nuclear energy [radioactivity, nuclear fission and fusion]
TRANSCRIPT
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Chapter 7: Nuclear Energy
7.1 Introduction
7.2 Radioactive decay
7.3 Different radioactive decay processes
7. !a"s of radioactive decay
7.# $alf life
7.% &verage or 'ean lifeti'e
7.7 (ass Defect and )inding energy
7.* Energy due to +ission
7., Nuclear Chain Reactions
7.1- Controlled Nuclear +ission
7.11 i'escales of nuclear chain reactions
7.12 Effective neutron 'ultiplication factor
7.13 Energy due to +usion
7.1 I'portant fusion reactions
7.1 Introduction
Nuclear energy is released by three exoenergetic (or exothermic) processes:
• Radioactive decay, where a neutron or proton in the radioactive nucleus decays
spontaneously by emitting either particles, electromagnetic radiation (gamma rays),
neutrinos (or all of them)
• Fusion, two atomic nuclei fuse together to form a heavier nucleus
• Fission, the breaking of a heavy nucleus into two (or more rarely three) lighter nuclei
he difference between the masses before and after the nuclear reactions representing the above
three nuclear processes corresponds to the reaction energy or nuclear energy, according to
the mass!energy relation "#mc$%
Nuclear energy was first discovered by French physicist &enri 'ecuerel in *+, when he found
that photographic plates stored in the dark near uranium were blackened like -!ray plates, which
had been .ust recently discovered at the time *+/%
E/a'ple01: +ind the energy euivalent of 1g' of 'atter.
0olution:
m# 1%11 kg
2age of 3
http://en.wikipedia.org/wiki/Exothermichttp://en.wikipedia.org/wiki/Radioactive_decayhttp://en.wikipedia.org/wiki/Radioactivehttp://en.wikipedia.org/wiki/Electromagnetic_radiationhttp://en.wikipedia.org/wiki/Neutrinoshttp://en.wikipedia.org/wiki/Nuclear_fusionhttp://en.wikipedia.org/wiki/Nuclear_fissionhttp://en.wikipedia.org/wiki/Francehttp://en.wikipedia.org/wiki/Henri_Becquerelhttp://en.wikipedia.org/wiki/Uraniumhttp://en.wikipedia.org/wiki/X-rayhttp://en.wikipedia.org/wiki/Exothermichttp://en.wikipedia.org/wiki/Radioactive_decayhttp://en.wikipedia.org/wiki/Radioactivehttp://en.wikipedia.org/wiki/Electromagnetic_radiationhttp://en.wikipedia.org/wiki/Neutrinoshttp://en.wikipedia.org/wiki/Nuclear_fusionhttp://en.wikipedia.org/wiki/Nuclear_fissionhttp://en.wikipedia.org/wiki/Francehttp://en.wikipedia.org/wiki/Henri_Becquerelhttp://en.wikipedia.org/wiki/Uraniumhttp://en.wikipedia.org/wiki/X-ray
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"#mc$ # 1%11x (4%1x1*)$ # +x14 .oule or watt!sec
36001000
109 13
x
x= # $%/x15 k6h
7.2 Radioactive decay
For a nucleus to be stable, the number of neutrons should in most cases be little higher than thenumber of protons% For example, oxygen has three stable isotopes
]10,9,8[,, 18
8
17
8
16
8 = N OOO and five known unstable (i%e% radioactive) isotopes
Oand OOOO 20
8
19
8
15
8
14
8
13
8 ,,, % 7n the case of the isotopes ]7,6,5[,, 15
8
14
8
13
8 = N OOO there
are not enough neutrons for stability while the isotopes ]12,11[, 20
8
19
8 = N OO have too many
neutrons%
7.3 Different radioactive decay processes
i )eta Decay:
Nuclei such as O15
8 which are lacking in neutrons, undergo 450decay% 7n this process one of the
protons in the nucleus is transformed into a neutron and a positron and a neutrino are emitted%
his transformation is written as
ν β + → + N O
15
7
15
8
where 89 signifies the emitted positron which in this context is called a 8!ray and denotes the
neutrino%
'y contrast, nuclei like O19
8 which are excessively rich in neutron, decay by −−β decay. 7n this
process one of the neutron in the nucleus is transformed into a proton and an electron and an
antineutrino are emitted% his transformation is written as
ν+ → −−β
+6 )+
+
)+
*
7t should be noted that in both 89!decay and −−β decay the atomic mass number remains the
same%
ii Electron capture:
; nucleus lacking in neutrons can also increase its neutron number by electron capture% 7n thisprocess, an atomic electron interacts with one of the protons in the nucleus and a neutron is
formed of the union% his leaves a vacancy in the electron cloud which is latter filled by another
electron% 3+ :
ie8 3+$$
1)
3+$4
→ +−
iii &lpha Decay:
;nother way by which some unstable nuclei undergo radioactive decay is by the emission of analpha particle ? He
4
2@% For example
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$e.h9 3
$
$43
+1
$4*
+$ +→
Aecay by 0e'ission is comparatively rare in nuclides lighter than lead, but it is common for the
heavier nuclei%
iv ;a''a Radiation:
he nucleus formed as the result of 8!decay (9 or !), electron capture or B!decay is often left in an
excited state following the transformation% he excited nucleus then decays by the emission of one or more independent of ti'eG% his constant is
called the decay constantG and is denoted by H%
Iet us suppose that at time t the number of radioactive nuclei which have not yet decayed is N(t)%
he rate at which these nuclei decay is therefore )(t N λ disintegrations per unit time% his decayrate is called the activity of the sample% ;ctivity is measured in 'ecuerel (') which is one
disintegration per second%
Iet us suppose that at the beginning of disintegrations i%e% at t#o, the number of radioactive nuclei
present in the sample is N1%
0ince )(t N λ dt nuclei decay in the time interval dt, it follows that the decrease in the number of
undecayed nuclei in the sample in time dt is
]1[,)()( dt t N t dN λ =−
his euation can be written as
]2[),()(
t N dt
t dN λ −=
From ?$@ we can write
" (4) shows that the number of
surviving nuclei at any time t decreases
exponentially with time%
From ?4@, we can write
2age 4 of 3
]3[,0
0
0
)(]/[log
0
t e N t N t N N
t d N
dN
dt N
dN
e
t N
N
λ
λ
λ
λ
−=∴−=
′−=∴
−=
∫ ∫
http://en.wikipedia.org/wiki/Cobalthttp://en.wikipedia.org/wiki/Nickelhttp://en.wikipedia.org/wiki/Beta_rayshttp://en.wikipedia.org/wiki/Cobalthttp://en.wikipedia.org/wiki/Nickelhttp://en.wikipedia.org/wiki/Beta_rays
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]4[,0
)(
0)(
t e At Aor
t e N t N
λ
λ λ λ
−=
−=
where ;(t) is the activity at time t and ;1 is the activity at time t#o%
7.# $alf life
7t is defined as the time interval during which one half of the total number of nuclei that were
present at the beginning of the time interval have decayed %
7f N nuclei are present at time t and one half that number 2/12 N N = have survived at time t$,
we can write
%,3-2
1221
212
212
2
1
2-2
1-1
.ln][/
ln][
][
t t lifehalf T
t t
t t
e N
N
t
t
e N N
e N N
7.% &verage or 'ean lifeti'e
7f there are N1 radioactive nuclei at time #1, the number that decay in some time interval dt at t is
dt t e N dt t N dN λ λ λ −==− 0][
7f we multiply this number by the life time t of these nuclei, sum over all the possible lifetimes from
t#1 to t#∞ , and divide by the total number of nuclei, we get the average or mean lifetime τ :
λ
λ λ λ
λ λ τ
λ
λ λ
λ λ
1
}1
]/{[
1
0
0
0
0000
==
+=
===
∫
∫
∫ ∫ ∫
∞
∞∞
∞∞∞
−
−−
−−
dt e
dt ete
tedt et dN t N
t
t t
t t
Now, we know
τ λ t
t e Ae At A
−−
== 00)(
7n can be seen from this euation that in one mean (average) life, the activity falls to (Je) of its
initial value%
E/a'ple02: Find the half!lives of a radioactive material if its activity drops to (J)th of its initial value
in 41 years%
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0olution: 6e know thatn N N )2/1(0= %
4
)2
1()
2
1(
16
1 4
0
=
===∴
nor
N
N n
herefore, &alf!life # otal time of disintegration J No% of half!lives # 41 years J 3 # 5%/ years%
E/a'ple03: 6hat percentage of initial amount of a radioactive material decays during the time, eual
to mean lifetime of this materialK
0olution: Number of radioactive nuclei left after one mean lifetime
# λ τ λτ λ
/1,/000 ==== −−
e N e N e N N t
%63100)7.2
11(100)
11(100
0
0 =−=−=−
=∴ x xe
x N
N N decayof Percentage
7.7 (ass Defect and )inding energy
; nucleus consists of proton and neutrons% &owever, the total mass of a nucleus is always less
than the sum of the masses of its constituents e%g% the mass of helium nucleus is 3%114** amu
(atomic mass unit) whereas the mass of $ protons and $ neutrons totals 3%144$ amu%
hus, the mass of a helium nucleus is 1%1$+$3 amu less than the sum of the masses of its
constituents% his difference is known as mass defect% he mass defect of a given nucleus can be
calculated by using the euation:
(ass defect ? @A 'p 5 & 0 A 'nB nuclear 'ass
6here, L # atomic number # number of protons in the nucleus
; # atomic mass number # total number of protons and neutrons in the nucleus
mp # mass of proton # %115*$/ amu
mn # mass of neutron # %11*/ amu
he energy euivalent of mass defect is called binding energy% ;n amount of mass eual to mass
defect has been converted into potential energy which holds the nucleus together%
E/a'ple0: ho" that a 'ass defect of 1a'u is euivalent to a=out ,31 (ev of energy.
0olution:
amu # %1/3x1!$5kg
"nergy # mc$ # %x1!$5x(4%1x1*)$ # %3+x1!1M # MeV x
x 931
1061
10491
19
10
=−
−
%
%
%
he binding energy per nucleon of different elements is different% 7n Figure below, we show the
binding energy per nucleon as a function of mass number for different elements% 7t is seen from
this figure that binding energy is highest at the centre of the periodic table% his means that if
lighter elements are fused together or heavier elements split, release of energy would take place%
his gives two fundamental ways of obtaining nuclear energy:
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(i) Fusion of light elements into heavier elements
(ii) Fission of heavy elements into lighter elements%
he nuclear fission process is used in nuclear power plants for generation of energy% he nuclear
fusion process has still not been exploited commercially%
Fig% : 'inding energy per nucleon as a function of mass number%
he peak at ;#3 corresponds to the exceptionally stable He4
2 nucleus which is alpha
particle% he binding energy per nucleon is a maximum for nuclei of mass number ; #/%
his figure suggests that we can liberate energy from the nucleus in two different ways% 7f we
split a heavy nucleus into two lighter nuclei, energy is released because the binding energy
per nucleon is greater for the two lighter fragments than it is for the original nucleus% his
process is known as nuclear fission% For example, if the uranium nucleus is broken into two
smaller nuclei, the binding energy difference per nucleon as about 1%* DeE% he total energy
given off is therefore
?1%* DeEJnucleon@?$4/ nucleons@#** DeE
;lternatively when we combine two light nuclei into a heavier nucleus, again, energy is
released when the binding energy per nucleon is greater in the final nucleus than it is in thetwo original nuclei% he process is known as nuclear fusion% For instance, if two deuterium?
H 2
1 @ nuclei combine to form a He4
2 helium nucleus, over $4 DeE is released% 7n fact,
nuclear fusion is the main energy source of the sun and other stars%
E/a'ple0#: +ind the average =inding energy per nucleon for a heavy hydrogen $$)
> =
9raniu' 9$4/+$
.
0olution: (a) For $$)
, atomic mass # $%13 amu
Dass defect # (%115*$/ x ) 9 %11*/ ($!) > $%13 # 1%11$4+ amu
otal binding energy # +4x 1%11$4+ DeE # $%$$/ DeE
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;verage binding energy per nucleon #number massatomic
energybindingtotal# ))$/)
$
$$/$.
.=
DeE
(b) For 9$4/+$ , atomic mass # $4/%134+ amu
Dass defect # (%115*$/ x +$) 9 %11*/ ($4/!+$) > $4/%134+ # %+/ amu
otal binding energy # +4x %+/ DeE # 5*$%+/ DeE
;verage binding energy per nucleon #number massatomic
energybindingtotal# /*55
$4/
)5*$%+/.= DeE
7.* Energy due to +ission
he fission of a heavy atom can be caused by bombarding
it with a thermal neutron% 7f a $4/< atom is bombarded by a
neutron, the nucleus splits to give nuclei of other elements%
ne possible fission reaction of $4/< is
$4/< 9 n # 3*Ia 9 */'r 9 4n
he mass euation of this reaction is
($4/%$3 9 %11+) amu # (35%+ 9 *3%+4* 9 4%11$5) amu
he mass of fission products is 1%$15 amu less than mass on the left hand side% hus this
reaction means a mass defect of 1%$15 amu, which is euivalent to
1%$15x+4 DeE # +$%55 DeE of energy
Cenerally, it is assumed that fission of $4/< causes a release of $11 DeE of energy as:
/ DeE
5 DeE
DeE
5 DeE
DeE
+ DeE
2-- (e8
O kinetic energy of fission products
O gamma rays
O kinetic energy of the neutrons
O energy from fission products
O gamma rays from fission products
O anti!neutrinos from fission products
DeE (million electron volts) # %1+ x 1 013 .oules
7f all the atoms of kg of pure $4/< ($/%3x1$4 atoms) were fissioned, the energy released would
be euivalent to that contained 4x1 kg of coal% Natural uranium contains only 1%5P $4/
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n Kr BanU 109436
13956
10
23592
3++→+
+ertile (aterial: ; fertile material is one that will capture a neutron, and transmute by
radioactive decay into a fissile material% Fertile isotopes may also undergo fission directly, but
only if impacted by a high energy neutron, typically in the DeE range% Fertile materials are$4$
h,$4*
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the emission of a neutron is often considered its Tbirth,T and the subseuent absorption is
considered its Tdeath%T For thermal (slow!neutron) fission reactors, the typical prompt neutron
lifetime is on the order of 1U3 seconds, and for fast fission reactors, the prompt neutron lifetime is
on the order of 1U5 seconds% hese extremely short lifetimes mean that in second, 1,111 to
1,111,111 neutron lifetimes can pass%
(ean generation ti'e
he 'ean generation ti'e, V, is the average time from a neutron emission to a capture that
results in fission% he mean generation time is different from the prompt neutron lifetime because
the mean generation time only includes neutron absorptions that lead to fission reactions (not
other absorption reactions)% he two times are related by the following formula:
7n this formula, k is the effective neutron multiplication factor, described below%
7.12 Effective neutron 'ultiplication factor
he ratio of neutrons available for fissioning in any one generation to the number available in thepreceding generation is called the effective multiplication factor , k eff , and is calculated by:
keff #generationpreceding>fissionsof Nu'=er
generationone>fissionsof Nu'=er
• k eff W (subcriticality): he system cannot sustain a chain reaction, and any beginning
of a chain reaction dies out over time%
• k ef f # (criticality): "very fission causes an average of one more fission, leading to a
fission (and power) level that is constant% Nuclear power plants operate with k eff #
unless the power level is being increased or decreased%
• k eff X (supercriticality): For every fission in the material, it is likely that there will be T
k eff T fissions after the next mean generation time. he result is that the number of
fission reactions increases exponentially% Nuclear weapons are designed to operate
under this state%
7n a nuclear reactor, k eff will actually oscillate from slightly less than to slightly more than , due
primarily to thermal effects (as more power is produced, the fuel rods warm and thus expand,
lowering their capture ratio, and thus driving k lower)% his leaves the average value of k at
exactly % Aelayed neutrons play an important role in the timing of these oscillations%
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http://en.wikipedia.org/wiki/Subcriticalityhttp://en.wikipedia.org/wiki/Criticalityhttp://en.wikipedia.org/wiki/Supercriticalityhttp://en.wikipedia.org/wiki/Subcriticalityhttp://en.wikipedia.org/wiki/Criticalityhttp://en.wikipedia.org/wiki/Supercriticality
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7.13 Energy due to +usion
Nuclear energy can also be released by
fusion of two light elements (elements
with low atomic numbers)% Nuclear fusion
reactors, if they can be made to work,
promise virtually unlimited power for the
indefinite future% his is because the fuel,
isotopes of hydrogen, are essentially
unlimited on "arth% "fforts to control the
fusion process and harness it to produce power have
been underway in the
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7.1 I'portant fusion reactions
i &strophysical reaction chains
he most important fusion process in nature is that which powers the stars% he net result is the
fusion of four protons into one alpha particle, with the release of two electrons, two neutrinos, andenergy, but several individual reactions are involved, depending on the mass of the star% For stars
of the siSe of the sun or smaller, the proton!proton chain dominates% 7n heavier stars, the YN
cycle is more important%
ii roton0proton cycle
0ince the sun is composed of ordinary hydrogen, rather than deuterium, it is first necessary to
convert the hydrogen to deuterium% his is done according to the reaction
ν ++→+ +e H H H 211
1
1
1
his process involves converting a proton to a neutron and is analogous to the beta!decay
processes discussed earlier% nce we have obtained $& (deuterium), the next reaction that can
occur is
γ +→+ He H H 321
1
2
1
followed by H He He He 114
2
3
2
3
2 2+→+
Note that the first two reactions must occur twice in order to produce the two
4
&e we need for thethird reaction% 6e can write the net process as
he net result is the fusion of four protons into one alpha particle, with the release of two
electrons, two neutrinos, and energy, but several individual reactions are involved, depending on
the mass of the star% 0ince the two positrons disappear in this process, the only masses
remaining are four hydrogen atoms and the one helium atom, and so
MeV u MeV uu xcmmQ f i 7.26)/5.931()002603.4007825.14()( 2
=−=−=
"ach fusion reaction liberates about $%5 DeE of energy% Iet us now try to calculate the rate at
which these fusion reactions occur in the sun% he power output from the sun may be shown to
be about 3x1$ 6, which corresponds to about 14* DeEJs% hus there must be about 14* fusion
reactions per second, consuming around 3x14* protons per second% For stars the siSe of the sun
or smaller, the proton!proton chain dominates%
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γ + ν++→
γ + ν+β++→+→
γ + ν+β+→
+
+
+
$$$3
$$$$,
4
3
$
)
)
)
)
3
$
4
$
4
$
)
)
4
$
)
)
e$e$
$$e$e$e$
$e$
http://en.wikipedia.org/wiki/Protonhttp://en.wikipedia.org/wiki/Alpha_particlehttp://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Neutrinohttp://en.wikipedia.org/wiki/Proton-proton_chainhttp://en.wikipedia.org/wiki/Protonhttp://en.wikipedia.org/wiki/Alpha_particlehttp://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Neutrinohttp://en.wikipedia.org/wiki/Proton-proton_chain
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iii Car=on cycle
7n heavier stars, the YN cycle is more important% ; more likely seuence of reactions in thecarbon cycle is shown below:
HeC H N
e N O
O H N
N H C
eC N
N H C
412115
1515
15114
14113
1313
13112
+→+
++→
+→+
+→+
++→
+→+
+
+
ν
γ
γ
ν
γ
Notice that the $Y plays the role of catalyst we neither produce nor consume any $Y in these
reactions, but the presence of the carbon permits this seuence of reactions to take place at a
much greater rate than the previously discussed proton!proton cycle% he net process is still
described by 3
&→
3
&e, and of course the Z value is the same% 0ince the coulomb repulsionbetween & and Y is larger than the Youlomb repulsion between two & nuclei, more thermal
energy and a correspondingly higher temperature are needed for the carbon cycle% he carbon
cycle probably becomes important at a temperature of about $1 - 1 =, while the 0un[s interior
temperature is only / - 1 =%
6hen all of the hydrogen has been converted to helium, the 0un will contract and its temperature
will increase until helium burning occurs, by processes such as
4 3&e → $Y
wo &e nuclei have a larger mutual Youlomb repulsion than two & nuclei, so helium fusion needs
more thermal energy than hydrogen fusion%
6hen the helium is used up, a still higher temperature will allow carbon fusion to make even
heavier elements, for example, $3Dg% 0uch processes will continue until /Fe is reached beyond
this point no further energy is gained by fusion%
6!8ED EF&(!E
ro=le' No. 1: 0how that after 1 half!lives a radioactive material is reduced to J111 part
approximately%
0olution: From radioactive decay law, one can show that a radioactive material, after n half!lives, will
decay ton N N )2/1(0= %
1000/1024/)2/1(00
10
0 N N N N ≅==∴
ro=le' No. 2: he half!life of radium is 11 years% ;fter how much time (J) th part of radium will
remain un!disintegrated in the sampleK
0olution: Civen 16/0 N N = , therefore,n
N N )2/1()2/1()16/1(/ 4
0 ===
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4=∴ n
herefore, time of disintegration # No% of half!lives x half!lives # 3 x 11 # 311 years%
herefore, &alf!life # otal time of disintegration J No% of half!lives # 41 years J 3 # 5%/ years%
ro=le' No. 3: here is a stream of neutrons of kinetic energy of 1%1$/ eE% 7f the half!life of neutron is
511 seconds, what fraction of neutrons will decay, before they travel a distance 1 mK
0olution: 6e know that
sm x x
x x x
m
E xK vmv E K /1019.2
1067.1
106.1025.2.2
2
1.
3
27
192 ===⇒=
−
−
ime needed to travel a distance of 1 m # 1 m J ($%+ x1 4) # 3%/5x1!4 seconds%
6
0
)700/1057.4(
0
1053.61
99999347.03
−
−−
=−=∴
=== −
x N
N
decayed fraction
ee N
N xt λ
ro=le' No. : ; sample of uranium is a mixture of three isotopes U and U U 238
92
235
92
234
92 , , present
in the ratio 1%11P, 1%5P and ++%$*3P respectively% he half!lives of these isotopes are $%/x1 /
years, 5%x1* years and 3%/x1+ years respectively% Yalculate the contribution to activity (P) of each
isotope in sample%
0olution: No% of U 234
92 nuclei in the mixture #0
24 N #234
1002.6006.0 23 x x(aking total mass of
11gm)
No% of U 235
92 nuclei in the mixture #0
25 N #235
1002.671.0 23 x x
No% of U 238
92 nuclei in the mixture #0
28 N #238
1002.6284.99 23
x x
he relative contribution of the isotopes in the activity (P) would be
45.46:%13.2:%41.51926.0:0425.0:02.110926.0:100425.0:1002.1
105.4238
1002.6284.99:
101.7235
693.01002.671.0:
105.2234
693.01002.6006.0::
101010
9
23
8
23
5
23
28
0
2825
0
2524
0
24
⇒⇒
⇒
−−−
x x xor
x x
x x
x x
x x x
x x
x x x N N N λ λ λ
ro=le' No. #: he half!lives ofU
235
92 andU
238
92 are 5%x1
*
years and 3%/x1
+
years respectively%
oday the isotopic abundance of U 235
92 and U 238
92 are respectively 1%5$P and ++%$*P% ;ssuming
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that initially these isotopes were in eual abundance and no isotopic separation has occurred,
calculate the age of these elements on the earth%
0olution:
t
t
e N N
e N N
28
25
0
2828
0
2525
λ
λ
−
−
=
=
7n the beginning,0
28
0
25 N N = %
years x xt
year per x x
year per x x No!
t or
t or
e
et
t
910
109282/128
10825
2/125
2825
2528
1099.510]54.176.9/[927.4
1054.1105.4/693.0/693.0
1076.9101.7/693.0/693.0
]/[927.4
927.488.137ln][
88.13772.0
28.99
25
28
=−=∴
===
===
−=
==−−
==∴
−
−
−
−
−
λ
λ
λ λ
λ λ
λ
λ
ro=le' No. %: Yalculate the amount of energy reuired to remove a neutron from40
20Ya
nucleus%Civen the masses of40
20Ya#4+%+$/*+ amu,
39
20Ya#4*%+51+ amu, mn#%11*/
amu%
0olution:
nCaCa +→ 392040
20
Dass defect # (mass of Ya!4+ 9n) > mass of Ya!31 # 4*%+51+ 9 %11*/ > 4+%+$/*+
= 0.016767 amu = 0.016767 amu × 931 !"/amu = 15.61 !"
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