-decay: X Y + AZ N N-2

A-4 42 2Z-2

Conservation of Energy TTcmmmYYX 2)(

QFor a parent AX nucleus at rest:

m

p

m

pQ

Y

Y22

22

ppY

YYm

pmmTmmQ /1/12

2

AYm

m

QQT

411

Y

mmTQ /1

Note: for heavy nuclei QT

to within ~98% accuracy, anyway

We’ll see from a few examples that typically T 4-5 MeV

Is Pu unstable to -decay?23694

Pu U + 23694

23292

42 + Q

Q = (MPu – MU M)c2

= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u

= 5.87 MeV > 0

Repeating an OLDIE but GOODIE from Lecture 13 on “Radiation”:

Some (especially the heaviest) nuclei are unstable with respect to the emission of heavy particles

•essentially the break up of a nucleus.

In one extreme: the emission of a single nucleon

but it includes the far more common alpha emission

and fission of the original nucleus into smaller, approximately equal sized nuclei.

Table 8.1 Energy Release (Q value) for various modes of decay of 232U

Emitted particle Energy Released (MeV)

n -7.26 MeV1H -6.122H -10.703H -10.243He -9.924He +5.51 MeV5He -2.596He -6.196Li -3.797Li -1.94

attractivenuclear potential

repulsiveCoulomb potential

The calculation of the kinetic energy of an alpha particle emitted by the nucleus 238U.

The model for this calculation is illustrated on the potentialenergy diagram at the right.

Let’s follow:

stepping through the details:

The mean binding energy per nucleon B/A for 238U

(from the Semi-empirical mass formula)is 7.5 MeV.

This potential energy curve combines a nuclear well

of radius 7.75 fm (from R = 1.25 x A1/3 fm)

and the Coulomb potential energy

of an alpha in the electric field of the daughter 234Th nucleus.

Thus to remove 4 average nucleons would require 30 MeV.

Compare to using thesemi-empirical mass formula

to calculate the energy required to remove

2 protons and 2 neutrons from the highest 238U

energy levels.

assumes they are the last two particles of each type added to the 234Th nucleus.

24.4 MeV

For the alpha particle m= 0.03035 u which gives 28.3 MeV binding energy!

The model for alpha emission proposes that the alpha particle is preformed inside the nucleus.

protons 2 1.00728 u

neutrons 2 1.00866 u

Mass of parts 4.03188 u Mass of alpha 4.00153 u

1 u = 1.66054 10-27 kg = 931.494 MeV/c2

Alphaparticle

NN

NN

An alpha particle with positive energy is created inside the nucleus

where it is trapped by the potential barrier.

According to quantum mechanics it has a finite probability of escape.

The binding energy released (28.3 MeV) appears in part as kinetic energy of the alpha.

Let’s see how well quantum mechanicsand our model of the potential

can calculate that probability (decay rate)

Nuclear potential

Coulomb potential

finite (but small) probability of being found outsidethe nucleus at any time

Tunneling

always some probability of a piece of the nucleus escaping

the nuclear potential

with a STATIC POTENTIAL this probability is CONSTANT!

Let’s examine this through a simple model: an forms inside the nucleus and then escapes through quantum mechanical barrier penetration.

The potential seen by the is spherically symmetric, so we can start by first separating the variables -

the functions Ylm are the same spherical harmonics

you saw for the wavefunctions for the hydrogen atom.

),()()( lmnl YrRr

02

)1()(/2/)(

2

2222

nlnl rR

mr

llrVEmdrrRd

Then the equation for the radial function Rnl(r) can be written as

For states without orbital angular momentum ( l = 0) this reduces to an equation like that for a 1-dimensional barrier.

eX where drErVm

)(2

2

In this case the inner limit of the integral is effectively the nuclear radius R, and the outer limit is taken as the point at which the ’s kinetic energy is equal in magnitude to its potential energy.

02

)1()(/2/)(

2

2222

nlnl rR

mr

llrVEmdrrRd

The transmitted part of the wave function X is of the form

The integral is carried out over the range of the potential barrier.

A solution can be found by approximating the shape of the potential as a succession of thin rectangular barriers.

xkxk DeCex 22)(II

In simple 1-dimensional case

E

I II III

V

)(2

22 EVm

k

where

222222II |)(| rkeDrx

In simple 1-dimensional case

E

I II III

V

probability of tunneling to here

x = r1 x = r2

R

E

Where

2

2

0

)2(2

4

1

r

eZT

r2

So let’s just write

r

eZrV

2

0

)2(2

4

1)(

Tr

rrV 2)(as

the point at which the ’s kinetic energy is equal to its potential energy.

hence drErVm

)(2

22

22 cos/ rr

drdr cossin2 2

1tan 2 r

r

then with the substitutions:

with E=T becomes drr

rmT1

22 2

2

drmT

drmT

222

22

sin22

2

)cossin2)((tan2

2

2221

0

2

2/1//cos

4

)2(2

22 rRrRrR

T

eZmT

and for R << r2 the term in the square brackets reduces to

2/22

rR

Performing the integral yields:

22 cos/ rr

2

2

0

)2(2

4

1

r

eZT

a

axxdxax

4

2sin

2sin2

2

0

2

2/2

24

)2(2

22 rR

T

eZmT

into which we can again substitute for r2 from

00

2

4

)2(82

2

)2(

ZmRe

T

meZ

2

2

0

)2(2

4

1

r

eZT

T

eZr

2

02

)2(2

4

1

and get

)2()2(

exp|)(| 2 ZRBT

ZAerX

When the result is substituted into the exponential the expression for the transmission becomes

The decay probability is = f X where f   is the frequency with which the alpha particle hits the inside of the barrier. Thus

T

ZAZRBf

)2()2(lnln

f can be estimated from crude time between striking nuclear barrier

vv

rA )fm 6(2)(2 03/1

of 4-8 MeV “pre-formed” alpha

second/1022 timesf

Easily giving estimates for = 106/sec – 10-21/sec

Some Alpha Decay Energies and Half-lives

Isotope T(MeV) 1/2 (sec-1)

232Th 4.01 1.41010 y 1.61018

238U 4.19 4.5109 y 4.91018

230Th 4.69 8.0104 y 2.81013

238Pu 5.50 88 years 2.51010

230U 5.89 20.8 days 3.9107

220Rn 6.29 56 seconds 1.2102

222Ac 7.01 5 seconds 0.14216Rn 8.05 45.0 sec 1.510

212Po 8.78 0.30 sec 2.310

216Rn 8.78 0.10 sec 6.910

T

ZAZRBf

)2()2(lnln

should be compared with the emperical Geiger-Nuttall law

DTC lnln

this quantum mechanically-motivated relation

The dependence of alpha-decay half-life on the kinetic energy of the alpha particle.Values are marked for some isotopes of thorium.

232ThQ=4.08 MeV=1.4×1010 yr

218ThQ=9.85 MeV=1.0×10-7 sec

For each series of isotopes theexperimental data agree (1911)

The potential seen by an electron in the hydrogen atom

is spherically symmetric (depends only on r, not its direction)!

r

erU

2

04

1)(

),,(),,()(sin

1sinsin

sin

1

2 2

22

2

2

rErrUr

rrr

To solve we apply a separation of variables: )()()(),,( FPrRr

Recognizing that we write Schrödinger’s equation in spherical polar coordinates

r

erU

2

04

1)(

),,(),,()(sin

1sinsin

sin

1

2 2

22

2

2

rErrUr

rrr

)()()(),,( FPrRr with

)()()()()()()(

sin

1)()(sin)()(sin)()(

sin

1

2 2

22

2

2

FPrERFPrRrU

FPrR

PFrR

r

Rr

rFP

r

ErUF

F

P

Pr

Rr

rrRr

)(sin

1

)(

1sin

)(

1

)(

sin

sin

1

2 2

22

2

2

ErUF

rF

P

Prr

Rr

rrRr 222

2

2222

2

2)(

2

sin

1

)(

1sin

)(sin

1

)(

1

ErUF

rF

P

Prr

Rr

rrRr 222

2

2222

2

2)(

2

sin

1

)(

1sin

)(sin

1

)(

1

ErUF

FrYrX 22

22 2)(

2

)(

/),(),(

),()(

/ 22

rZF

F

= K (some constant)

ErUKr

P

Prr

Rr

rrRr 222222

2

2)(

2

sin

1sin

)(sin

1

)(

1

ErUQrr

Rr

rrRr 2222

2

2)(

2)(

1

)(

1

)())((2

)(

12

22

QrUEr

r

Rr

rrR

= K2 (also some constant)

Then the problem becomes finding solutions tothe separate “stand alone” equations

each of which uniquely constraints the wavefunction:

A solution to the radial equation can exist only when a constant arising in its

solution is restricted to integer values (giving the principal quantum number)

Similarly, a constant arises in the colatitude

equation giving an orbital quantum number

Finally, constraints on the azimuthal equation

give a magnetic quantum number

Let’s examine this through a simple model: an forms inside the nucleus and then escapes through quantum mechanical barrier penetration.

The potential seen by the is spherically symmetric, so we can start by first separating the variables -

the functions Ylm are the same spherical harmonics

you saw for the wavefunctions for the hydrogen atom.

),()()( lmnl YrRr

02

)1()(/2/)(

2

2222

nlnl rR

mr

llrVEmdrrRd

Then the equation for the radial function Rnl(r) can be written as

x

x

y

y

z

z

x

y

z

y

z

x

y

z

x

y

z

x

x'

y'

x'

y'

z'

PARITY TRANSFORMATIONS ALL are equivalent to a reflection (axis inversion) plus a rotation

The PARITY OPERATOR on 3-dim space vectors

every point is carried through the originto the diametrically opposite location

Wave functions MAY or MAY NOT have a well-defined parity(even or odd functions…or NEITHER)

xcos xx cos)cos(P P = +1

xsin xx sin)sin(P P = 1

but the more general

xx sincos xx sincosP

However for any spherically symmetric potential, the Hamiltonian:

H(-r) = H(r) H(r)→ →

[ P, H ] = 0

So they bound states of such a system have DEFINITE PARITY!

That means, for example, all the wave functions of the hydrogen atom!

azrea

z /23

100

1

aZrea

Zr

a

Z 2/23

200 21

2

1

iaZr erea

Z

sin

8

1 2/25

121

cos2

1 2/25

210aZrre

a

Z

iaZr erea

Z sin

8

1 2/25

211

aZrea

rZ

a

Zr

a

Z 3/2

2223

300 21827381

1

108

6

4

3

2

1

-dE

/dx

[ MeV

·g-1cm

2 ]

Muon momentum [ GeV/c ]0.01 0.1 1.0 10 100 1000

1 – 1.5 MeVg/cm2

Minimum Ionizing:

-dE/dx = (4Noz2e4/mev2)(Z/A)[ln{2mev2/I(1-2)}-2] I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV

The scinitillator responds to the dE/dx of each

MIP track passing through

A typical gamma detectorhas a light-sensitive

photomultiplier attachedto a small NaI crystal.

If an incoming particle initiates a shower,each track segment (averaging an interaction length)will leave behind an ionization trail with about the same energy deposition.

The total signal strength Number of track segments

Basically avg

MIPtracksmeasuredENE

Measuring energy in a calorimeter is a counting experiment governed by the statistical fluctuations expected in counting random events.

Since E Ntracks and N = N

we should expect E E

and the relative errorE E 1

E E E =

E = AE

a constant that characterizes the resolution

of a calorimeter

(r)=(r)ml () the angular part of the solutions

are the SPHERICAL HARMONICS

ml () = Pm

l (cos)eim

Pml (cos) = (1)msinm [( )m Pl (cos)]

(2l + 1)( lm)! 4( l + m)!

d d (cos)

d d (cos)Pl (cos) = [( )l (-sin2)l ] 1

2l l!

The Spherical Harmonics Y ,ℓ m(,)

ℓ = 0

ℓ = 1

ℓ = 2

ℓ = 34

100 Y

ieY sin

8

311

cos4

310

Y

ieY 2

2

sin2

15

4

122

ieY cossin

8

1521

2

12cos2

3

4

1520

Y

ieY 3

3

sin4

35

4

133

ieY 2

cos2

sin2

105

4

132

ieY 12cos5sin

4

21

4

131

cos

2

33cos2

5

4

730Y

then note r r means

x

y

z

and: Pml (cos) Pm

l (cos()) = Pml (-cos)

(eim=(1)m

so: eimeimeim

but d/d(cos) d/d(cos)

(-sin2)l = (1cos2l

ml () = Pm

l (cos)eim

Pml (cos) = (1)m(1-cos2m [( )m Pl (cos)]

(2l + 1)( lm)! 4( l + m)!

d d (cos)

d d (cos)Pl (cos) = [( )l (-sin2)l ] 1

2l l!

So under the parity transformation:

P:ml () =m

l (-)=(-1)l(-1)m(-1)m m

l ()

= (-1)l(-1)2m ml () )=(-1)l m

l ()

An atomic state’s parity is determined by its angular momentum

l=0 (s-state) constant parity = +1l=1 (p-state) cos parity = 1l=2 (d-state) (3cos2-1) parity = +1

Spherical harmonics have (-1)l parity.

In its rest frame, the initial momentum of the parent nuclei is just its

spin: Iinitial = sX

and: Ifinal = sX' + s + ℓ

1p1/2

1p3/2

1s1/2

4He

S = 0 So |sX' – sX| < ℓsX' + sX

the parity of the emitted particle is (1)ℓ

mY ~

Since the emitted is described by a wavefunction:

Which defines a selection rule: restricting us to conservation of angular momentum and parity.

If P X' = P X then ℓ = even

If P X' = P X then ℓ = odd

If the 2p, 2n not plucked from the outermost shells(though highest probability is that they are)

then they will leave gaps (unfilled subshells) anywhere:Excited nuclei left behind!

EXAMPLE:

If SX = 0

|sX' – sX| < ℓsX' + sX

ℓ=sX' (conservation of angular momentum)

0 3 nuclear transition would mean ℓ=3 so PX' = PX

i.e. 0+3 is possible, but

0+3 is NOT possible

0 2do not change the parity of the nucleus

0 4so PX' = PX

So 0+2

0+4would both be impossible