ligbt of frequency /o aud is receding from a stationary ob-server with velocity u. Nou-reletivistically, the observer sees light of frequency / givenby

t= tof r t .n b!*itti.Aty, time dilation of the source clock must be included so that /s becomeshtf t - azlC and / is now given by

' ' . : 1 '

For an approadinE source change u + -u.

11.12 The wiatl- on tbe plot of the observed line shape of Fig. 11.12(b) is about 3.5 mm.Ths conesponds tb a velocity, l, of.0.47 mm s-1. For a photon of 14.4 keV, tbe Dopplershift due to this velocity is 14400 xufc= 2.3 x 10-6 eV. This width is the couvolu.tionof two ideutical liorentzians: this means the required width is one half this result, that is1.1 x 10-E eV. (The convolution of two identical Loreutzians of width I is a Lorentziansof width 21,)

Exerclgeg *

11 .1 1 Find thCJoimula for the relativistic Dopple? bffecr and compare ir to rhat for non.relativistic velocltles. :

. , , ' ' ' . '11.12 Given that the width of th6 M6ssbauer absorprion lino in Fig. 1 1 .1 2(b) is entirely dueto the convolution'ot the idontical line widths at emission and absorption, estimate that linswidth in eV. ,

11 .1 3 Use the information given in Fig. 1 1 .1 2 on the !!Fe M6ssbauer transition to calctlatetho quadrupole splitting in bifenocenyl and the isomer shift between this nucleus in achromium foil and in biferocenyl. Give the results in electfonvolts. Why is it that tho €nergyof tho €xch6d state ol flFe is split and not that of the ground itate? Why is the result of thosplitting a doublet?

1 1 .1 4 A 1 4.4 keV photon from !!Fe is red.shifted as it rises from a source 8t ground level roan absorber foil at a height of 10m. What volocity of the absorber foil is required tocompensat€ the red shift. and in which direction?

Clnpter ll 85

11.13 The splitting A& is 23 mm on the plot and that corresponds to a velocity of2.3 mm s-l, This gives au enerry splitting of l.l x 10-? eV. The centre of the lines is moved? mm which corresponds to a velocity of 0.71 mm s-r and an energy shift of 3.4 x l0-EeV.

The grouod state of ![Fe ha.s spin 1/2 and can only have zero electric quadrupolemoment and, therefore, there is no ground state splitting in any electric 6eld. The excitpdstate hss spin 3/2 a.nd therefore can have and has a Don-zero quadrupole moment. Ofthe four magletic substates, j, = *312, *1/2, the energr shift caused by the electricfield gradient is the same for the pair * 312 and - 312 afi is tbe same but oppositesign for the pair + Il2 afi - l/2. Thus the splitting gives a doublet, with centre at theuuperturbed energy, aud the isomer shift is the 3.4 x 10-8 eV.

11.14 The behaviour in a gravitational field acting downwards (acceleration 9) is thesame 8n that in a system moving upwards with acceleration g, Thus between the time thephoton leaves the source aad arrives at the detector (t) it is as if the latter had acquiredan upward velocity gt. Thus at the detector the photon energy E is red-shifted by aoenerry AE = Egtlc. But ! = h/c wbere I is the height of the detector above the source.Hence AE = Ect/&, A downward velocity u of the absorber foil is required to compensatewhere

Thereforef a = s h / c

For g = 9.8 m s-2, h = 10 m we have

o = 3 . 3 x 1 0 - ? m s - l= 3.3 x 10-{ mm s-r

^a ECILD- = ----;-

a c "

, n, l, l ' 1

1 d . f l .

Solutions to exercises in Chapter L2

' Excrclrcr

12'1 Assigntheleptongenerationeubscrlpranddlrt lnguichantlncutr lnogfromnoutr inoclnthe following reactlons and decayr_Use tho symbols v,,i., v,,-v,, v,,1,,

t t ' - d * e ' * y

l t ' - c ' * v + v

l t - - c ' + v + v

K ' - zoa c ' * v

Ko -zo *c - * y

! ' - n * l - * v

I ' - A o * c ' * y

D0 { K . +n0+c r + y

y + p { n + c {

v+flClrf lAr+c'y + p { l - + p + r t ry + n + c ' + p

l H < l H c + c - * vn ' < p ' * vn ' 4 c - + v

r ' r r ' * r o * y ,

12.2 Draw Feynman diagrame for the foilowing decayr: do this at tho quark rever ror thooeInvolvlng h6dronr,

t ' - c ' * 4 * y ,

K o - r " * c . * y .

D ' - K o + t ' + v

t ' q n ' * 9 ,

[ + p + c . + i .

E -+Ao+

r -

K ' - n ' * n * z '

12.3 Draw Feynman diagrams for the foilowing reactions: do this at the quark level forthose involving hadrons.

v . + C ' { v . + C -

e - + P { n + Y .

l ' * c - - i r * v .

v , f p - p ' a 6 ' '

12'4 Check rhat rhs consarvation l€ws given in the box of Fig, 12,4 arcobeyed by all theexamples in this figure.

12,5 Show that tho following decays cannot occur by first order weak intsractions:

D - K ' * z + f + r {D ' - go . ro '

Do< K. * l I - * i ,

l(o + z+* c. * 9..

Chqtet 12 87

Soliitioni12.1 ',l

I p+ - e+ lDr+ %':":lp'-r'r{ rp +r.' ' .i 'i

K+ 1ro +p|.f /.i:;R o - i , r + + e _ + r r , "D-*n+p -+wD+ * Ao + e+-+ v..., .Do -r K- + no + e+.ir.

1 2 . 2 r - 4 e - + V . + 4 . ' , 1

vj +ilol -ilAfTt''4+p 1tr- {ifTi+vr * r -r e- +lp:::', '

iH 4!He+e- +r. 'r+ - p+ * yr '.",,i"tr- { e- f D.,r ' , . .*, , ' .

t' - r' I'r9"iV:-'

I ( o + n - * o + * u .

D + - R o r p + + v u

,_, *-:*,.

,6'

/Wt -t+lt

t ( o - . - , -^ i + l , r

/u1. vl74vt

^ a Y - a r .D' i - - i ' c

t+ + tt+ +D,

d tr'

2". w ' t '

vi

'ililii

lLil1,il l i l

88 Chapter 12

A + p + e - + t

K + + r + + r - + r +

" ,n':40',U + gd + - + d

E tl-

/..d'J

" , Y " t , u8 - 8<l-r-d

N.

u-'41! "'

" , . - ' t ' \ ;

" .

h all the ha.drons there are gluon interactiou keeping the quarks bound and in the lastdiagran, gluon luteractiols qeate the dd pair. In the last two c8ses, other diagrams cancontribute.

12,3 v.+ e- + yr + e-

vr--r,-tsv'

\ *9- -*- - e-

ai

and

Chapter 12 89

equivaleat to a W+

Ve ------t-€

wl€-4 -v "

In the last and in the following diagrams a W- one way is entirelypropagating tbe otber way.

e - + p + n + Y .

e-+-*- v'

\ . -1.w

U + + do l-:-:-! "

L f + e - - D p * u o

, , , -g-VY

,-'*#vr

u,t 4' 11 -t /r- .l' l\++

12.5 First order weak interaction means one single W excbange between the coustitueDtquarks and leptons of the initial and final states, that js two Wf1f2 vertices (see Fig. f2.4)'

In fact the first two decays can occur in first order but are strongly inhibited because

in each case both W vertices change quark gqnerations. Iu both decays tbese inhibited

changes a,re \ , , ' .\ r .

c _ d *w l ,W + + u + s , o r W + + u * s .

The decayDo - K+ + tt- +r,p

requires that one vertex of the W exchange qeates lhe 1t-Or. The other vertex involvesquarks aod there is no way that one suc"h vertex can cbange tbe c and I quarla iu tbe D0

into the E and u quarks in tbe K+.Similarly the decay

Ko * r* + e- +i,"

vrr-+r.-t

\ fv'd -r-l-F

U #U #

F-

uu a *U

90 Chapter 12

requires the cbange I and d in the K0 into u and il in the n+, something a single w verrexcannot do.

Solutions

12.6 _To test the proposal that these decays have u ( 4o, form the qu*tity ,/Otsince T-o, the maximum kinetic energy of the emitted electin, is very close to e.

Q MeY a, s-r ulQo s-'MeY8 v

D+T+n

1.8173

4.0830.782

5.04 x 10-J2.53 x105

0.39l. l4 x 10-3

2.59 x l0-{L.22 x L0-13.40 x 10-{3.90 x l0-3

Although tbe numbers in tbe w/e' column vary over an order of magnitude, that is asmall range compared with that of_the values of.u, 2xl0s. Thus these d-ecays ao aoiro*approximately to the rule @ ( ?**,

Exercises

1 2'6 The following weak inleraction decays have rhe e-varues and lransition rates given:

, r lO+rfN.*c. *r, . , g= l .gl McV, qr=5.04x l0-rs-r.2 '+1+e '+v , , '

0 -73MeV, @=2.53x l0 rs - r ,x'- t f*e'*v,, 0=4.083MeV, ro=0.39s-r,n -p . |e - * i , 0=0.782 MeV, @= l . l4x lO- !s - ' ,

Show that lhess decays approximately conform to the rule that ar€ f*...

12.7 Summarizetheevidencsthat, innucrear/-decay,atransmutationtakespracebeti i ieena neutron and a proton and that a neutrino (having spin-! and zero rost maisl is emitted.. The following tabte gives the half-life t,ztor threi pl-t n-ririonr. wuctea spins ar;;;in brackets and r-. is the maximum kineiic energy of the /-particre in megaerectronvorrs.

!He(o) r !Li(l) iiH(i) { lHe(})r

r!Bc(O) * ' !B(3) 1Explain briefly ths faclors you couldmagnitudes of these liferimes.

. 7i,.. (MeV) ,,,r(s)3.5 o.8r0.018 4 x l0lo'55 5 x 16tr

expect to be important in accounting for the

(Adapted from rho 1 973 examination of the Finsl Honours School in Natural Scienco, physics, Univemityof Oxford.)

12 .8 Es t imate themaanr i feo f D 'mesong iven lhar ther .meanr i fe is2 .2xr0- !s . (Assumethat the wcs verlex has tho same strength as the wev. vertex and rhat the wcd verlex isnegligibly woak.)

. ,..,;..

. " . ' '

12.7 We'refer ouly to p- decay. ?he evidence ia p+ decay,, "oo.j';i"j;:,:r:.

nt

t' *Tilr.?r.1',of

the ratio of charge to mass esrablished thar the particres emitted

2' Chemical studies showed that the nucleus changed its atomic number by *1.3. Point 2 lmplies that e neutrgn changes to a protoa within the nucleus.4' The kinetic etrelsr spectrun gf electrons emi*ed odended ..;;;; from zero roa ma:cimum Tro, which is difrerent for each r"iiu. ou.t.* and equrr,l to the mass-ene-rgy diference between pa,rent and daughter nuclei. ,.ri5' Pauli proposed, as ao exprenation ofpoint 4, that a neutral, light particre was emittedsimultaneously with the electron. Fermi poi t-li, iau" onto a quantitative, successfurbasis with his sinpre tneo.v

toj /;ae1v. Tir-ugrr p*iae is calJ-a ".,irj"i. rsycoavention, anti_neutrino ia d-_decay.)

6, (Anti-)neutrinos from the..decay of fission produch,-which are niutrou rich, weredetected about 20 years after pauli .ra. ifJ pr.pl"f.7' To conserve angular momentum in p-decay the neutrino must trave half-odd_integerspin' Moderu relativistic theories oi p-d""ov;;;., auccessfullyr"tbat the ueutriuoepiu is |.8' Measuremeut of the shape oithe erectrol kinetic eierry sp-ectrum aear the ma.:dmumindicate that the neutrino mass is less than about z0 ev/c2. There is no evidence tbatit is not zero.

I

The main factors influencing the [a[-life are

IIi

ii

l. The Q value for the decay. :, :.;:. "

2. The change in nuclear spin aia iripy frorr the parent to daughter nucleus.3' The matrix element between tbe wivefunctions of the pare.rt anddaughter uucrei.Factor I has its effect through increasing the decqQ(=?.*)ril;-il;.,sLe.prined--by;.-fi1 jri"l'T..'#=h!1yp.]"il,,r|t:n:

transition rate is due to tm aio.'."i i;: w.-..i;.fr]!..#ffiUi:-*' . 1

- Decay . . r t1 x zll* (s Mevs) ..

lHe + $Li 4'.25 x t02, !H+ lHe Z;86 x 10_r

{Be * tfB 2.52 x t1tz

Factor 2 determines the degree of forbiddeaness: the decay of lfBe is second or thirdforbidden which in*eases its rau ure tul"ti". ;;;"t of fH which is a euper-allowed trau_sition. Factor 3 accounts for the efect .f th; ";;;;;;ates

involved.,tlii.riiore diferentare their space w4vefuaction.the greater the u6 rirc, *a viie versa. This is exemplifedby the rfBe decay; this parent anJits a-"usu,ur n"u;lp;s 0 and 3 respectivery, indiia.tinga large difference in their nucrear wavefuaitioo, *itl iiu cousequence tb&t tbe halfrife isgreatly increased from what would otherwise be the case.A general discussion ofthese issues is given in i*tion fZ.S.

l'llillilliill i

t i

lii

/

92 Chapter 12

12.8 The general D+ decay diagram is

../ '

t1 '

D'

where the choice of I and 2 are

(throo tlm€s for three colours)

et

or F'o r a

v.vDu

The choices r+u; aud sc are energy forbidden and su is unlikely compared with du(Fi& 12.5). Tbe guestion also iodicates that c + d instead of c -r-s at the lower verrexis negligible. Thus there a^re five basic decay medanisme each having the sa-e strengthasp+decay: ' ' : :: '

\

Eech has e decay rate whicb ls that of the muon scsled up by the fifth power of theeuergy release. If we aeglect the mass of the final state particles, the meau life of tbe D+is

I / m , , \ 67b = f,x s " t i l /2.20xt0-6 / 105.6 \6

5 \1869.3/ -

= 2.53 x 10-13 s.

However, ia this model that neglects c + d quark cha,uge, D+ decay must give among thefnal state hadrons, one K-meson. The lighkst of these is the Kj and Ki doublet withEass Dear 494 MeV. Thus tbe mardmum energy dvailable to distribute as kinetic euergya.mong the decay products is 1869 - aga MeV. If we repeat the calculatiou with lgz5instead of 1869 MeV the result is

nr = 11.8 x 10-13 e.

The measured lifetime is 10 ? + 0.2 x l0-r3 s .

C+l-+- S

,-T,

2

Solutions12'9 The final states of the n+ro system catr be specified by the relative orbital mo-mentum, /' since the pions lave no spin the total angurar momentum quantum num-9.:J,

=./. The odd parity bf the two o'.roo, *u*r'that the total system parit' is(-t)(-t11-11t = (-l)r. Thus the allowed total spin_firity srstes are:

Chapter 12 93

This model predicts that branching fractioas for the inclusive modc

e+uo * hadrons, p+ uy *hadrons, and hadrons alone,

will be in ratio l : I :3. The only fraction that has been determined is lz * 2% for e+plus anything.

Exercises

12.9 By 1 955 tho symbor d' had bccn assigncd to a particlc which was obscwcd to decay

0, <v, q ro,

and which had a mass of about 500 MeV/6:. ws now know that it was, in fact, theK''meson and one of hs major decay modes: crassify the possibre states of spin .na p.iitv orthe final state and stata ths restrictions these crassiiications prace on the spin and parity orthe d' under the various assumptions about the conservation laws that apply.

12'1 0 Explain qualitatively why all of tha following processes can be anributed to the weaxforce despite the very ditferent lifetimes involved:

( r r r= 1 .5 x l0 -os , c= 105 Mev) ;

I

(rr :i= 60? s, Q = 0.782 MeV);

(r,r=4.f x l0os, Q=6.36Y.U,.

(r, ,r- I .4 x l0-. s, Q= 92 Y.Y;

Describe an experiment which shows that pariry is not conssrved in woak interactions,explaining carefully why h shows that parity ii noi conserved,

Thb decay

r!O.(7r- 2' ; - "g17'-6i; + chas a panial width of ordsr 10-rosV. what can be inferred about the strong force?(Ad8ptsd from tho 1 985 examination of the Finsl Honours School of N6rirral science, physics, universiryol Oxford.) 'r

(a) muon decay

l l ' + c ' * v r l i t

(b) neutron decay1 + p * c - * i .

(c) otomic electron capture

lBe+c- - l l i+ r .

(d) muqn capture in muonic aromsrlC*p- - ' !B * v,

0+ , 1 - , 2+ , 3 - , . . .

94 Chapter 12'

Ftom the pre.lg56 view, the decay

0+ - r+ +ro

conserved angular momertum and parity and the spin-parity of the 0+ had to be one ofthese values. If parity is not conserved then all thai.ro be gaid is that tbe existence ofthis decay places no reetrictions on.the spin and parity of the d+,

. In the case that ellurar momentum is not conserved, irrespective of parity conser-

vatioo or non-conservatioa, no restrictions on the spin-parity of the d+ foilow irom theexistence of this decay mode.

l-2.10 All these decays iouoi* thu emission of one or more neutrinos, a process thatthe weak interaction alone can cause. A large range ofharflives is expecied io.ilitu.gurange of Q'values. we expect tv12es or hrz Tlno to have values that are closer than arethe values of f172.

Note that for the p+ decay the three final state particles are all relativistic so that e isnot equal to ?.o.

For the two capture reactions the situation is difierent:1) The two-body 6nal state has a density ofstates (Table 12.6) that is proportional to the

square of the centr+of-mass momentum; for these reactions, that means proportionalto Q2.

2) The transition rate is proportional to the probability of finding the e- (or p-) at thenucleus, that is to lrlt(r = 0)12, wbere r/(r) is the e-(p-) waiefunction. i.{egtectiag

. screening tbis probability is proportional to Zsms (m'= rn or rnr).3) There are Z protons available for capturing the lepion.Thus we expect \12 Q2 21 nf to have similar values for the two capture reactious:

Decay aMeV

'l lnu

MeVhp Q's MeVs

tVzTios MeVs

p ' + e + + v . + r Fn + p + e - + %

r050.782

OJ

0.782 1.8 x 102l0{Xo1 6.3 x 102

I.8 x 102

Reactiou t 1p Z 'mo Qzs MeVs

{Be+e- + !L i *u jtlC + p- -r?B + z,

1.2 x 10!1.8 x l0?

These results are satisfactorily close in each pair and show that tbe underlying interactionis the same in each pair. More complete caLulations would show that the interactiou isquantitatively the same in all four processes.

. .Tf. two original experiments that separately showed parity non_conserrration in-volved the decay of !!Co and the r+ - p+ - e+ decay chain. The former is describedin sectiou 12.9. we brie8y outline the second. This experiment used electronic detec-fors-and is described by Segr6, E. (1972). Nuclei and, particles (2ud edn), pp ?4O_1.Benjamin/cummings, fuading, u.s.A. Heie we use the pictures of this decay cloio, sho*oin Fig. 9.14, to describe the priuciplcs,

If d is the angle between the positron momentum (p.) and the direction of themomentum (Pr) of tbe p+ just before coming to rest, then ii was fouud that the frequency

distribution of eveuts in cos d was uotiuuiform but

/V(cos d) o< (l + c cos d) ,

with o - -1/3. we say the decay was forward/back asymmetric. Now there are threefacts:

a) Anlnpolarise{ sa,m.ple ofparticles cannot give anything but an isotropic distributiouof dec&y products (in its rest frame).

b) A polarised sample ofspin l/2 particles cannot give anything but au isotropic di+tribution ofdecay products if the decay interaction is parity ciose*ing.

--''

c) A polarized sample of uustable particles of spin > | can decay to give an anisotropicdistribution of products with respect to tbe a:ds of polarlzation.-Ho*uu.r, this dis-tributioa cannot be forward/back asymmetric unless parity couservation is violatedin the decay,

Thus the observed asymmetry gave two conclusious:

1. The p* TT:y

are polarised with respect.to their direction df motion; they havenon-zero helicity (Section 12.10).

2. Parity is not conserved in their decay.

The decay asymmetry means that the decay probability ", ":juuo

0 depeuds on cos g,that is a dependence on the pseudoscalar p..s, where s, is.the epio of the nuon. Rdecay rate depending on a pseudoscaler always tndicates parity. ion-cons.roiioo"

-

. Tl. tron-zero helicity of the muon also means that parity is not coneerved in the*ot_".* : tt* + z, (Sectiou 11.10). Again, a decay rate deleuds oo

" pseodoscatar,

P, .S, iu this case.The decay

tfo. *tNc + o

bas produi! ground state ou"j1:iUat ire evea-even aud therefdre both heve jp = 0+.The excited'o.rygen has jP = 2-,'lf. angular momentum is conserved, the rciative orbitalaugular mombntum, l, in the fual state must bave | = 2. But such a state has evenparity. The deiay is parity changing. The observed width, 1g-ro eV, is e mean life ofabout 10-7 s. In tbese light nuclei, coulomb barrier effects.are negligible and the meanlife is very long compared witb the meau life expected from a strong inte..action decay,nanely < 10-20 s. The couclusion is that the strong interaction cirinot significantly violateparity consewation. In fact, the decay rate is probably consisttiit witb weak interactiooeffects, either directly iu the decal itself (it is a nouJeptoiric dr:cay;section 12.1), or bymixing of opposite parity states at a very low level (probably l0:l{) in the nuclear statesinvolved, or bbth.

I

I

96 Chapter 12

Solutions12.11 The (ant!) neutrinos emitted in neutron decay:are;961 handed (helicity +l).

n i p + e - + D . .

A mirror image of that decay shows a left banded (helicity -1) antineutrino (Fig. l2.S).Thus the mirror image is not an observed decay. That is e cousequence ofnon-conservationof parity' or what is equivalent, of non-invariance under the parity transformation (p).However uature is, exc€pt at a very low level, invariant under cp (section 12.9), Thismeans a charge conjugated, mirror image of the aeutron decay is equally likely. This isthe decaY

n + F+e+ +yo,

with a left handed (helicity -1) neutrino.

12.12 consider the unpolarized source emitting electrons with helicity -1. Its mirrorimage is a source emittiag electrons with belicity +1. If parity were to be conserved, thesetwo processes would have equal transitiou rates and the electrons could not be polarizedon the average. Thus the fact that they are poluized means unequal trausition rates andparity non-conserrration.

12.13 Right circularly polarized photons have helicity +1, left circularly polarized pho-toqs have belicity -1. As in tbe case of neutrinos, Don-zero average helicity is possibleoaly if tbe hteractiotr producing the photons does not conserve parity. Tbus parity con-serving electromagaetism cannot cause the emission of circularly polarized pbor,oos froman unpolarized source (Section 12.10).

The answer to the secoud part of tbe problem is no; parity is conserved. The directioaof tbo photonr ia defined with reepect to the direction of the applied magnetic field, B, NowB ie an axial vector beceuse it is caused by au electric current flowing in a clockwise or inan autidodcwise directibn around a space arcis. If s, is the photoa spin, then the quantity

Exercises

12.11 Showthat, i fr ight-handsd(hel ici ty+1)neutr inosaraemirtsdinthe/-.decayofthonoutron. h is lefi.handed neutrinos that are emitted in p.-decay of the anti-nsutron.

12.12 Show that, if 8n unpolarized source emits electrons which are spin-polarized withhelicity -1, then parity is not consewed in the decay. ,

12.13'The electromagnetic intoraction consrr"r, p.lity. Show that this means rhat anunpolarized source of /-rays cannot emit chcularly polarized photons.

An elearic discharge is esrablished in low-preisure helium gas in a magnetic field. Rodlight emined along the'diroction ol the'magnetic lield is found to consist of two lines with asmall wavelength separation dopsndont on tho strength of the magnetic field. Thes6 twolines are oppositely circularly polarized, Does this mean parity is not conservod in thetransitions teoding io thii emission of these photons? Erplain your answer,

Chapter 12 97

Sr. B is the scalar product of two a:<ial vectors and therefore is itself a scalar quantity.Iu a parity conserving situation, observables such as transition rate or wavelength candepend on scalar quantities. Thus a right hand circularly polarized photoa emitted in tbedirection of B (Sr. B > 0) can have a diferent wavelength from a left haaded photoncmitted in the same direction (S,, .B < 0), without parity non-conservation.

Exerclses

1-2.1 4 Beams of high-energy muon neutrinos can be obtained by generating intense beamdbf r'-mesons and,allowing them to decay while in flight. What fraction of the a'-mesons ina beam of momentum 200 GeV/c will decay while trav€lling a distance of 300 m?

At lhe end of the decay path (an evacuated tunnel) the beam is a mixture o{ r'-mesons,muons and noutrinos, What distinguishes thsse particles in their interactions with matter,and how is a neutrino beam frse of contamination by z-mesons and muons obtained?

[z'-meson mean-l i fe: 2.6x 10-t s.]

(Adapted trom the 1984 examination of the Final Honours School in Natural Science,Physics, University of 0xford.)

12.15 Antineutrinos of 2.3MeV from the fission product decay in a reactor have a lotalcross-section with protons ( i .+prer+n) of 6x10-'Em2. Calculate the mean free path ofthose nouvinos in water. Assume the antineutrinos are able to interact only with the freeprotons.

Estimate as well as you can ths mean free path of an electron neutr ino of 1.0 MeV in thematerial of a neutron star assuming nuclear density and tho absenca of protons andelectrons.

d(v.+n - P+e- ) = 10-" m2 at 1 '0 MeV'

12.16 The muon nEutrino-nucleon inelastic scattering cross.section is proportional to thenoutrino onergy. lt is I x 1 g-4 r: at 1 GeV. What is the mean free path of 1 00 GeV v, in steelassuming all the nucleons aro potsntial targets? (Steel: density=7.9x 1 93 kg m-r). A bubblachamber consisting of a right cylinder, cross-sectional_area 1 m2 contains 1 000 kg of liquidpropan6. lt is exposed to a bsam of 10 GeV muon neutrinos which is incident normally on acircular facQ of the cylinder, The neutrinos arrive in bursts from the accelerator used toproduce thein (how?). The bubble chamber is expanded in synchronization with the burstand any neutrino intoraction in the liquid that produces:iharged secondaries can beobserved. How many neurinos per burst are raquired to give an average of one interactionper 1 00 expansions?

1 2.1 7 Show that tho event of Fig. 12.18 in which a charmod meson was producod is duo tothe neutrino-{uark rsaction

v " * d + p - * c ,

and is unlikely to bo due to the reaction

y r + s { l ' + c ,

whore tha s is from the sea of qq pairs that a nucleon contains.

98 Chapter 12

Solutions12,.14 Reletlvisticslly, particles of mass M, having momeDtumE(= 1/fiiz7@p,1have a velocity o given by

-- --o '

PC'= T 'The time required to move d,rdistance .L is, to observers at rest inand P are measured is given by

P, aud total enerry

the frame in wbich .6

. L L Et = - = -

a P c 2 'The proper time interrral ? corresponding to t is t/,y where ,r = E/Md, Hence

MC LE ML'=-T fu=-F.Tbus if ,L - 300 m, P = 200 eey f c and, M = L39.6 Mey / c2 we have

- _ Mr, , 0.1396 300- = V; ;= lbo-^3;Tdt

= 6 .98 x l0 - ro s .

Then the fraction / of pions decaying is given by tbe radioactive decav law

f = r - e x p ( - T / r ) .

where r = 2.6 x l0-8 s. Then'

t = 2 . 6 5 x 1 0 - 2 .

The answers to the second-part of the question are found by reference to Fig, 12.3,which describes the production of a neutrino beam. The present problem concerns energiesabout ten to fifteen times greater thau those in that figure so that the a..iy p"[ -aabsorber thickness are that much greater than described there.

72'15 The number of free protons per cubic meter of water (1000 kg) is n = 1000 x6.022 x 1026 x (2/t8). The mean free path ,l is given Uy

^ = * 'where o = 6 x 10-'t8 m2 is the antineutrioo-proton total cross-section. Hence

) = 2.49 x l0r8 m.

Nuclear matter has a nucreon density that gives / nucleons a spherical radius R =7.2 Ar/3 fm. Thus the density is in, nucleons per im2

= 0.138 nucleons fm-3 .

Thc I McV clcctron ncutrino intcraction cross-section with ncutrons is given as l0-.7 mr -l0-r? fm2' Tbus the mean free path in neutron matter of nucle", a.oiitfi, giu.o uy

3 4 34nRt 4n(t .2) l

' i r' ! i ' i ' i '

: ' : " 1 ; , ,

Cheptorll? 99i1. i . j : - l i ; ! l l , ; i

. 1 I : - l . .^=G = d.iSETfb:ii'

= 2.25 x 10r? fm, , , , . , , .= 726r-. l.

12.16 We regret that there is a misprint for the.muon neutrino.lgtal eoss-section. Thecorrect figure is ? x 10-{3 m2 at I GeV. The problem tells us that this cross-sectioo isproporllggalto neutiinriehergy 8d-'thbt ir is ? x lb-{.m2 at tgg-g*;.Tbe uumber densityof ir6n'dibni, assuiriiiifitiiel iidTo0To troi;ii

:. r': :er. , .' "' ;, :, z.sI ror x6.o22xlbz6 . ";i;i,,

--' 5534'/ --* ., '';*'Wu.*o *soru, wlthlsufficient precision, that the isotopic coilf6dition of irou ls auchthat tbe average number ofnucleong per nucleus is 55.8a?. Thii'ide number density ofnulleons, n, is just tho numerator of tbis frsction. Tbeo the meui'free path I is given by

. 1A = - =

' n o'=

3.00 x 10e m ,

tbat is, 3 million kilometres.

7.9 x 10r x 6.022 x 1026 x 7 x l0-.1

We use the Case 1 definitid,n for the operational use of diies.section from Table2.7 to calculate the neutrino nu'mbers in the bubble chanber pio[lem. A f000 kg ofpropane provides, with euficientlprecision for the purpose of thiirquestiou, a ta^rget of10t x 6.02 x 1026 uucleons. The ocpected cross-gection at 10 GeV ig 7 x 10-{, m2. Ifthe target is immersed in a uniform neutrino beam, the prolabilty any uucleon guffers acollision for eacb neutrino incident per square metre is the product of these two numbers,tbat is 4.2 x l0-r2. To obtain atr average of one c6llision pequiles a bea,m whid has(4.2 x to-tz;-t = 2,4 x l0l1 neutriuos per square metre. If tbese neutrinos anive inbursts and 100 bursb are required io obtain one collision then bhere must be 2.4 x t}eneutrinos in each sqirare metre per burst. The most economical way of immersiug thetarget in such a bean is to have the beam iacident normally on one of the end faces of thecylindrical target. This has an area of 1 m2 so the required number of neutrinos per burst is2.4 x 10e.

16s lfow' of the productionrof a muou neutrino beam is giveu in Fig. 12.3. Tberea.re'[ow tecbniques for enhancing neutrino over anti-neutrino f,ux, or vic+vema. For moredetails, interested readers are referred to Fisk, H., and Sciulli, F. (L982). Annual Revieuof Nuclear andxParlicle Science, Vol. 32, pp 499-573.

12.1?' The muon produced is negative so the collision is a charged current interectionof a muon neutrino (not antineutrino) with a negatively charged quark in a proton, whichbecomes a positively charged quark. The charmed particle is a D'+ so contains a c quarkrot a e. Thus the collision is with a quark not an.aatiquark, Po'ssible reactioas a^re

l , v r l d . + p - * c2 , v p + l - l t - + c ' . : ' " '

If the reaction was 2 then the e querk has to come from the sea.of qq patrs that ecistsin the proton, Its conversion to a c would leave an unpaired $ amoDgst the remaining

/

100 Chapter 12

hadrons. That in turn would meau s stralge (S = +l) antibaryon or an ,S = *l meson.There is uo evidence of the former having bee,n produced. There is a possibility tbat tbe 5became a constituent of a I(0 whic,h either decayed to n0r0 or left the chamber, Howevertbere is no missing transverse momentum and this possibility is considered unlikely. Thustbe most likely reaction is 1.

Exerclse :

12.18 Supposetharthonaut; inosemittedbyrhosupernovaSNlSSTAwere.al l ganeratedand loft the cofo.in a period ol one second. A large fraction of the twenty neutrinos detectod160 000 years lator were all observed in a period of 2 s, with a mean snergy of about 8 MeV.Make a rough estimato of an upper limit on the neutrino mass.

(At12), then (puitiiij Pc equal to tbese energies) we bavd"' :"" "l '

Solution12.18 A po,rticle of total energy E, mometrtum P ord mass M has a velocity u = P/Ewhere E2 = P2d + m2ca. Suppose & zero mass particle, velocity c, ta,kes to seconds tocorrer the distance ftom SN1987A to the Earth. Then a particle of mass M takes a timets9lPc and the d4ay At in its arrival with respect to t6 is given by

At = to f * - l \ '\rc /

= 6(1 '6-11 .\ rc /

Suppose MC <<Pc expand tle square root binomially. Hence': '"

' ll' ' u't\L '. . 1 . ' ' \ : ' ' l r = rol(,.ffi) _,.| ,

:. ' . =tl l . Iry+ -1l ,: l ' L '

:-. to lufcr. . " , . ; , . . L t = ;V . .

. . :One approach is io*iiia.. tu. aelay for 4 MeV neriiiinos (Atr) and for 12 MeV neritrinos

. , ! . 4 ; * ,; i . : i ! r f+ r : r : r_

at{.-alu = , M'c' in MeV2? *'t(i- #)**u.

Now t6 = 160,000 years and we take Atr - Atrr = 2s. This gives

Since the neutrinos are emitted over a period of ls ard we do not know tbe extremes ofenergy this result would be a rougb upper limit.

Chapter 12 l0I

''Another approach includes the assumption ofa one second production aad emissionperiod. Thus one second is the ma:rimum dclay at tbe earth. Since the delay is dominatedby the lowest energy particles observed we could assumc this encrgy wa.s 4 MoV, say, thcu

tn M2caA r a : 1 s = ,

W .

This givesMt =2.5 x 10-6 Mev = 2.5 eV .

In fact, the calculation of an upper limit bas to take into account the energy sPectrum

of the neutrinos, the time distribution of emission and how that might vary with energy,

and tbe detection efrciency of the detectors. There are various calculations of this kind

and the upper limit is somewhat greater tban that given by our simple calculation. I

Solutions to exercises in Chapter 13Exercises

' , 1

13.1 consult Perkins (1 986) or Gottfried and Weisskopf (1 986) to exploro tho meaning ofthe charge-conlugation quantum number for neutral systems such is the zo meson and the3d and rSo states of positronium. Show that the decays

no -y]- l ,

Ps(|So)+y+y,Ps(lrS,) - 7* 7+ y,

conserve charge conjugation quantum number.

13'2 lf the neulron has an electric dipole momontp it must be parallel (or opposite) tolhespin

P = d S

whare d is a constant. The energy U of this dipole in an electric field E is

g = - p . E = - a S . E

By considering what happens to S.g under

(1 ) a parity transformation and(2) timo reversal

separately, show that d is zero unless the interactions responsible for the forces between theconstituent quarks in the neutron contain an element of parity non-invariance andsimultaneously of t ime-reversal non-invariance.

[The present value of the electr ic dipole moment of the neutron is -0.3+g.5x1g-rr "

nt(9=charge on proton), consistent with zero.l

13,3 Indicate, with a briof explanation. whether the following reactions or decays canproceed through the strong, the electromagnetic, or th6 weak intaraction. In each caseestimate as well as you can the order of magnitude of the cross-section or transition rate, asappropriate.

( l ) n + p * d + 7 ,( 2 ) E - - A + r( 3 ) p + p + r ' + d ,( 4 ) D ' - K - * z * * z '(5 ) K- +p* Ko+n,(6 ) z -+p+Ko+A,( ? ) c ' * e - + W ' * W - ,(8 ) Zo-p '+p- ,(9) n- -Ko+K-.

(d here stands for deuteron, not d quark.)

ChaPta.l3 103

Solutions13.1 A neutral system consisting of a fermion plus its antiferaion in a bouud state is

an eigemtate of well-defined charge parity (the e,harge conjugation quartum rumber).That charge parity, C, is (-t1l+' where 0 is the quantum number.for the relatirre orbital

angular momentum and s is the total spin quantum number; Thui'Ps lSo has / ='0' s = 0

uod C = *1 and Pss5r has I = 0,a = 1 and C = -L.The n0 is-ii 1.96 state of u[ + ddquarks and has C = *1. The photon has C = -li n photons have Q = (-1)". Thus the

electromaguetic decay of a tso ffsystem can only give an even'.number 6f pbotons, the

decay of 3Sr state can only give an odd number of photons, if C is conserved. The decaysgiven couform to this result'

13.2 The energy' U' of an electric dipole in in electric freld is given by

u = - p . E . I '

N o w i f t h e p = a s ' w h e r e S i s t h e s p i n o f a p a r t i c l e , t h e n : : " o : .

L I = - o S . E . r a 1 : ; 1 . , ; . r , , . , -

Now S is an ar<ial vector snd E a vector so that s'E is a pseudoscala^r. II the system is

invariant under the parity traosformation, an observable such'aslU cannot depend oB a

pseudoscalar and we conclude o = 0 and hence p = g'

Under time reversal E remains unchanged but S c"hanges sign (tbe pslticle rotatloD

reverses). If the system is lnveriatrt under time revers&l then U cannot change sign and

we conciude, by a separate a,lgument' thet a = 0 and p = 0'

Thus unless the inrrariances under time reversal and under the pa,rity transformation

are both violated in the internal motiotr of the neutron, it must have zelo electric dipole

momenl.

1B.g (1) D + p + d + ?. This is obviously electromaguetic. The cross-sectior dePetrds

strongly on the neutron energy. It falls from 333 mb for thermal Beutrors to 30 pb for

Deutrons of 14.4 MeV.( 2 ) 3 - - A + z r - . T h e r e i s a q u a r k f l a v o u r c b a n g e , s - + l l r s o t h i s E u s t b e a

weak iJeraction proce$. The value of P (see Table 13.5) is 139 MeV/c so we expect a

trausition rate of about (rss;6 x a factor (- p.2') for the unfavoured quark cbange' That

gives 2 x 10e s-r. Tbe observed value is 6'1 x 10e s-r'(3 )o+p+d*r r+ .T f , i s invo lveshadronsa loneandthere isuoquark8avour

.l-g.. if itS,tn.tefore a stronginteraction process with a ma:<imum coss-sectior of orde!

10 mb (obseh'ed 3 mb) but decreasing with energy'

t,il D+ ) K- +n+ +n+, There is a quark flavour change,"C + s and P is 845 MeV

so the transition rate i8 exPected to be less than 4.3 x 101{ s-..1'.The observed value is

about 8 x 1010 s-r. Tbis difiereuie is not. surprising ln view of the simplicity of the recipe

given in Table 13.5., , .|

--

6) K: I O - fo t n. dis reaction has AS = +2 (5 =- f.lr.anseaas) and.therefore

requirL'a secg-ni ordgl wga.k iqteraction. That neans a cross-;$|!3,that is unotiservably

small (< 10-56 m2). .,.+;-*: iOt ,r-+ D d K0+ A. At{uark flavours ere conserved but tlis associa,ted ploduction

'eans tfiat an s5 quark pair has'to be produced. This inhibits.the leactiogr a,nd the cross-

iiction has a ma.:<imuiii bf aborit I mb.

,

104 Chapter 13

(7) e++ e- - w++ w-. This reactiou has e threehord at a centre-of-mass energy of160 Gev. It is electromagnetic so, ebove threshold it has a cross-section oforder lonijrTswhere s ls lhe gquare of the centr+of-mass etrergy. At 200 GeV this gives g x 1gi{ ,rrz.

(8) 70 - p+ + p-. Tbis is electroweak. Fton Table 12.s, the observed partial widthis 85 MeV. This gives the transition rate to be 1.2 x 1023 s-r.

(9) o- - K0 + K-. This process violates baryou number couservation. If that rareprocess does occur,.theo the decay would have 8 transition rate comparable to that forproton decay. None [as,been obeerved errd, indepeudent ofdecay modes, the proton decayrate has an uPper limit.of about 2 x 10ls s-r. Particulu modes have transition ratei less

I

{

than 3 x t0-ro r-t'(Section 18.6).

Exerclso

13.4 You are planning to communicate with the inhabitants of a distant galaxy and wish todetErmino whether thoir onvironment is of particlos by our reckoning, or of antiparticles.Invbstigatc the stateinent that, although the wo8k lnteractlon vlotates charge+onjugationlnvariance, there is no weak lnteractlon experiment that can be described in *orJ. *hirnpermits a resolution of that question, other than experiments using the Ko-Ro system,

Solution13.4 we assume frstly that nature is invariant under the cp transformation(section 12.9). u the distant inhabitants tell us that the charged particles emitted intbe decay of neutroas have belicity fl that canaot tell ur uoytuiog ouout their matteror antimatter status. The reason is that helicity is defined by the angular momentum ofthe spin aloag the direction of motion, and we do aot kaow whether ihey define positive

Tsol.r Bonentum by a clockwise or anticlockwise motion around a given ucis, oo. .uotbey tell us,

The experiment that can convey information concerns the K0-K0 system. The quan-tum mechanics of this two-state system is such'that one half of a beam which is pure K0or pure R0 initially, dicays quickly leaving a long-lived beam of particles .ou.i xr. ruo,have omong otbers, two pa^rticular decay modes:

_ K r , + u r - * e * * v o ,

K l + r * * e - + i , . .

A manifestatioa of cP non-invariance is that these two modes have unequal transitionratee. The frgt exceeds the second by O,Bvo, The description of the experimeut to measuretbis cherge asymmetry needs no distinction between lefl and right. If the siga of the chargeof the leptou emitted in the faster decay is the same as the charge of the leptor in theiratgms, then the conclusion is that they live in a world of antimatler.

Clutptcr IJ 105

Exercises

13.5 Use Huyghen's construction to show that Cerenkov radiation is emined at an angle

cos-t(c/nv, to tho direction of a charged particle moving with velocity v through a

transparent medium of refractive index n.

1 3.6 Considar how a water Corenkov detector for proton decay responds to the following

two possible decay modes:

(a ) p+p ' *Ko,(b ) p -V.+2. .

List toaturos of the detected light which could be used to distinguish these decays ftom

other processes which might imitats them such as

v r * p + y + * n * K o ,

v * p r v * n i z ' ,

respcrtivcty.

Solutions13.5

A d

If tbe radiating particle, velocity r.,r mov€s from A to B in time t emitting Cerenkov

radiation, then the direction of the wavefrgnt of light emitted between A and B is at an

angle d, where CB is tbe $,aveftont and A0B is i right angle. Now AB : ut and AC =

cl/a, whe,re n, is the refractive index of the medium. Theq

\ 0 = cos-r(c/trn) . i : '

Tbe three-dimensional wavefront is found by rotation around the line of flight of the

particle.

13.6 A proton decay or a neutrino interaction will finally give a number of visible pho

tons that read the boundary ald a fraction of these are detected by the photomultipliers

stationed tbere, A given photomultiplier can give the relative arrival time of a pboton that

it may detect with a precision of I to 2 ns and a pulse signal proportional to the number

of photons arriving at its sensitive surface. The problem is t0 look at tbe pattern of hit

photo6ultipliers, the signal timiags aud energies to obtain ma:<imum information about

any event, For a pa,rticular protoD decay mode there may be several routes from decay

/

106 Chapter 13

to pbotomultlpller, The same is true from a neutrlno luduced event, Thus the polnts toconsider aroll. The kinematics of tbe primary event,2. The possible branching in secondary, or even subsequent decays, or interactions.3. How each particle loses energy and the distance trevellgd, before it comes to rest or

iDteracts,4. The path lengtb and"r'eiocity of charged particles to determine the emount and spatiel

distribution of tbe Cerenkov light emitted.5. The relative timing of the arrival time of light at the boundary.6. The effects of qoo-coutainment of the decay products.7. The characteristics of the neutrino 0ux at the detector.

For example, the neutrino from

P + t + ? r '

is undetectable but the pion always has a kinetic energy of 340 MeV. If a nuclear in-teraction does not intervene it travels a distance of about 150 cm water emitting lightinto a cone decreasing from half angle of 38o to 0o, The amount of detected light is thesame, within Suctuations, for all pions of this energy. If the pion is moving normally to aboundary, all the Cerenkov light strikes the boundary within a time interval that is muchshorter than the flight time of the pion. That interval is spread for non-normal angles butin a way contrected with tbe light distribution, Thus this decay pion gives a very charac-teristic pattern of detected photons. There will also be some photons from the flight ofthepositronintbefinaldecayof thechainzr+ - p+ - e+,normallyhappeningafterthepion has come to rest. However, these positrons are delayed by a time characteristic ofthe mean life of the p+ meson, 2.2 ps. The interactions

v * D - u + D + 1 t +

will give pions of a wide range of energies, but otherwise behaving as do the decay pions,The neutrinos are produced in many of the decay modes of K and r mesons pro-

duced in the upper atmospbere by primary cosmic radiation (nainly protons). Even deepunderground these neutrinos come from all directioas aud aay siugle pion cannot be elim-inated because it is in a direction a.ssociated with some near unidirectional neutrino 8ux.Thus tbe experiment searching for p + v*r* has to rely on observing events il an enerrypeak standing on a background with a broad energy distribution. The enerry resolutionavailable by measuring the Cerenkov light is not good, which makes it hard to see a peakon a continuous background. None of the water detectors ha.s observed &n event that canbe identified as a Droton decav.

ill rq,, '

14,1 Estimite the flux of pp chalri solar neutiinos at the Earth giuei;e sot31 tuminosrty l'"

(Table 14,2) 6nd thd tsct that two neutrinos 8re producod in evcry.conversion of four

iioio* to oi. o.parricle 8nd that.this ionverslon ptoduces 26.72 M-By..Assumo that thc pp

ihain is solely responsiblo lor ths solar energy production. lf3% of thc,energy is lostthis way

what is the averago nsutrlno snsrgy? :

1 4.2 The Oavis detector contaln3 600 ronnos of CrCl. with the chiorine containing 24.47%

bv number of llCl. The cross-3ectlon for the reactlon flCl+v.*c'+f!At svcrlgod ovor thc

over.threshotd'p8rt of tho neutrino spectrum expectod from th9 decay !B*!B+c'+r'. is

I x 10{! mr, Given that the productlon of flAr ls ono ovety two d8y3, calculato tho csptuto

rato in SNU 8nd tho flur of detectabte neutrinos'

14.3 Investigare how the fOllOwing teactions would be used in otder to dotect neutrinos:

Exerclrcr

llGa + v.+llGe+ c' -0.236 MeVrlfln i v.+r$Sn'r + e- - 0.120 McV

I

t ! !Sn. +7+0.1l6 McV

Itgsn + y+ 0.498 McV.

[For the gallium det€c\or sea Kksten (1979) and for an indium detector see Raghavan

(1976). The " and ' indicato exched statesl

Solutions14.1 Flom Table 14'2, the total euergy output of the sun is

^L6 = 333x1026Js- r= 2'39 x 103t MeV s-l '

' . r : l ' ; i

Each o-particlb produced means the getreratiotr o126.72 MeV. Therefore tbe uumber of

o-particles produced isN' = 8'94 * 193? t-t '

Doubling this gives the number of ueutrinos. Therefore the neutrino flux at the earth is

;NJi;td.d bi lzrp2, wben R is the Ea,rth-Sun distarce 1.5 x.10E km. The result is

6.33 x 1ol{ m-2 s-t "

The average neutritro energy is given by 3Vo of2632MeV ghai& by two neutriaos. Tbe

*"fi i; O,lO MeV. [n fact-the fraction is more like 2% leading to an average erergy ot

0.26 MeV.l

i08 Chapter t4

14.2 The 600 tonnes of C2C! (molecular weigbt 165.83) contains4 x 6 x l ( ) s

t633l_x6.022x1026atoms of chlorine of which a fraction 24.47v0 are ffcr giving 2.r3 x l03o atoms of thatisotope. An average capture rate of one euery't*ro ily, i, s.zs x 10-6 s-r or 2.zl x 10-16fnturl p-er eecond per target nucreus: trai is z.zt ir.ru. rni, yierd is the cross-sectiontiaes the flux;.thus the detectable neutrino A* i, I

--

? .? l x t0_s6 -_ : ^fiioT-

=2'7 x loroY m:2 s-r .

14'3 (1) The gallium method 1s

t\ yne in priucipa.r as thar used by Davis (section14'?). The difrerence is that the ta^rget is suiliu'o il-t-; of chlorine, which makes for adiffereat chemistry of sctraction. n iaaitiJo, ;h;fi; captule' neutrinos from the ppcycle, which chlorine does not,The target flGa is stable ard 39.920 of naturar gallium. The product llGe is un-stable (mean life 16.4 days),decaying. t liC" Uy-.f-..tion capture (e = 2BZ keV). Twoexperiments are under way (1993) uriog i?

"oa go iiio.r respectivery ofgallium. Afteran o(peure of about 21 days, a,ny germanium (- I atom per 2 tonnes) is chemically ex_tracted, converted to the Sas !e1t=an9, oriog i.ur.i;r, introduced into a counter wherethe electron captures are detected by the sen;tivitio iu. eugu, electrons, as in Davis,sequipmeat.

The stardard solar model predicts a capture rate of 128 sNU of which two thirdsis due to pp neutrinos. preliminary.results'are *rrrri.r, with the expected flux of ppneutrinos plus the less-than-expected flux from other solar cycres, * ii..r".a uy o*i,(and corfrmed by the Kamiokaude e*purimeiq CJr. ri.8).For an upto-date reoort, interested t."auo

"ti r.r.rred to the proceedings of the

9t:l:.."i:": co$91e1ce (uuuirioo e, (tr$) ;;; r'rnyr;"r, (proc. Suppl.) Vol. 3r,pp. lll-124, North Holland.):(2) Indiun: the eaergr level diagram is shown in the figure.

614k€V r=4 .85x lo ts

498 kev

'l3ln 'lisnt=7,4 x l9l. years95.7 % of naturallndlum

AlomlcmasskeV

vo capture

Chapter 14 109

- The neutrino capture reaction in tlfln has a threshold of l2g kev. A capture leadsthe emission of an electron, a delay characieristic of a mean lifc of 4.g5 ps, a gamma ray of116 keV followed by a secoud ,y-ray of4gg kev. The intcrvar between the first and second'y-rays is too short to be seen by any practical detector. The originar proposal was thatthis sequence of events could be found in a suitably designed, indiumloaded detector.

About 2 tonues of indium wourd be required, instiumented suitably (an unsolvedproblem). The capture rate is about ?s0 sNU. unfortunately for every neut.ino eveatthere would be about 6 x r0r0 decays tlfln - rlfsn + e- + ,.. Finding the requiredevents in that background is a challenge too great lor existiag technology.

Exercisa

14,4 Draw the Feynman diagrams for the processes:

( a ) c ' +c - <v , * i , ,

( b ) c ' + c - - v . * 4 ,( c ) 7+c r - v . *4 te ' ,

(d) c' + (Z,A) - e' * (Z,A) I v,.r -v,,

Solutiont4.4(")

g*\ 4 / 'u

\ - _ _ _ _ - / -

"- / \ou

(b) It is the same as for (a) except that the neutrinos are labelled v" and. io.(.)

<;e-

For the reaction with a positrou target, reverse the arrows on the electron line and relabelas e+.

ll0 Chapter t4

(d)

<;" . ' r i

- g - |

For the reaction with an incident positroD, reve*e the *ro*, oo Jha erectron line andrelabel as e+.

^"^.^]t:-"3 "f (:l *9 (d) there is anorher diagram: the posirion of the real or (virtual)pnoton vertex on the electron line may be exchanged with that of the Zo vertex.

rr*iira. ". -, .,_r-"*:ei; "".i

Clapter Ii ill

Exercircl

14.5 show that tho potontial enorgy due to gravitational attraction ol a spherical mass ,r/ ofuniform density and redius .R is

where O is tho gravitational constant.

14.6 csfculate tho pot'ntiar energy in joules for one sorar mass 11M.l otmstorisl olunlform denslty and radius (a) t light yoar, (b) one solar radius (X'" iniaUfe f +.i1, tij100km, and (d ) i0km,It is likely that lhe donsity is greater at the contro. How wilt thls change th6 potentisl

energy?

1 4,7 Suppose that the iron core of a pre-supernova star has a mass of 1 ,4 M^ and a radiusof l00.km and that it coltapses to I uniform sphere of neutrons of radim io iil *.rr.ii"ith' vifial theorom of the form of equation (r 4.2) hotds. carcurare tho energy .onrrr.J'inneutronization ond the number of olactfon neutrinos produced. Given thai the remaininoenergy is radiated as neutrino-antineutrino pairs of all kinds of averago ohsrgy r z*i z rrruv]calculato the total numbor radiated

14'8 Given that the supernova of probrem r4.7 is at a distance of 163000 right yean,calculata the total numbor of neutrinos of all types anlving at each squa16 m6trs atihe iarth.Also sstimat6 the number of reactions

i * P - n + c '

that will occur in 1000 tonnes of water. Assume that the cross-section is given by

^ -4P'E'G1" --iF,T'

whe re P. and d are tha positron momentum and energy respectiueii and o. is the Fermicoupling constant. Assume only one-sixth of the neutrinos ars eiactidn antineutrinos.

14.9 Modifv,tha nucleai sami-empirical mass formula iry rhe addition of a term giving ihemecnanrcar potenrial energy due td gravitationsl anraction and apply.it.to a neutron sta,(z=01. Find tho smallost radius'r8 noutron star can have assumlng nuciear densitythroughout and the absence of any other constitu6nts.

Solutions14.5 The dessity, p, of the body is given by

3Mp=m

ll2 Chapter 14

Suppose we build this body layer by layer. on a part of way through this process wherewe have reached a radius r, msss Mr! f Rt, we add a laycr thickness dr and mass 4rr2 pd,r,This layer is sdded at a radius wbere the gravitational potential is -6Mrr/83 so thattbe contributioo, dU, to the eaergr is

du = -(#) gtrr2pdr)

, 3GM2ra ,= -7_or r

Thus tbe mechanical potential enerry of the whole'assembly is

u=tau = -loesg{4''di3GM2= --E- '

14.6 In this pioblem M = 1.99 x 1030 kg, hence'Y

=1.S9 x l05o J m.

a) 8= 1 l ightyear= 9.46 x l0r5 m U= -1.68 x 103{ J.b ) R = R " = 6 . 9 6 x 1 0 8 mc) .R = 100 kmd ) . R = 1 0 k m

U = -2.28 x 10{r J.U = -1.59 x l0{5 J.U = -1.59 x 10{6 J.

If the density is greater towards the centre then the material is more concentratedat regions of more negative gravitational potential energy; the total potential ener6y mustthen be more negative than found for the uniform density case.

14.7 The gravitational potential energr, O, is giveu by (problem l4.S)

e= -3cM2/5n.

Il M = l.l'11'[a= 1.4 x 1.988 x 1030 kg, then, since G = 6,628 x l0-rr ms kg-r s-2, weh a v e

Q = - 3 . 1 o l x l o 5 o / g .

For .R = 100 kn, O = -3.101 x 10{5 J.For rB = 10 km, O = -3,101 x 10{6 J.Heace, for the collapse from 100 to l0 km

A Q = - 2 . ? 9 1 x 1 0 { 6 J .

The time averaged internal kinetic enerry and tbe gravitational potential satisfy (Eq. la.2)

2 ? * O = 0 .

Therefore, in this collapse

AT = -6912

= 1.398 x 10{6 J ,aod the increase in binding energr AB is

Aa = -411+0) -A? = 5.99 x 1057.

Chapter14 113

= 1.395 x 10{6 J .

Tbis energy becomcs availablc as tbe collapsc occurs and is uscd in thc neutronization ofthe iron a,ld in radiating neutrinos.

The energy required for neutronization has two parts:

1. the enerry, .91, required dismantle the iron a-ssumed to be atomic, ilto its co[stituent

nucleons and electrons,2, tbe energr, 82, consumed in the reaction

e - + p + t r * t / c r 0 . 7 8 2 M e V .

Tbe total number of iron nuclei Np. is given by the ma.ss of the core and the atomic weight

of iron:1.4 x 1.989 x 1030 x 6.022xL026

NF" = = 3.001 x 1055 .55.847

we assume the iron bu z = 26, A = 56 so that the total number of nucleons, N, is given

bY N = 56 N'" = 1.681 x 105? '

The binding energy per nucleon is 8.? MeV (Fig. a.2) so the energy required to dismantle

the iron is& = 8.7N = L.462 x 1056 MeV ,

where we have ueglected the binding energy of the atomic electrons. The number of protons

(and electrons) to be converted to Deutrons is 26 Nr" and each conversion produces one

electron neutrino so the number thus is

26Np. = ?.803 x 1056 .

Each conversioo requires 0.782 MeV' giving an enerry requirement

' 82= 6'102 x 1056 MeV '

Thus the total energl required to neutronize the 1.4 M6 of iron is

Ey* E2 = 1.523 x 1058 MeV ,= 2.440 x l0{5 J .

Tbis energy comes from the increase in the core's binding energy, AB. Thus the energy

to be radiated is -

' AB - (4 + &) : 1.39s x 10'6 - 2.440 x 10{5 J

= 1.151 x 10{6 J

= ?'18f x 1058 MeV '

This eneigr is radiated as neutrino-antineutrino pairs of average energy 12 * 12 MeV and

therefore ihe number ofpairs radiated is 2.994 x 105? or 5.998 x 10s7 neutrinos. Including

the neutronization electron neutrinos the total number is 6.77 x 1057.

Summarizing:

Energr consumed in neutronization = 2.44 x l}as J'

Number of electron neutrinos.produced = ?.80 x 1056'

Number of neutrinos and aotfieutrinosproduced as pairs l

ll4 Chapter J4 '

Notice that in this calculation we allowed no kinetic euergr for the electron neutrinosproduced in neutronization.-A.more conviucing result would-be to share the energy tobe radiated among these and the ueutrinos of ihe pairs of a[ generatious, each neutrinohaving 12 Mev. This changes tbe last number from s.99 to s.21 x 1057.

14'8 The distance in light years has to be converted into metres. Dach year has365| days, each day 24 horjrs and go on. Hence

f63 000 light years = 163 000 x 365.28 x 24 x g600 x 2.998 x 108 m=- 1 ,54 x l02 l m.

we toke the number of neutrinos produccd to be 5.9g x r0r? eo thot the numbor crossingono squore metre el the earth is

5.99 x 1057

GJAilgqtG = 2'01 x 10" neutrinos m-2 '

The reaction

! 1 * p + s a g +

e = _(M" _ M, + m")c2 = _1.804 MeV ,and an effective cross-section given by (Table 12,9)

o = AP"E:G2FnhaC

we neglcct the recoil momentum and energy of the neutron so that the average neutrinogivcs a positron of kinctic cnergy 10.20 MeV and hcnce E" = 10.?l MeV, wh'ich in turnhas

P"c= r[gg@= lo.zo MeV.

siving

, = nnip€tr = 7 .73 x16-.6 *z .r(hc)aThe cxpected number of events is the product of the aumber of free protons in the targer,the cross-section and the number of eiectron antineutrinos crossing one square metre

= #F

x 106 x 6.022 x 1026 x Z.?3 x 10-a6 x I

x z.Or x l0r{ = 1.23.

of course, wc can only obscrvc an integcr numbcr of events, but this number can be usedin tbe Poisson formula to give the probability of observiag 0, l, 2... events.

14'9 , _

we use the semi-empirical mass formula from Table 4.r. For a neutron star, treatedas u-niform density sphere, z = 0, A = N, wbere lf is the number of neutrons. Thus thebinding energy appears to be

B = a,N - a.N2l3 - o,r.N ,l{owcvcr, we tnust now include a term for thc gravitational binding energy. The coulombtcrm that wc have dropped is, for a uniformly charged sphere of odiu, E, given by

22 322e2" Attt 2lresR'

Chaptal4 115

This is a negative contribution to the bioding enerry. The gravitational potential will bea positive contribution to the binding energy and, using tbe results of problems 4.1 and14.4, we make the replacement

_ 322e2 - +3G(/VM")2

20rcsB 5R

We assume uuclear density so that R = Ra/{t/r = l,llrltl3 x l0-l! m and we have for theneutron star binding energy

B= (av - oe)ff - /'.Nztt *%Itl

we expect N to be so large thot N > /v2/3 and we neglect the surface term, For nucleiov = 15.56 MeV, o1 = 23.29 MeV (Table 4.1) so that, for small .lV, the binding energr isnegative and we expcct no nuclei composed oolely of Deutrons, However, at large /v thegravitational term can dominate and at some value of N, B is zero and for larger iv ispositive. The critical value of N gives the minimum size possible for a neutron star whichmust have a positive binding energy. The critical N is given by

o=(av-aa)No+"*ry*

*2t3 (o1 -'oy)sit4t \ o = - .

Therefore

Substituting numbers we obtain

with a radius

N o = 4 . 8 1 x 1 0 5 5

r ? = 4 3 4 5 m .