Nuclear Energy [Radioactivity, Nuclear Fission and Fusion]

8/17/2019 Nuclear Energy [Radioactivity, Nuclear Fission and Fusion]

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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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7/14

;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/


8/14

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*


9/14

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%

2age + of 3

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


10/14

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


11/14

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%

2age of 3

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


12/14

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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http://en.wikipedia.org/wiki/CNO_cyclehttp://en.wikipedia.org/wiki/CNO_cycle


13/14

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


25 N #235

1002.671.0 23 x x

No% of U 238


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

2age 4 of 3


14/14

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 !"

2age 3 of 3

Documents

Nuclear Energy [Radioactivity, Nuclear Fission and Fusion]