Chap 40 SM

301

Chapter 40 Nuclear Physics Conceptual Problems 1 • Isotopes of nitrogen, iron and tin have stable isotopes 14N, 56Fe and 118Sn. Give the symbols for two other isotopes of (a) nitrogen, (b) iron, and (c) tin. Determine the Concept Two or more nuclides with the same atomic number Z but different N and A numbers are called isotopes. (a) Two other isotopes of 14N are:

15N, 16N

(b) Two other isotopes of 56Fe are:

54Fe, 55Fe

(c) Two other isotopes of 118Sn are: 117Sn, 119Sn

2 • Why is the decay chain A = 4n + 1 not found in nature? Determine the Concept The parent of that series, 237Np, the longest-lived member of this decay series, has a half-life of 2 × 106 y that is much shorter than the age of Earth. There is no naturally occurring Np remaining on Earth. 3 • A decay by α emission is often followed by β decay. When this occurs, it is by β– and not β+ decay. Why? Determine the Concept Generally, β-decay leaves the daughter nucleus neutron rich, i.e., above the line of stability. The daughter nucleus therefore tends to decay via β – emission which converts a nuclear neutron to a proton. 4 • The half-life of 14C is much less than the age of the universe, yet 14C is found in nature. Why? Determine the Concept 14C is found on Earth because it is constantly being formed by cosmic rays in the upper atmosphere in the reaction 14N + n → 14C + 1H. 5 • What effect would a long-term variation in cosmic-ray activity have on the accuracy of 14C dating? Determine the Concept It would make the dating unreliable because the current concentration of 14C is not equal to that at some earlier time.

Chapter 40

302

6 • Why does an element that has Z = 130 not exist? Determine the Concept An element with such a high Z value (the Coulomb- repulsion force is proportional to Z2) would either fission spontaneously or decay almost immediately by α emission (see Figure 40-3). 7 • Why is a moderator needed in an ordinary nuclear fission reactor? Determine the Concept The probability for neutron capture by the fissionable nucleus is large only for slow (thermal) neutrons. The neutrons emitted in the fission process are fast (high energy) neutrons and must be slowed to thermal neutrons before they are likely to be captured by another fissionable nucleus. 8 • Explain why water is more effective than lead in slowing down fast neutrons. Determine the Concept The process of ″slowing down″ involves the sharing of energy of a fast neutron and another nucleus in an elastic collision. The fast particle will lose maximum energy in such a collision if the target particle is of the same mass as the incident particle. Hence, neutron-proton collisions are most effective in slowing down neutrons. However, ordinary water cannot be used as a moderator because protons will capture the slow neutrons and form deuterons. 9 • The stable isotope of sodium is 23Na. What kind of beta decay would you expect of (a) 22Na and (b) 24Na? Determine the Concept Beta decay occurs in nuclei that have too many or too few neutrons for stability. In β decay, A remains the same while Z either increases by 1 (β − decay) or decreases by 1 (β + decay). (a) The reaction is: β01

2210

2211 NeNa ++→ ⇒ +β decay

(b) The reaction is: β01

2412

2411 MgNa −+→ ⇒ −β decay

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303

10 • What is the advantage of a breeder reactor over an ordinary reactor? What are the disadvantages of a breeder reactor?

Advantages

Disadvantages

The reactor uses 238U, which, by neutron capture and subsequent decays, produces 239Pu. Thus plutonium isotope fissions by fast neutron capture. Thus, the breeder reactor uses the plentiful uranium isotope and does not need a moderator to slow the neutrons needed for fission.

The fraction of delayed neutrons emitted in the fission of 239Pu is very small. Consequently, control of the fission reaction is very difficult, and the safety hazards are more severe than for the ordinary reactor that uses 235U as fuel.

11 • True or false: (a) In a breeder reactor, fuel can be produced as fast as it is consumed. (b) The atomic nucleus is composed of protons, neutrons, and electrons. (c) The mass of a 2H nucleus is less than the mass of a 1H nucleus plus the mass

of a neutron. (d) After two half-lives, all the radioactive nuclei in a given sample have

decayed. (a) True (given an unlimited supply of 238U). (b) False. The nucleus does not contain electrons. (c) True. (d) False. After two half-lives, three-fourths of the radioactive nuclei in a given sample have decayed. 12 • Why is it that extreme changes in the temperature or the pressure of a radioactive sample have little or no effect on the radioactivity? Determine the Concept Exponential time dependence is characteristic of all radioactivity and indicates that radioactive decay is a statistical process. Because each nucleus is well shielded from others by the atomic electrons, pressure and temperature changes have little or no effect on the rate of radioactive decay or other nuclear properties.

Chapter 40

304

Estimation and Approximation 13 • We found in Chapter 25 that the ratio of the resistivity of the most insulating material to the resistivity of the least resistive material (excluding superconductors) is approximately 1022. Few properties of materials show such a wide range of values. Using information in the textbook or other resources, find the ratio of largest to smallest for some nuclear properties of matter. Some examples might be the range of mass densities found in an atom, the half-life of radioactive nuclei, or the range of nuclear masses. Picture the Problem There is no table of half lives in the text although the information is mentioned in the alpha particle discussion for alpha decay (about 15 orders of magnitude). Mass density in an atom ranges roughly as the cube of the radius of an atom to that of the nucleus, also about 15 orders of magnitude. Nuclear masses only range 2 orders of magnitude.

Material property

Ratio (order of magnitude)

Mass density

1015

Half life

1015

Nuclear masses 2 14 •• According to the United States Department of Energy, the U.S. population consumes approximately 1020 joules of energy each year. Estimate the mass (in kilograms) of (a) uranium that would be needed to produce this much energy using nuclear fission and (b) deuterium and tritium that would be needed to produce this much energy using nuclear fusion. Picture the Problem The mass of 235U required is given by ( ) ,235A235 MNNm = where M235 is the molecular mass of 235U and N is the number of fissions required to produce 1020 J. The mass of deuterium and tritium required can be found similarly. (a) Relate the mass of 235U required to the number of fissions N required: 235

A235 M

NNm =

where M235 is the molecular mass of 235U.

Because fission per

annual

EEN = :

fussionper A

235annual235 EN

MEm =

Nuclear Physics

305

Substitute numerical values and evaluate m235:

( )( ) kg105

eVJ101.602MeV200

molnuclei10022.6

g/mol235J101 619

23

20

235 ×≈

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××⎟

⎠⎞

⎜⎝⎛ ×

×=

−m

(b) Relate the mass of 2H and 3H required to the number of fusions N required:

HHA

HH 3232 ++= M

NNm

where HH 32 +

M is the molecular mass of 2H + 3H.

Because fission per

annual

EEN = :

fission perA

HHannualHH

32

32 ENME

m ++

=

Substitute numerical values and evaluate

HH 32 +m :

( )( ) kg103

eVJ101.602MeV18

molnuclei10022.6

g/mol5J101 619

23

20

HH 32 ×≈

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××⎟

⎠⎞

⎜⎝⎛ ×

×=

−+m

Properties of Nuclei 15 • [SSM] Calculate the binding energy and the binding energy per nucleon from the masses given in Table 40-1 for (a) 12C, (b) 56Fe, and (c) 238U. Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 12C, Z = 6 and N = 6. Add the mass of the neutrons to that of the protons:

u940098.12u6651.0086u825007.1666 np =×+×=+ mm

Subtract the mass of 12C from this result:

( ) u9400.098u12u940098.1266

Cnp 12 =−=−+ mmm

Chapter 40

306

Multiply the mass difference by c2 and convert to MeV:

( ) ( ) MeV2.92uMeV/5.931u940098.0

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV68.712MeV2.92b ==

AE

(b) For 56Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the protons:

u400463.56u6651.00803u825007.1263026 np =×+×=+ mm

Subtract the mass of 56Fe from this result:

( ) u4580.528u942934.55u400463.563026

Cnp 12 =−=−+ mmm


( ) ( ) MeV492uMeV/5.931u4580.528

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV79.856MeV492b ==

AE

(c) For 238U, Z = 92 and N = 146. Add the mass of the neutrons to that of the protons:

u990984.239u6651.008146u825007.19214692 np =×+×=+ mm

Subtract the mass of 238U from this result:

( ) u207934.1u783050.238u990984.23914692

Unp 238 =−=−+ mmm


( ) ( ) MeV1802uMeV/5.931u207934.1

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV57.7238

MeV1802b ==AE

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307

16 • Calculate the binding energy and the binding energy per nucleon from the masses given in Table 40-1 for (a) 6Li, (b) 39K, and (c) 208Pb. Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 6Li, Z = 3 and N = 3. Add the mass of the neutrons to that of the protons:

u470049.6u6651.0083u825007.1333 np =×+×=+ mm

Subtract the mass of 6Li from this result:

( ) u3480.034u122015.6u470049.633

Lnp 6 =−=−+i

mmm


( ) ( ) MeV0.32uMeV/5.931u3480.034

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV33.56MeV0.32b ==

AE

(b) For 39K, Z = 19 and N = 20. Add the mass of the neutrons to that of the protons:

u975321.39u6651.00802u825007.1192019 np =×+×=+ mm

Subtract the mass of 39K from this result:

( ) u2680.358u707963.38u975321.392019

Knp 39 =−=−+ mmm


( ) ( ) MeV334uMeV/5.931u2680.358

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV56.839MeV334b ==

AE

Chapter 40

308

(c) For 208Pb, Z = 82 and N = 126. Add the mass of the neutrons to that of the protons:

u440733.209u6651.008126u825007.18212682 np =×+×=+ mm

Subtract the mass of 208Pb from this result:

( ) u804756.1u636976.207u440733.20912682

Pbnp 208 =−=−+ mmm


( ) ( ) MeV1636uMeV/5.931u804756.1

222

b =⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ=

cccmE

and the binding energy per nucleon is MeV87.7208

MeV1636b ==AE

17 • Use the radius formula R = R0A 1/ 3 (Equation 40-1), where R0 = 1.2 fm, to compute the radii of the following nuclei: (a) 16O, (b) 56Fe, and (c) 197Au. Picture the Problem The nuclear radius is given by 31

0 ARR = where R0 = 1.2 fm. (a) The radius of 16O is: ( )( ) fm0.316fm2.1 31

O16 ==R

(b) The radius of 56Fe is: ( )( ) fm6.456fm2.1 31Fe56 ==R

(c) The radius of 197Au is: ( )( ) fm0.7197fm2.1 31

Au197 ==R 18 • During a fission process, a 239Pu nucleus splits into two nuclei whose mass number ratio is 3 to 1. Calculate the radii of the nuclei formed during this process. Picture the Problem The nuclear radius is given by 31

0 ARR = where R0 = 1.2 fm. The radii of the daughter nuclei are given by:

310 ARR = (1)

Because the ratio of the mass numbers of the daughter nuclei is 3 to 1:

(239)43

1 =A and A2 =14

(239)

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309

Substitute in equation (1) to obtain: ( ) fm8.6

42393fm1.2

1/3

1 =⎟⎠⎞

⎜⎝⎛ ×

=R

and

( ) fm7.44

239fm1.21/3

2 =⎟⎠⎞

⎜⎝⎛=R

19 •• [SSM] The neutron, when isolated from an atomic nucleus, decays into a proton, an electron, and an antineutrino as follows: n → 1H +e− +ν . The thermal energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In both joules and electron volts, calculate the energy of a thermal neutron at 25ºC. (b) What is the speed of this thermal neutron? (c) A beam of monoenergetic thermal neutrons is produced at 25ºC and has an intensity I. After traveling 1350 km, the beam has an intensity of 1

2 I. Using this information, estimate the half-life of the neutron. Express your answer in minutes. Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from I to I/2, is the half-life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed. (a) The thermal energy of the neutron is: ( )( )

meV7.25

J101.602eV1J1011.4

J1011.4

K27325J/K10381.1

1921

21

23

thermal

=

×××=

×=

+×=

=

−−

−

−

kTE

(b) Equate thermalE and the kinetic energy of the neutron to obtain:

2n2

1thermal vmE = ⇒

n

thermal2m

Ev =

Substitute numerical values and evaluate v:

( ) km/s22.2kg101.673J104.112

27

21

=××

= −

−

v

(c) Relate the half-life, ,21t to the speed of the neutrons in the beam:

vxt =21

Chapter 40

310

Substitute numerical values and evaluate :21t

min1.10

s60min1s086

km/s2.22km 1350

1/2

=

×==t

20 • Use the radius formula R = R0A 1/ 3 (Equation 40-1), where R0 = 1.2 fm, for the radius of a spherical nucleus to calculate the density of nuclear matter. Express your answer in grams per cubic centimeter. Picture the Problem We can use the definition of density, the equation for the volume of a sphere, and the given approximation to calculate the density of nuclear matter in grams per cubic centimeter. Express the density of a spherical nucleus:

Vm

=ρ

The approximate mass is:

( )Am kg1066.1 27−×=

The volume of the nucleus is given by:

( ) ARARV 303

433103

4 ππ ==

Substitute for m and V to obtain: ( )

( )30

27

303

4

27

4kg1066.13

kg1066.1

R

ARA

π

πρ

−

−

×=

×=

Substitute numerical values and evaluate ρ:

( )( )

314

317

3

27

g/cm103.2

kg/m1029.2fm2.14

kg1066.13

×=

×=

×=

−

πρ

21 •• [SSM] In 1920, 12 years before the discovery of the neutron, Ernest Rutherford argued that proton–electron pairs might exist in the confines of the nucleus in order to explain the mass number, A, being greater than the nuclear charge, Z. He also used this argument to account for the source of beta particles in radioactive decay. Rutherford’s scattering experiments in 1910 showed that the nucleus had a diameter of approximately 10 fm. Using this nuclear diameter, the uncertainty principle, and that beta particles have an energy range of 0.02 MeV to 3.40 MeV, show why the hypothetical electrons cannot be confined to a region occupied by the nucleus.

Nuclear Physics

311

Picture the Problem The Heisenberg uncertainty principle relates the uncertainty

in position, Δx, to the uncertainty in momentum, Δp, by ΔxΔp ≥ 1

2h.

Solve the Heisenberg equation for Δp:

xp

Δ≈Δ

2h

Assuming that electrons are contained within the nucleus, the uncertainty in their momenta must be:

( )m/skg1025.5

m10102sJ1005.1

21

15

34

⋅×=

×⋅×

≈Δ

−

−

−

p

The kinetic energy of the electron is given by:

pcK =

Substitute numerical values and evaluate K for a nuclear electron:

( )( )MeV10

J101.602eV1J1058.1m/s10998.2m/skg1025.5 19

12821

≈×

××=×⋅×= −−−K

This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of 1 to 1000 eV. Hence the premise on which it is based (that electrons are contained within the nucleus) must be false. 22 •• Consider the following fission process:

92236 U +n → 37

95 Rb + 55137 Cs + 4n . Determine the electrostatic potential energy, in

MeV, of the reaction products when the surfaces of the 95Rb nucleus and the 137Cs nucleus are just touching immediately after being formed during the fission process. Picture the Problem The separation of the nuclei when they are just touching is the sum of their radii. The radius of a nucleus whose atomic number is A is given by 31

0 ARR = , where R0 = 1.2 fm. The electrostatic potential energy of the system is given by:

( )( )CsRb

CsRb21

RReZeZ

kRqq

kU+

==

where R is the distance from the center of the 95Rb nucleus to the center of the 137Cs nucleus.

The radii of the nuclei are: 1/3RboRb ARR = and 1/3

CsoCs ARR =

Chapter 40

312

Substitute for RRb and RCs and simplify to obtain:

( )( )

( )( )( )1/3

Cs1/3Rb

CsRb

o

2

1/3Cso

1/3Rbo

CsRb2

AAZZ

Rke

ARARZZ

keU

+=

+=

Substitute numerical values and evaluate U:

( )( )( )

( )( )( )GeV25.0

J101.602eV1J10026.4

371595537

m101.2C101.602/CNm108.988

1911

1/31/315

219229

=×

××=

+×××

=

−−

−

−

U

Radioactivity 23 • Homer enters the visitors’ chambers, and his Geiger beeper sounds. He shuts off the beeper, removes the device from his shoulder patch, and holds it near the only new object in the room—an orb that is to be presented as a gift from the visiting Cartesians. Pushing a button marked ″monitor, ″ Homer reads that the device is reading a counting rate of 4000 counts/s above the background counting rate. After 10 min, the counting rate has dropped to 1000 counts/s above the background rate. (a) What is the half-life of the source? (b) How high will the counting rate be counting above the background counting rate 20 min after the monitoring device was switched on? Picture the Problem The counting rate, as a function of the number of half-lives n, is given by ( ) 02

1 RR n= , where R0 = 4000 counts/s. (a) The counting rate after n half-lives is:

( ) 021 RR n= ⇒

( )( )2

10

lnln RRn =

Substitute numerical values and evaluate n: ( ) 2

lncounts/s4000counts/s1000ln

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛

=n

Because there are two half-lives in 10 min:

min521 =t

(b) At the end of 4 half-lives: ( ) ( ) Bq250counts/s40004

21 ==R

24 • A certain source gives 2000 counts/s at time t = 0. Its half-life is 2.0 min. How many counts per second will it give after (a) 4.0 min, (b) 6.0 min, and (c) 8.0 min?

Nuclear Physics

313

Picture the Problem The counting rate, as a function of the number of half-lives n, is given by ( ) .02

1 RR n= (a) When t = 4.0 min, two half-lives will have passed and n = 2:

( ) ( ) Bq500counts/s2000221 ==R

(b) When t = 6.0 min, three half-lives will have passed and n = 3:

( ) ( ) Bq250counts/s2000321 ==R

(c) When t = 8.0 min, four half-lives will have passed and n = 4:

( ) ( ) Bq125counts/s2000421 ==R

25 • The counting rate from a radioactive source is 8000 counts/s at time t = 0, and 10 min later the rate is 1000 counts/s. (a) What is the half-life? (b) What is the decay constant? (c) What is the counting rate after 20 min? Picture the Problem The counting rate, as a function of the number of half-lives n, is given by ( ) 02

1 RR n= and the decay constant λ is related to the half-life by ( ) .2ln21 λ=t

(a) Relate the counting rate at time t = 10 min to the counting rate at t = 0:

( ) 021

min10 RR n= ⇒( )

( )21

0min 10

lnln RRn =

Substitute numerical values and evaluate n: ( ) 3

lncounts/s8000counts/s1000ln

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛

=n

Therefore, 3 half-lives have passed in 10 min:

min103 21 =t ⇒ s20021 =t

(b) The decay constant λ is related to the half-life by:

( )λ2ln

21 =t ⇒( )21

2lnt

=λ

Substitute for t1/2 and evaluate λ: ( ) 13 s105.3s200

2ln −−×==λ

(c) Six half-lives will have passed in 20 min:

( ) ( ) Bq125Bq8000621

min20 ==R

Chapter 40

314

26 • The half-life of radium is 1620 y. Calculate the number of disintegrations per second of 1.00 g of radium and show that the disintegration rate is approximately 1.0 Ci. Picture the Problem We can use teNR λλ −= 0 to show that the disintegration rate is approximately 1.0 Ci. The decay rate is given by:

teNR λλ −= 0 where N0 is the number of nuclei at t = 0.

The decay constant λ is related to the half-life by:

( )λ2ln

21 =t ⇒( )21

2lnt

=λ

Substitute for t1/2 and evaluate λ:

( )

111

14

s10356.1Ms31.56

y 1y1028.4y1620

2ln

−−

−−

×=

××==λ

The number of radioactive nuclei N 0 at t = 0 can be found from Avogadro’s number NA, the mass mRa of the sample, and the molar mass M of the sample:

Mm

NN Ra

A

0 = ⇒M

mNN RaA0 =

Substitute numerical values and evaluate N0:

( ) nuclei 10664.2mol/g226g 00.1nuclei/mol10022.6 2123

0 ×=⎟⎟⎠

⎞⎜⎜⎝

⎛×=N

Substitute numerical values for λ and N0 and evaluate R:

( )( ) ( ) ( )

Ci0.1s107.3

s1061.310664.2s10356.1110

1100s10356.121111 111

=×≈

×=××=−

−×−−− −−

eR

27 • A radioactive piece of silver foil (t1/2 = 2.4 min) is placed near a Geiger counter and 1000 counts/s are observed at time t = 0. (a) What is the counting rate at t = 2.4 min and at t = 4.8 min? (b) If the counting efficiency is 20 percent, how many radioactive silver nuclei are there at time t = 0? At time t = 2.4 min? (c) At what time will the counting rate be about 30 counts/s?

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315

Picture the Problem We can use ( ) 021 RR n= to relate the counting rate R to the

number of half-lives n that have passed since t = 0. The detection efficiency depends on the probability that a radioactive decay particle will enter the detector and the probability that upon entering the detector it will produce a count. If the efficiency is 20 percent, the decay rate must be 5 times the counting rate. (a) When t = 2.4 min, n = 1 and:

( ) ( ) Bq500counts/s1000121

min4.2 ==R

When t = 4.8 min, n = 2 and:

( ) ( ) Bq250counts/s1000221

min8.4 ==R

(b) The number of radioactive nuclei is related to the decay rate R, and the decay constant λ:

NR λ= ⇒ λRN = (1)

The decay constant is related to the half-life: 13

21

s10813.4

s144693.0

min4.2693.0693.0

−−×=

===t

λ

Calculate the decay rate at t = 0 from the counting rate:

10 s5000counts/s10005 −=×=R

Substitute in equation (1) and evaluate N0 at t = 0:

613

10

0 100.1s104.813

s5000×=

×== −−

−

λRN

Calculate the decay rate at t = 2.4 min from the counting rate:

1min4.2 s0052counts/s5005 −=×=R

Substitute in equation (1) and evaluate N2.4 min: 5

13

1min4.2

min4.2

102.5

s104.813s2500

×=

×== −−

−

λR

N

(c) The time at which the counting rate will be about 30 counts/s is the product of the number of half-lives that will have passed and the half-life:

21ntt = (2)

Chapter 40

316

The counting rate R after n half-lives is related to the counting rate at t = 0 by:

( ) 021 RR n= ⇒

( )( )2

10

lnln RRn =

Substitute numerical values and evaluate n:

( )( )

059.5ln

counts/s1000counts/s30ln

21

=

=n

Substitute numerical values for n and t1/2 in equation (2) and evaluate t:

( )( ) min12min4.2059.5 ==t

28 • Use Table 40-1 to calculate the energy release, in MeV, for the α decay of (a) 226Ra and (b) 242Pu. Picture the Problem Knowing each of these reactions, we can use Table 40-1 to find the differences in the masses of the nuclei and then convert this difference into the energy released in each reaction. (a) Write the reaction: HeRnRa 4222226 +→ Use Table 40-1 to find ΔE:

( )

MeV87.4

uMeV/5.931u603002.4u571017.222u403025.226

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=Δ

ccE

(b) Write the reaction:

HeUPu 4238242 +→

Use Table 40-1 to find ΔE:

( )

MeV98.4

uMeV/5.931u603002.4u783050.238u737058.242

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=Δ

ccE

29 •• [SSM] Plutonium is very toxic to the human body. Once it enters the body it collects primarily in the bones, although it also can be found in other organs. Red blood cells are synthesized within the marrow of the bones. The isotope 239Pu is an alpha emitter that has a half-life of 24 360 years. Because alpha particles are an ionizing radiation, the blood-making ability of the marrow is, in time, destroyed by the presence of 239Pu. In addition, many kinds of cancers will

Nuclear Physics

317

also develop in the surrounding tissues because of the ionizing effects of the alpha particles. (a) If a person accidentally ingested 2.0 μg of 239Pu and all of it is absorbed by the bones of the person, how many alpha particles are produced per second within the body of the person? (b) When, in years, will the activity be 1000 alpha particles per second? Picture the Problem Each 239Pu nucleus emits an alpha particle whose activity, A, depends on the decay constant of 239Pu and on the number N of nuclei present in the ingested 239Pu. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of 239Pu. Finally, we can use the dependence of the activity on time to find the time at which the activity will be 1000 alpha particles per second. (a) The activity of the nuclei present in the ingested 239Pu is given by:

NR λ= (1)

Find the constant for the decay of 239Pu:

( )( )( )

1132/1

s10016.9

Ms/y56.31y24360693.02ln

−−×=

==t

λ

Express the number of nuclei N present in the mass mPu of 239Pu ingested: Pu

Pu

A Mm

NN

= ⇒Pu

APu M

NmN =

where MPu is the atomic mass of 239Pu.

Substitute numerical values and evaluate N:

( )

nuclei10039.5

g/mol239nuclei/mol10022.6g0.2

15

23

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×= μN

Substitute numerical values in equation (1) and evaluate A:

( )( )/s105.4/s10543.4

10039.5s10016.933

15113

αα

α

×=×=

××= −−R

(b) The activity varies with time according to: teRR λ−= 0 ⇒

λ−

⎟⎟⎠

⎞⎜⎜⎝

⎛

= 0

lnRR

t

Substitute numerical values and evaluate t:

( )

y103.5

yMs56.31s10016.9

s/10543.4s/1000ln

4

113

3

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×−

⎟⎠⎞

⎜⎝⎛

×=−−

αα

t

Chapter 40

318

30 •• Consider an alpha-emitting parent nucleus ZA X initially at rest. The

nucleus decays into a daughter nucleus Y and an alpha particle as follows: ZA X→ Z−2

A−4 Y+ 24α +Q . (a) Show that the alpha particle has a kinetic energy of

(A – 4)Q/A. (b) Show that the kinetic energy of the recoiling daughter nucleus is given by KY = 4Q/A. Picture the Problem We can use conservation of energy and conservation of linear momentum to relate the momenta and kinetic energies of the nuclei to the decay’s Q value. (a) Express the kinetic energies of the alpha particle and daughter nucleus:

α

αααα m

pvmK2

22

21 == (1)

and

Y

2Y2

YY21

Y 2mpvmK == (2)

Solve equations (1) and (2) for 2

αp and :2

Dp

ααα Kmp 22 = and YY2Y 2 Kmp =

From the conservation of linear momentum we have:

fi pp rr=

or, because the parent is initially at rest, Y0 pp −= α and Ypp =α .

Because the momenta are equal: αα KmKm 22 YY =

Solving for KY and simplifying gives: αα

α KA

KmmK

44

YY −

==

Because the daughter nucleus and the alpha particle share the Q-value:

α

ααα

α

KA

A

KA

KKA

KKQ

⎟⎠⎞

⎜⎝⎛

−=

⎟⎠⎞

⎜⎝⎛ +

−=+

−=

+=

4

14

44

4Y

Solve for Kα to obtain:

QA

AK ⎟⎠⎞

⎜⎝⎛ −

=4

α

(b) Substituting for Kα in the expression for Q yields:

QA

AKQ ⎟⎠⎞

⎜⎝⎛ −

+=4

Y

Nuclear Physics

319

Solving for KY gives: AQQ

AAQK 44

Y =⎟⎠⎞

⎜⎝⎛ −

−=

31 • [SSM] The fissile material 239Pu is an alpha emitter. Write the reaction that describes 239Pu undergoing alpha decay. Given that 239Pu, 235U, and an alpha particle have respective masses of 239.052 156 u, 235.043 923 u, and 4.002 603 u, use the relations appearing in Problem 30 to calculate the kinetic energies of the alpha particle and the recoiling daughter nucleus. Picture the Problem We can write the equation of the decay process by using the fact that the post-decay sum of the Z and A numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations from Problem 30 is given by Q = −(Δm)c2. 239Pu undergoes alpha decay according to:

Q++→ α4223592

23994 UPu

The Q value, in MeV, for the decay is given by:

( ) ( )[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

uMeV/931.5 2

2UPu

ccmmmQ α

Substitute numerical values and evaluate Q:

( ) ( )[ ]

MeV24.5

u931.5MeV/u6034.002u923235.043u156239.052

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

ccQ

From Problem 30, the kinetic energy of the alpha particle is given by:

QA

AK ⎟⎠⎞

⎜⎝⎛ −

=4

α

Substitute numerical values and evaluate Kα:

( )

MeV5.15

MeV5.24239

4239

=

⎟⎠⎞

⎜⎝⎛ −

=αK

From Problem 30, the kinetic energy of 235U is given by:

AQK 4

U =

Substitute numerical values and evaluate

U235K : ( ) keV2.89

235MeV24.54

U235 ==K

Chapter 40

320

32 • Through a friend in the security department at the museum, Angela obtains a sample of a wooden tool handle that contains 175 g of carbon. The decay rate of the 14C in the sample is 8.1 Bq. How long ago was the wood in the handle last alive? Picture the Problem We can find the age of the sample using ( ) 02

1 RR nn = to find

n and then applying .21ntt =

Express the age of the bone in terms of the half-life of 14C and the number n of half-lives that have elapsed:

21ntt = (1)

The decay rate Rn after n half-lives is related to the counting rate R0 at t = 0 by:

( ) 021 RR n

n = ⇒ ( )( )2

10

lnln RRn =

Because there are 15.0 decays per minute per gram of carbon in a living organism:

Bq75.43

g175s60

min1gmin

decays15.00

=

××⋅

=R

Substitute numerical values for R and R0 and evaluate n:

( ) 433.2ln

Bq43.75Bq1.8ln

21

=⎟⎟⎠

⎞⎜⎜⎝

⎛

=n

Substitute numerical values in equation (1) and evaluate t:

( )( ) y000,14y5730433.2 ≈=t

33 • A sample of a radioactive isotope is found to have an activity of 115.0 Bq immediately after it is pulled from the reactor that formed the isotope. Its activity 2 h 15 min later is measured to be 85.2 Bq. (a) Calculate the decay constant and the half-life of the sample. (b) How many radioactive nuclei were there in the sample initially? Picture the Problem We can solve teRR λ−= 0 for λ to find the decay constant of

the sample and use λ

2ln21 =t to find its half-life. The number of radioactive nuclei

in the sample initially can be found from 00 NR λ= . (a) The decay rate is given by:

teRR λ−= 0 ⇒tRR

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

= 0

lnλ

Nuclear Physics

321

Substitute numerical values and evaluate λ:

1h133.0h25.2Bq0.115Bq2.85ln

−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=λ

The half-life is related to the decay constant:

( ) ( ) h20.5h133.0

2ln2ln121 === −λ

t

(b) The number N0 of radioactive nuclei in the sample initially is related to the decay constant λ and the initial decay rate R0:

00 NR λ= ⇒ λ

00

RN =

Substitute numerical values and evaluate N0:

6

10 1011.3

s3600h1h133.0

Bq115×=

×=

−N

34 •• A 1.00-mg sample of substance that has an atomic mass of 59.934 u and emits β particles has an activity of 1.131 Ci. Find the decay constant for this substance in reciprocal seconds and find the half-life in years. Picture the Problem The decay constant λ can be found from the decay rate R and the number of radioactive nuclei N at the moment of interest and the half-life, in turn, can be found from the decay constant. The decay rate R is related to the decay constant λ and the number of radioactive nuclei N at the moment of interest:

NR λ= ⇒ NR

=λ (1)

The number of radioactive nuclei N at the moment of interest can be found from Avogadro’s number, the mass m of the sample, and the molar mass M of the sample:

Mm

NN

=A

⇒MmNN A=


( ) nuclei 10004.1mol/g59.934

mg1.00nuclei/mol10022.6 1923 ×=×=N

Chapter 40

322

Substitute numerical values in equation (1) and evaluate λ:

19

19

10

s1017.4nuclei 10004.1Ci

Bq103.7Ci131.1

−−×=

×

××

=λ

Find the half-life from the decay constant:

( ) ( )

y27.5

Ms31.56y1s1067.1

s1017.42ln2ln

8

1921

=

××=

×== −−λ

t

35 •• [SSM] Radiation has been used for a long time in medical therapy to control the development and growth of cancer cells. Cobalt-60, a gamma emitter that emits photons that have energies of 1.17 MeV and 1.33 MeV, is used to irradiate and destroy deep-rooted cancers. Small needles made of 60Co of a specified activity are encased in gold and used as body implants in tumors for time periods that are related to tumor size, tumor cell reproductive rate, and the activity of the needle. (a) A 1.00 μg sample of 60Co, that has a half-life of 5.27 y, that is used to irradiate a small internal tumor with gamma rays, is prepared in the cyclotron of a medical center. Determine the activity of the sample in curies. (b) What will the activity of the sample be 1.75 y from now? Picture the Problem We can use NR λ=0 to find the initial activity of the sample and teRR λ−= 0 to find the activity of the sample after 1.75 y. (a) The initial activity of the sample is the product of the decay constant λ for 60Co and the number of atoms N of 60Co initially present in the sample:

NR λ=0

Express N in terms of the mass m of the sample, the molar mass M of 60Co, and Avogadro’s number NA:

Mm

NN

=A

⇒ ANMmN =

Substituting for N yields: M

mNR A0

λ=

The decay constant is given by:

( )1/2

2lnt

=λ

Nuclear Physics

323

Substitute for λ to obtain:

( )MtmNR

1/2

A0

2ln=

Substitute numerical values and evaluate R0:

( )( )

mCi 13.1

s 103.7Ci 1s 10183.4

molg 60

yMs 31.56y 27.5

molnuclei106.022g101.002ln

11017

236

0

=

×××=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎠⎞

⎜⎝⎛ ××

= −−

−

R

(b) The activity varies with time according to:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

− == 5.27y0.693

00

tt eReRR λ

Evaluate R at t = 1.75 y: ( )

mCi898.0

mCi 1.13 5.27y1.75y0.693

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−

eR

36 •• (a) Show that if the decay rate is R0 at time t = 0 and R1 at some later time t = t1, the decay constant is given by ( )10

11 ln RRt −=λ and the half-life is

given by t1/2 = t1 ln(2)/ln(R0/R1). (b) Use these results to find the decay constant and the half-life if the decay rate is 1200 Bq at t = 0 and 800 Bq at t1 = 60.0 s. Picture the Problem We can solve Equation 40-7 for λ to show that λ = t1

−1ln(R0/R1). (a) The half-life as a function of the decay constant λ is given by:

( )λ2ln

21 =t (1)

From Equation 40-7 we have: 1

01teRR λ−= ⇒ ( )10

11 ln RRt −=λ

Substituting for λ in equation (1) and simplifying gives:

( )( )

( )( )10

1

101

121 ln

2lnln

2lnRR

tRRt

t == −

(b) Substitute numerical values for t, R1, and R0 and evaluate λ:

1s00676.0Bq800Bq1200ln

s0.601 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛=λ

Use the decay constant to find the half-life:

( ) ( ) s103s00676.0

2ln2ln121 === −λ

t

Chapter 40

324

39 •• A wooden casket is thought to be 18 000 years old. How much carbon would have to be recovered from this object to yield a 14C counting rate of no less than 5 counts/min with a detection efficiency of 20 percent? Picture the Problem Because the detection efficiency is 20 percent, the decay rate must be five times the counting rate. The required mass is given by M = (5 counts/min)/R, where R is the current counting rate per gram of carbon. We can use the assumed age of the casket to find the number of half-lives that have elapsed and ( ) 02

1 RR n= to find the current counting rate per gram of 14C. The mass of carbon required is:

( )R

M counts/min55= (1)

Because there were about 15.0 decays per minute per gram of the living wood, the counting rate per gram is:

( ) ( ) ( )gcounts/min1521

021 ⋅== nn RR

We can find n from the assumed age of the casket and the half-life of 14C:

141.3y5730y000,18==n

Substitute for n and evaluate R:

( ) ( )gcounts/min70.1

gcounts/min15141.321

⋅=

⋅=R

Substitute for R in equation (1) and evaluate M:

( )( ) g15

gcounts/min70.1counts/min55

≈⋅

=M

38 •• A sample of radioactive material is initially found to have an activity of 115.0 decays/min. After 4 d 5 h, its activity is measured to be 73.5 decays/min. (a) Calculate the half-life of the material. (b) How long (from the initial time) will it take for the sample to reach an activity of 10.0 decays/min? Picture the Problem We can use the decay rate equation teRR λ−= 0 and the expression relating the half-life of a source to its decay constant to find the half-life of the sample. Solving the decay-rate equation for t will yield the time at which the activity level drops to any given value. (a) The half-life of the material is given by:

( )λ2ln

21 =t (1)

The decay rate is given by: teRR λ−= 0 ⇒ ( )t

RR0ln=λ (2)

Nuclear Physics

325

Substitute for λ in equation (1) and simplify to obtain:

( )( )

( )( ) tRR

tRR

t00

21 ln2ln

ln2ln

==

Substitute numerical values and evaluate t1/2:

( ) ( )

h156

h101

decays/min5.73decays/min115ln

2ln21

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=t

(b) Solving equation (2) for t yields:

( )λ−

= 0ln RRt

Express λ in terms of t1/2:

( )21

2lnt

=λ

Substitute for λ in the expression for t to obtain:

( )( ) 21

0

2lnln tRRt−

=

Substitute numerical values and evaluate t:

( ) ( )

d9.22h24

d1h550

h1562ln

decays/min115decays/min10ln

=×=

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=t

39 •• The rubidium isotope 87Rb is a β– emitter that has a half-life of 4.9 × 1010 y. It decays into 87Sr. This nuclear decay is used to determine the age of rocks and fossils. Rocks containing the fossils of early animals have a ratio of 87Sr to 87Rb of 0.0100. Assuming that there was no 87Sr present when the rocks were formed, calculate the age of these fossils. Picture the Problem We can use the decay rate equation teRR λ−= 0 and the expression relating the half-life of a source to its decay constant to find the age of the fossils. The decay rate is given by: teRR λ−= 0 ⇒

( )λ−

= 0ln RRt

Express λ in terms of t1/2:

( )21

2lnt

=λ

Chapter 40

326

Substitute for λ in the expression for t to obtain:

( )( ) 21

0

2lnln tRRt−

=

or, because the activity at any time is proportional to the number of radioactive nuclei present,

( )( ) 21

Rb0,Rb

2lnln

tNN

t−

= (1)

The number of 87Sr nuclei present in the rocks is given by:

RbRb,0Sr NNN −= ⇒ RbSrRb,0 NNN +=

We’re given that: RbSr 01.0 NN = ⇒ 01.0

Rb

Sr =NN

Express the ratio of N0,Rb to NRb:

01.1101.0

1Rb

Sr

Rb

RbSr

Rb

Rb,0

=+=

+=+

=NN

NNN

NN

Substitute numerical values in equation (1) and evaluate the age of the fossils: ( ) ( ) y100.7y109.4

2ln01.11ln

810 ×≈×−

⎟⎠⎞

⎜⎝⎛

=t

40 ••• Consider a single nucleus of a radioactive isotope that has a decay rate equal to λ. The nucleus has not decayed at t = 0. The probability that the nucleus will decay between time t and time t + dt is equal to λe−λtdt . (a) Show that this statement is consistent with the fact that the probability is 1 that the nucleus will decay between t = 0 and t = ∞. (b) Show that the expected lifetime of the nucleus

is equal to 1/λ. Hint: The expected lifetime is equal to tλe−λtdt0

∞

∫ divided by

λe−λtdt

0

∞

∫ . (c) A sample of material contains a significant number of these

radioactive nuclei at time t = 0. What will have been the mean lifetime of these nuclei some time later, after they have all decayed? Picture the Problem (a) Integrating the probability function between 0 and ∞ will show that its value is 1. (b) We can evaluate this integral by changing variables to obtain a form that we can find in a table of integrals. (a) The probability that the nucleus will decay between t = 0 and t = ∞ is given by:

∫∫∞

−∞

− ==00

dtedteP tt λλ λλ

Nuclear Physics

327

Integrating yields: ] 10

0

=−=⎥⎦

⎤−

=∞−

∞−t

t

eeP λλ

λλ

(b) The expected lifetime of the nucleus is given by:

∫

∫∞

−

∞−

=

0

0

dte

dtet

t

t

λ

λ

λ

λτ

or, because 10

=∫∞

− dte tλλ ,

∫∞

−=0

dtet tλλτ

Change variables by letting tx λ= . Then:

dtdx λ= , λdxdt = , and

λxt =

Substitute to obtain:

∫∫∫∞

−∞

−∞

− ===000

1 dxxedxexdtet xxt

λλλ

λλτ λ

From integral tables: 1

0

=∫∞

− dxxe x

Substitute in the expression for τ to obtain:

λτ 1=

(c) The mean lifetime after all the nuclei have decayed is λ1 , the same as the

expected lifetime. Nuclear Reactions 41 • Using Table 40-1, find the Q values for the following reactions: (a) 1H + 3H → 3He + n + Q and (b) 2H + 2H → 3He + n + Q. Picture the Problem We can use ( ) 2cmQ Δ−= to find the Q values for these reactions.

Chapter 40

328

(a) Find the mass of each atom from Table 40-1:

u8251.007H1 =m

u049016.3H3 =m

u029016.3He3 =m

u665008.1n =m

Calculate the initial mass mi of the incoming particles:

u874023.4u049016.3u8251.007i

=+=m

Calculate the final mass mf: u694024.4

u665008.1u029016.3f

=+=m

Calculate the increase in mass:

u820000.0u874023.4u694024.4

if

=−=

−=Δ mmm

Calculate the Q value:

( )

( )

MeV764.0

MeV/5.931u820000.02

2

2

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Δ−=

ucc

cmQ

(b) Proceed as in (a) to obtain: ( )

MeV27.3

uMeV/5.931u510003.0 2

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛= ccQ

Remarks: Because Q < 0 for the first reaction, it is endothermic. Because Q > 0 for the second reaction, it is exothermic. 42 • Using Table 40-1, find the Q values for the following reactions: (a) 2H + 2H → 3H + 1H + Q, (b) 2H + 3He → 4He + 1H + Q, and (c) 6Li + n → 3H + 4He + Q. Picture the Problem We can use ( ) 2cmQ Δ−= to find the Q values for these reactions. (a) Find the mass of each atom from Table 40-1:

u102014.2H2 =m

u049016.3H3 =m

u825007.1H1 =m

Nuclear Physics

329

Calculate the initial mass mi of the incoming particles:

( )u204028.4

u102014.22i

==m

Calculate the final mass mf: u874023.4

u825007.1u049016.3f

=+=m


u330004.0u204028.4u874023.4

if

−=−=

−=Δ mmm


( )

( )

MeV03.4

MeV/5.931u330004.02

2

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

Δ−=

ucc

cmQ

(b) Proceed as in (a) to obtain: ( )

( )

MeV4.18

uMeV/5.931u703019.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

Δ−=

cc

cmQ

(c) Proceed as in (a) to obtain: ( )

( )

MeV78.4

uMeV/5.931u135005.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

Δ−=

cc

cmQ

43 •• [SSM] (a) Use the values 14.003 242 u and 14.003 074 u for the atomic masses of

14C and 14 N , respectively, to calculate the Q value (in MeV)

for the β-decay reaction 614C → 7

14 N + e− + ve . (b) Explain why you should not add the mass of the electron to that of atomic 14 N for the calculation in (a). Picture the Problem We can use ( ) 2cmQ Δ−= to find the Q values for this reaction. (a) The masses of the atoms are: u242003.14

C14 =m u074003.14

N14 =m

Chapter 40

330


u168000.0u242003.14u074003.14

if

−=−=

−=Δ mmm


( )

( )

MeV156.0

uMeV/5.931u168000.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

Δ−=

cc

cmQ

(b) The masses given are for atoms, not nuclei, so for nuclear masses the masses are too large by the atomic number multiplied by the mass of an electron. For the given nuclear reaction, the mass of the carbon atom is too large by e6m and the mass of the nitrogen atom is too large by e7m . Subtracting e6m from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron) is mathematically equivalent to explicitly subtracting e1m from the right side of the equation. 44 •• (a) Use the values 13.005 738 u and 13.003 354 u for the atomic masses of 7

13N and 613C , respectively, to calculate the Q value (in MeV) for the β-

decay reaction 713N → 6

13C + e+ + ve . (b) Explain why you need to add twice the mass of an electron to the mass of 6

13C during the calculation of the Q value for the reaction in (a). Picture the Problem We can use ( ) 2cmQ Δ−= to find the Q values for this reaction. (a) The masses of the atoms are: u738005.13

N13 =m

and u354003.13

C13 =m

For β+ decay: ( )

( ) 2e

2fi

2efi

2

2

cmcmm

cmmmQ

−−=

−−=

Calculate mi − mf: u384002.0

u354003.13u738005.13fi

=−=−mm

Nuclear Physics

331


( ) ( ) MeV20.1MeV511.02MeV/5.931u384002.02

2 =−⎟⎟⎠

⎞⎜⎜⎝

⎛=

uccQ

(b) The atomic masses include the masses of the electrons of the neutral atoms. In this reaction the initial atom has 7 electrons and the final atom has 6 electrons. Moreover, in addition to one electron not included in the atomic masses, a positron of mass equal to that of an electron is created. Consequently, one must add the rest energies of two electrons to the rest energy of the daughter atomic mass when calculating Q. Fission and Fusion 45 • Assuming an average energy of 200 MeV per fission, calculate the number of fissions per second needed for a 500-MW reactor. Picture the Problem The power output of the reactor is the product of the number of fissions per second and energy liberated per fission. Express the required number N of fissions per second in terms of the power output P and the energy released per fission Eper fission:

fissionper EPN =


119

198

s1056.1

MeV200J101.602

eV1sJ1000.5

MeV200MW500

−

−

×=

×××

=

=N

46 • If the reproduction factor in a reactor is 1.1, find the number of generations needed for the power level to (a) double, (b) increase by a factor of 10, and (c) increase by a factor of 100. Find the time needed in each case if (d) there are no delayed neutrons, so that the time between generations is 1.0 ms, and (e) there are delayed neutrons that make the average time between generations 100 ms.

Chapter 40

332

Picture the Problem If k = 1.1, the reaction rate after N generations is 1.1N. We can find the number of generations by setting 1.1N equal, in turn, to 2, 10, and 100 and solving for N. The time to increase by a given factor is the number of generations N needed to increase by that factor times the generation time. (a) Set 1.1N equal to 2 and solve for N:

( ) 21.1 =N ⇒ ( ) ( )2ln1.1ln =N

and ( )( ) 3.7

1.1ln2ln

==N

(b) Set 1.1N equal to 10 and solve for N:

( ) 101.1 =N ⇒ ( ) ( )10ln1.1ln =N

and ( )( ) 24

1.1ln10ln

==N

(c) Set 1.1N equal to 100 and solve for N:

( ) 1001.1 =N ⇒ ( ) ( )100ln1.1ln =N

and ( )( ) 48

1.1ln100ln

==N

(d) Multiply the number of generations by the generation time:

( )( ) ms3.7ms0.127.712 === Ntt

( )( ) ms24ms0.12.24110 === Ntt

( )( ) ms48ms0.13.481100 === Ntt

(e) Multiply the number of generations by the generation time:

( )( ) s73.0ms10027.712 === Ntt

( )( ) s4.2ms1002.24110 === Ntt

( )( ) s8.4ms1003.481100 === Ntt 47 •• [SSM] Consider the following fission reaction:

92235U +n→ 42

95Mo+ 57139 La +2n +Q . The masses of the neutron, 235U, 95Mo, and

139La are 1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u, respectively. Calculate the Q value, in MeV, for this fission reaction. Picture the Problem We can use Q = −(Δm)c2, where Δm = mf − mi, to calculate the Q value. The Q value, in MeV, is given by: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛Δ−=

uMeV/ 5.931 2

2 ccmQ

Nuclear Physics

333

Calculate the change in mass Δm:

( )( )

u068223.0u6651.008u923235.043

u6651.0082u348138.906u842905.94if

−=+−

++=−=Δ mmm

Substitute for Δm and evaluate Q:

( ) MeV208uMeV/931.5u068223.0

22 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

ccQ

48 •• In 1989, researchers claimed to have achieved fusion in an electrochemical cell at room temperature. Their now thoroughly discredited claim was that a power output of 4.00 W was produced by deuterium fusion reactions in the palladium electrode of their apparatus. The two most likely reactions are 2H + 2H → 3He + n + 3.27 MeV and 2H + 2H → 3H + 1H + 4.03 MeV Of the deuterium nuclei that participated in these reactions, assume half of the deuterium nuclei participated in the first reaction and the other half participated in the second reaction. How many neutrons per second would we expect to be emitted in the generation of 4.00 W of power? Picture the Problem We can find the number of neutrons per second in the generation of 4 W of power from the number of reactions per second. The number of neutrons emitted per second is:

NN 21

n = where N is the number of reactions per second.

The number of reactions per second is:

112

19

s10841.6

MeV4.03MeV27.3J10602.1

eV1sJ00.4

2

−

−

×=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+×

×=N

Chapter 40

334

Substitute for N and evaluate Nn: ( )neutrons/s1042.3

s10841.612

11221

n

×=

×= −N

49 •• A fusion reactor that uses only deuterium for fuel would have the two reactions in Problem 50 taking place in the reactor. The 3H produced in the second reaction reacts immediately with another 2H in the reaction 3H + 2H → 4He + n + 17.6 MeV The ratio of 2H to 1H atoms in naturally occurring hydrogen is 1.5 × 10–4. How much energy would be produced from 4.0 L of water if all of the 2H nuclei undergo fusion? Picture the Problem We can use the energy released in the reactions of Problem 50, together with the 17.6 MeV released in the reaction described in this problem, to find the energy released using 5 2H nuclei. Finding the number of D atoms in 4.0 L of H2O, we can then find the energy produced if all of the 2H nuclei undergo fusion. Find the energy released using 5 2H nuclei:

MeV9.24MeV17.6MeV4.03MeV27.3

=++=Q

The number of H atoms in 4.0 L of H2O is:

AOH

H g/mol182 2 N

mN ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Substitute numerical values and evaluate NH:

( ) 10676.2atoms/mol10022.6g/mol18

kg0.42 2623H ×=×⎟⎟

⎠

⎞⎜⎜⎝

⎛=N

The number of D atoms in 4.0 L of H2O is:

( )( )( )

22

264H

4D

1001.4 10676.2105.1

105.1

×=

××=

×=−

− NN

The energy produced is given by: QNE

5D=

Nuclear Physics

335

Substitute numerical values and evaluate E:

( )

J102.3eV

J101.602MeV10997.1

MeV9.245

1001.4

10

1923

22

×=

×××=

×=

−

E

50 ••• The fusion reaction between 2H and 3H is

3H + 2H → 4He + n + 17.6 MeV

Using the conservation of momentum and the given Q value, find the final energies of both the 4He nucleus and the neutron, assuming the initial kinetic energy of the system is 1.00 MeV and the initial momentum of the system is zero. Picture the Problem We can use the conservation of momentum and the given Q value to find the final energies of both the 4He nucleus and the neutron, assuming that the initial momentum of the system is zero. Apply conservation of energy to obtain:

nHe

2nn2

12HeHe2

1MeV6.18KK

vmvm+=

+= (1)

Apply conservation of momentum to obtain:

0nnHeHe =+ vmvm (2)

Solve equation (2) for vHe:

He

nnHe m

vmv −= ⇒ 2n

2

He

n2He v

mmv ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Substitute for 2

Hev in equation (1): 2nn2

12n

2

He

nHe2

1MeV6.18 vmvmmm +⎟⎟

⎠

⎞⎜⎜⎝

⎛=

or

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

He

nn

He

n2nn2

1

1

1MeV6.18

mmK

mmvm

Solving for Kn yields:

He

nn

1

MeV6.18

mmK

+=

Chapter 40

336

Substitute numerical values for mn and mHe and evaluate Kn:

MeV9.14

MeV86.14

u002603.4u008665.11

MeV6.18n

=

=+

=K

Use equation (1) to find KHe:

MeV7.3

MeV86.14MeV6.18MeV6.18 nHe

=

−=−= KK

51 ••• Energy is generated in the Sun and other stars by fusion. One of the fusion cycles, the proton–proton cycle, consists of the following reactions:

1H + 1H → 2H + e+ + ve

1H + 2H → 3He + γ

followed by

1H + 3He → 4He + e+ + ve

(a) Show that the net effect of these reactions is

41H → 4He + 2e+ + 2ve + γ

(b) Show that 24.7 MeV is released during this cycle (not counting the additional energy of 1.02 MeV that is released when each positron meets an electron and the two annihilate). (c) The Sun radiates energy at the rate of approximately 4.0 × 1026 W. Assuming that this is due to the conversion of four protons into helium, γ-rays, and neutrinos, which releases 26.7 MeV, what is the rate of proton consumption in the Sun? How long will the Sun last if it continues to radiate at its present level? (Assume that protons constitute about half of the total mass, 2.0 × 1030 kg, of the Sun.) Picture the Problem Adding the three reactions will yield their net effect. We can use (Δm)c2 to find the rest energy released in the cycle and find the rate of proton consumption from the ratio of the Sun’s power output to the released per proton in fusion. (a) Add the three reactions to obtain: 1H + 1H + 1H + 2H + 1H + 3He → 2H + β + + νe + 3He + γ + 4He + β + + νe Simplify to obtain: γνβ +++→ +

e41 22HeH4

Nuclear Physics

337

(b) Express the rest energy released in this cycle:

( ) ( ) 2ep

2 44 cmmmcm −−=Δ α

Use Table 40-1 to find the masses of the participants in the reaction and evaluate (Δm)c2:

( ) ( )[ ] ( )

MeV7.24

MeV511.04uMeV/5.931u603002.4u825007.14

222

=

−×−=Δcccm

(c) Express the rate R of proton consumption: E

PR = (1)

where E is the energy released per proton in fusion.

Find N, the number of protons in the Sun:

( )

56

27

3021

p

sun21

1096.5

kg101.673kg1099.1

×=

××

== −mm

N

where we have assumed that protons constitute about half of the total mass of the Sun.

The energy released per proton in fusion is:

( )

J1007.1eV

J101.602MeV675.6

MeV675.6MeV7.26

12

1941

−

−

×=

××=

==E

Substitute numerical values in equation (1) and evaluate R: 138

13812

26

s107.3

s1074.3J1007.1

W100.4

−

−−

×≈

×=××

=R

The time T for the consumption of all protons is:

y100.5

Ms31.56y1s1059.1

s1074.31096.5

10

18

138

56

×≈

××=

××

== −RNT

Chapter 40

338

General Problems 52 • (a) Show that ke2 = 1.44 MeV⋅fm, where k is the Coulomb constant and e is the magnitude of the electron charge. (b) Show that hc = 1240 MeV⋅fm. Picture the Problem We can use the values of k, e, h, and c and the appropriate conversion factors to show that ke2 = 1.44 MeV⋅fm and hc = 1240 MeV⋅fm (a) Evaluate ke2 to obtain:

( )( )

fmMeV44.1eV10

MeV1m10

fm1meV1044.1

meV1044.1J10602.1

eV1mJ10307.2

mN10307.2C10602.1C/mN10988.8

6159

919

28

2282192292

⋅=××⋅×=

⋅×=×

×⋅×=

⋅×=×⋅×=

−−

−−

−

−−ke

(b) Evaluate hc to obtain:

( )( )

fmMeV1240eV10

MeV1m10

fm1meV101240

meV101240J101.602

eV1mJ1099.1

mJ1099.1m/s10998.2sJ10626.6

6159

919

25

25834

⋅=××⋅×=

⋅×=×

×⋅×=

⋅×=×⋅×=

−−

−−

−

−−hc

53 • [SSM] The counting rate from a radioactive source is 6400 counts/s. The half-life of the source is 10 s. Make a plot of the counting rate as a function of time for times up to 1 min. What is the decay constant for this source? Picture the Problem We can use the given information regarding the half-life of the source to find its decay constant. We can then plot a graph of the counting rate as a function of time. The decay constant is related to the half-life of the source:

( ) ( )

1

1

21

s069.0

s0693.0s102ln2ln

−

−

=

===t

λ

The activity of the source is given by:

( ) ( )tt eeRR1s0693.0

0 Bq6400−−− == λ

Nuclear Physics

339

The following graph of ( ) ( )teR1s0693.0Bq6400−−= was plotted using a spreadsheet

program.

0

1000

2000

3000

4000

5000

6000

7000

0 10 20 30 40 50 60

t (s)

R (B

q)

54 • Find the energy needed to remove a neutron from (a) 4He and (b) 7Li. Picture the Problem The energy needed to remove a neutron is given by

( ) 2cmQ Δ= where Δm is the difference between the sum of the masses of the reaction products and the mass of the target nucleus. (a) The reaction is:

nHeHe 34 +=+Q

The masses are (see Table 40-1): u603002.4He4 =m

u029016.3He3 =m

u6651.008n =m

Calculate the final mass: u694024.4

u6651.008u029016.3f

=+=m


u091022.0u603002.4u694024.4

if

=−=

−=Δ mmm

Chapter 40

340

Calculate the energy Q needed to remove a neutron from 4He:

( )

( )

MeV6.20

uMeV/5.931u091022.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ=

cc

cmQ

(b) The reaction is: nLiLi 67 +=+Q

The masses are (see Table 40-1): u004016.7

Li7 =m u122015.6

Li6 =m u6651.008n =m

Calculate the final mass: u787023.7

u6651.008u122015.6f

=+=m


u783007.0u004016.7u787023.7

if

=−=

−=Δ mmm

Calculate the energy Q needed to remove a neutron from 4He:

( )

( )

MeV25.7

uMeV/5.931u783007.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ=

cc

cmQ

55 • The isotope 14C decays according to

14C→14N + e− + v e . The atomic mass of 14N is 14.003 074 u. Determine the maximum kinetic energy of the electron. (Neglect recoil of the nitrogen atom.) Picture the Problem The maximum kinetic energy of the electron is given by

( ) .2NCmax 1414 cmmQK −==

The maximum kinetic energy of the electron is the Q value for the reaction:

( ) 2NCmax 1414 cmmQK −==

Find the mass of each atom from Table 40-1:

u24214.003C14 =m

u07414.003N14 =m

Nuclear Physics

341

Calculate :NC 1414 mmm −=Δ

u168000.0u07414.003u24214.003

=−=Δm

Calculate the maximum kinetic energy of the electron:

( )

( )

keV156

uMeV/5.931u168000.0

22

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Δ=

cc

cmQ

56 • The density of a neutron star is the same as the density of a nucleus. If our Sun were to collapse to a neutron star, what would be the radius of that object? Picture the Problem We can use the definition density to find the radius of the neutron star. Relate the mass of the neutron star to the mass of the sun M, the volume V of the star and the nuclear density ρ:

334 RVM πρρ == ⇒ 3

43πρMR =

where R is the radius of the star.

In Problem 20 it was established that:

317 kg/m10174.1 ×=ρ

Substitute numerical values and evaluate R:

( )( )km9.15

kg/m10174.14kg1099.13

3317

30

=

××

=π

R

57 •• [SSM] Show that the 109Ag nucleus is stable and does not undergo alpha decay, 47

109 Ag → 24 He + 45

105Rh +Q . The mass of the 109Ag nucleus is 108.904 756 u, and the products of the decay are 4.002 603 u and 104.905 250 u, respectively. Picture the Problem We can show that 109Ag is stable against alpha decay by demonstrating that its Q value is negative. The Q value for this reaction is:

( )[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=

uMeV/ 5.931

22

AgRhccmmmQ α

Chapter 40

342


( )[ ]

MeV88.2

uMeV/5.931u756904.108u250905.104u603002.4

22

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=

ccQ

Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, DKKQ α += . Kinetic energy cannot be negative; hence, alpha decay cannot occur unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, DP mmm α +> . Alpha decay cannot take place unless the total rest mass decreases. 58 •• Gamma rays can be used to induce photofission (fission triggered by the absorption of a photon) in nuclei. Calculate the threshold photon wavelength for the following nuclear reaction: 2H + γ → 1H + n. Use Table 40-1 for the masses of the interacting particles. Picture the Problem We can use thresholdthresholdthreshold λhchfE == , where Ethreshold

is the binding energy of the deuteron, to find the threshold wavelength for the given nuclear reaction. Express the threshold energy of the photon: threshold

thresholdthreshold λhchfE ==

Solve for the threshold wavelength:

thresholdthreshold E

hc=λ (1)

The threshold energy equals the binding energy of the deuteron:

( )[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

uMeV/5.931 2

2npDBthreshold

ccmmmEE

Substitute numerical values and evaluate Eth, the energy that must be added to the deuteron that will cause it to fission:

( )[ ]( )

J1055.3eV

J101.602MeV22.2

MeV/u 5.931u665008.1u825007.1u102014.2

1319

threshold

−−

×−=×

×−=

+−=E

Nuclear Physics

343

Substitute numerical values in equation (1) and evaluate λthreshold:

( )( ) pm560.0m1060.5J1055.3

m/s10998.2sJ10626.6 1313

834

threshold =×=×

×⋅×= −

−

−

λ

59 • The relative abundance of 40K (potassium 40) is 1.2 × 10–4. The isotope 40K has a molar mass of 40.0 g/mol, is radioactive and has a half-life of 1.3 × 109 y. Potassium is an essential element of every living cell. In the human body the mass of potassium constitutes approximately 0.36 percent of the total mass. Determine the activity of this radioactive source in a student whose mass is 60 kg. Picture the Problem The activity of a radioactive source is the product of the number of radioactive nuclei present and their decay constant. The activity of the isotope 40K in the student is:

( )21

4040

2lnt

NNR == λ (1)

Find N, the number of K nuclei in the student:

Mm

NN 0036.0

A

= ⇒M

mNN A0036.0=

where m is the mass of the student and M is the atomic mass of K.


( )( )( ) 2423

10326.3g/mol098.39

nuclei/mol10022.6kg600036.0×=

×=N

The number N40 of 40K nuclei in the student is the product of the relative abundance and the number of K nuclei in the student:

( )( )20

24440

10991.310326.3102.1

abundance Relative

×=

××=

×=−

NN

Substitute numerical values in equation (1) and evaluate R:

( ) ( )

Bq107.6

yMs31.56y103.1

2ln10991.3

3

9

20

×=

××

×=R

Chapter 40

344

60 •• When a positron makes contact with an electron, the electron–positron pair annihilate by way of the reaction β+ + β– → 2γ. Calculate the minimum total energy, in MeV, of the two photons created when a positron–electron pair annihilate. Picture the Problem We can find the energy released in the reaction

Q→+ −+ ββ by recognizing that a total of 2 electron masses are converted into energy in this annihilation. The energy released when a positron-electron pair annihilate is given by:

2e2 cmEQ ==


( )( )

MeV02.1

J101.602eV1J1064.1m/s10998.2kg10109.92 19

132831

=

×××=××= −

−−Q

61 •• The isotope 24Na is a β emitter and has a half-life of 15 h. A saline solution containing this radioactive isotope has an activity of 600 kBq and is injected into the bloodstream of a patient. Ten hours later, the activity of 1 mL of blood from this individual yields a counting rate of 12 counts/s at a counting efficiency of 20 percent. Determine the volume of blood in this patient. Picture the Problem We can use the fact that, after n half-lives, the decay rate of the 24Na isotope is ( ) 02

1 RR n= , where R0 is its decay rate at t = 0. The counting rate after n half-lives is related to the initial counting rate:

( ) 021 RR n=

Divide both sides of the equation by the volume V of blood in the patient:

( )VR

VR n 0

21=

We’re given that n = 2/3, R0 = 600 kBq, and, after n half-lives, the decay rate per unit volume is 60 Bq/mL:

( )VkBq600Bq/mL60 32

21=

Solving for V yields: ( )

L3.6

mL1030.6Bq/mL60

kBq600 33221

=

×=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Nuclear Physics

345

62 •• (a) Determine the distance of closest approach of an 8.0-MeV α particle in a head-on collision with a stationary nucleus of 197Au and with a stationary nucleus of 10B, neglecting the recoil of the struck nuclei. (b) Repeat the calculation taking into account the recoil of the struck nuclei. Picture the Problem We can solve this problem in the center of mass reference frame for the general case of an α particle in a head-on collision with a stationary nucleus of atomic mass M u and then substitute data for a stationary nucleus of 197Au and a nucleus of 10B. In the CM frame, the kinetic energy is: M

K

Mm

KK u411

lablabCM

+=

+=

α

At the point of closest approach:

( )( )min

2

minmin

21CM

22R

ZkeR

ZeekR

qkqK ===

or, because ke2 = 1.44 MeV⋅fm, ( )( )

minCM

2fmMeV44.1R

ZK ⋅=

Solve for Rmin to obtain:

( )( )CM

min2fmMeV44.1

KZR ⋅

= (1)

(a) Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab. Evaluate equation (1) for 197Au:

( )( )

fm28

MeV0.8792fmMeV44.1

min

=

×⋅=R

Evaluate equation (1) for 10B:

( )( )

fm8.1

MeV0.852fmMeV44.1

min

=

×⋅=R

(b) Find KCM for the 197Au nucleus:

MeV841.7

u197u41

MeV0.8CM =

+=K

Chapter 40

346

Substitute numerical values in equation (1) and evaluate Rmin:

( )( )

fm29

MeV841.7792fmMeV44.1

min

=

×⋅=R

Note that this result is about 2% greater that Rmin calculated ignoring recoil.

Find KCM for the 10B nucleus:

MeV714.5

u10u41

MeV0.8CM =

+=K

Substitute numerical values in equation (1) and evaluate Rmin:

( )( )

fm5.2

MeV714.552fmMeV44.1

min

=

×⋅=R

Note that this result is about 40% greater that Rmin calculated ignoring recoil.

63 •• Twelve nucleons are in a one-dimensional infinite square well of length L = 3.0 fm. (a) Using the approximation that the mass of a nucleon is 1.0 u, find the lowest energy of a nucleon in the well. Express your answer in MeV. What is the ground-state energy of the system of 12 nucleons in the well if (b) all the nucleons are neutrons so that there can be no more than 2 in each spatial state and (c) 6 of the nucleons are neutrons and 6 are protons so that there can be as many as 4 nucleons in each spatial state? (Neglect the energy of Coulomb repulsion of the protons.) Picture the Problem The allowed energy levels in a one-dimensional infinite

square well are given by ⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

22

8mLhnEn .

(a) The lowest energy of a nucleon of mass 1.0 u in the well corresponds to n = 1:

( )( )( )( )

MeV23

J101.602eV1J10678.3

fm0.3kg/u10661.1u0.18sJ10626.6

1912

227

234

1

=

×××=

×⋅×

=

−−

−

−

E

Nuclear Physics

347

(b) Because neutrons are fermions, there can be no more than two per state:

( ) ( )( ) GeV2.4MeV23182182

6543222

1

12

12

12

12

12

1654321

===

+++++=+++++=

E

EEEEEEEEEEEEE

(c) Find E for 4 protons and 4 neutrons:

( ) ( )( )

GeV3.1

MeV2356563244

1

12

12

1321

=

==++=++=

EEEEEEEE

64 •• The helium nucleus or α particle is a very tightly bound system. Nuclei with N = Z = 2n, where n is an integer (for example, 12C, 16O, 20Ne, and 24Mg), can be modeled as agglomerates of α particles. (a) Use this model to estimate the binding energy of a pair of α particles from the atomic masses of 4He and 16O. Assume that the four α particles in 16O form a regular tetrahedron that has one α particle at each vertex. (b) From the result obtained in Part (a) determine, on the basis of this model, the binding energy of 12C and compare your result with the result obtained from the atomic mass of 12C. Picture the Problem We can apply BE = (Δm)c2 to the model to find the binding energies and the binding energies/bond. (a) Find the binding energy BE for this model:

( )( )[ ]

( ) 2

2

2O

u497015.0u91515.994u603002.44

4BE 16

cc

cmm

=

−=

−= α

Because there is a bond between adjacent α particles (4) and a bond between diagonally opposite α particles (2), there are 6 bonds for the regular tetrahedron:

( ) ( )

MeV406.2

uMeV/c5.931u497015.0u497015.0BE

bondBE 2

2612

61

61

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=== cc

(b) 12C has 3 pairwise α particle bonds. Find the total BE for 12C with this model:

( ) ( ) ( )MeV406.23HeBE3CBE 412 +×=

Chapter 40

348

Calculate BE(4He):

( ) ( )[ ]( )[ ]

MeV30.28uMeV/5.931u603002.4u665008.1u825007.12

2HeBE2

2

2Henp

44

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−+=

cc

cmmm

Substitute numerical values and evaluate ( ):CBE 12

( ) ( ) ( ) MeV1.92MeV406.23MeV30.283CBE 12 =+= Use values in Table 40-1 to find ( ):CBE 12

( ) ( )[ ]( )[ ]

MeV2.92uMeV/5.931u000000.12u665008.1u825007.16

6CBE2

2

2Cnp

1212

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−+=

cc

cmmm

Note that this result is in good agreement with the model. 65 •• Nuclei of a radioactive isotope that has a decay constant of λ are produced in an accelerator at a constant rate Rp. The number of radioactive nuclei N then obeys the equation dN/dt = Rp – λN. (a) If N is zero at t = 0, sketch N versus t for this situation. (b) The isotope 62Cu is produced at a rate of 100 per second by placing ordinary copper (63Cu) in a beam of high-energy photons. The reaction is γ + 63Cu → 62Cu + n The isotope 62Cu decays by β decay and has a half-life of 10 min. After a time long enough so that dN/dt ≈ 0, how many 62Cu nuclei are there? Picture the Problem We can separate the variables in the differential equation dN/dt = Rp – λN and integrate to express N as a function of t. When dN/dt ≈ 0, Rp – λN∞ = 0, a condition we can use to find N∞. (a) Separate the variables in the differential equation to obtain:

dtNR

dN=

−λp

Integrate the left side of the equation from 0 to N and the right side from 0 to t to obtain:

∫∫ =−

tN

dtNR

dN

00 p

''

'λ

Nuclear Physics

349

Let u = Rp − λN′. Then:

'dNdu λ−= and

( )

( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

+−−=

−−=

−=−=− ∫∫

NRR

RNR

NR

uudu

N'RdN'

N

N

λλ

λλ

λ

λλ

λλλ

p

p

pp

0p

0 p

ln1

ln1ln1

'ln1

ln11 2

1

2

1

l

l

l

l

Because :'0

tdtt

=∫ tNR

R=⎟

⎟⎠

⎞⎜⎜⎝

⎛

− λλ p

pln1⇒ ( )te

RN λ

λ−−= 1p

The following graph of N(t) = (Rp/λ)(1 −e−λt) was plotted using a spreadsheet program. Note that N(t) approaches Rp/λ in the same manner that the charge on a capacitor approaches the value CV.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 1 2 3 4

N

pRλ

λ t

(b) When dN/dt = 0: 0p =− ∞NR λ ⇒

λpR

N =∞

Chapter 40

350

The decay constant is:

( )21

2lnt

=λ

Substitute for λ to obtain: ( ) 21

p

2lnt

RN =∞

Substitute numerical values and evaluate N∞: ( )

4

1

107.8

mins60min10

2lns100

×=

⎟⎠⎞

⎜⎝⎛ ×=

−

∞N

66 •• The total energy consumed in the United States in 1 y is approximately 7.0 × 1019 J. How many kilograms of 235U would be needed to provide this amount of energy if we assume that 200 MeV of energy is released by each fissioning uranium nucleus, that all of the uranium atoms undergo fission, and that all of the energy-conversion mechanisms used are 100 percent efficient?

Picture the Problem The mass of 235U required is given by ,235A

235 MNNm =

where M235 is the molecular mass of 235U and N is the number of fissions required to produce 7.0×1019 J. Relate the mass of 235U required to the number of fissions N required: 235

A235 M

NNm =

where M235 is the molecular mass of 235U.

Because fissionper

annual

EEN = :

fissionper A

235annual235 EN

MEm =

Substitute numerical values in equation (1) and evaluate m235:

( )( ) kg105.8

eVJ101.602MeV200

molnuclei10022.6

g/mol235J100.7 519

23

19

235 ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××⎟

⎠⎞

⎜⎝⎛ ×

×=

−m

67 •• (a) Find the wavelength of a particle in the ground state of a one-dimensional infinite square well of length L = 2.00 fm. (b) Find the momentum in units of MeV/c for a particle that has this wavelength. (c) Show that the total energy of an electron that has this wavelength is approximately E ≈ pc. (d) What is the kinetic energy of an electron in the ground state of this well? This

Nuclear Physics

351

calculation shows that if an electron were confined in a region of space as small as a nucleus, it would have a very large kinetic energy. Picture the Problem In the ground state of a one-dimensional infinite square well of length L the wavelength of a particle is 2L. We can use de Broglie’s equation to find p for the particle and the relationship 222

02 cpEE += with E0 << pc to show

that E ≈ pc. (a) In the ground state of a one-dimensional infinite square well of length L:

( ) fm00.4fm00.222 === Lλ

(b) Use de Broglie’s relation to obtain:

chchpλλ

==

Substitute numerical values and evaluate p: ( ) c

cp MeV/310

fm00.4nmeV1240

=⋅

=

(c) Relate the total energy of the electron to its rest energy and momentum:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+= 22

2022222

02 1

cpEcpcpEE

Because E0 << pc: 222 cpE ≈ ⇒ pcE ≈

(d) The kinetic energy of an electron in the ground state of this well is given by:

( ) MeV310MeV/3100

==

=≈−=

cc

pcEEEK

68 •• If 12C, 11B, and 1H have respective masses of 12.000 000 u, 11.009 306 u, and 1.007 825 u, determine the minimum energy, Q, in MeV, required to remove a proton from a 12C nucleus. Picture the Problem When a single proton is removed from a 12C nucleus, a 11B nucleus remains and we can use 2mcQ Δ= to determine the minimum energy required to remove a proton. The nuclear reaction is: HBC 1

1115

126 +→+Q

The minimum energy Q required is:

( ) 2CHB 12111 cmmmQ −+=

Chapter 40

352


( )[ ]

MeV0.16

uMeV/ 931.5u00012.000u8251.007u30611.009

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

ccQ

69 ••• [SSM] Assume that a neutron decays into a proton and an electron without the emission of a neutrino. The kinetic energy shared by the proton and the electron is then 0.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the proton must be equal and opposite the momentum of the electron. This determines the ratio of the kinetic energies of the two particles, but because the electron is relativistic, the exact calculation of these relative kinetic energies is somewhat challenging. (a) Assume that the kinetic energy of the electron is 0.782 MeV and calculate the momentum p of the electron in units of MeV/c. Hint: Use ( )22222 mccpE += (Equation 39-27). (b) Using your result from Part (a), calculate the kinetic energy p2/2mp of the proton. (c) Because the total kinetic energy of the electron and the proton is 0.782 MeV, the calculation in Part (b) gives a correction to the assumption that the kinetic energy of the electron is 0.782 MeV. What percentage of 0.782 MeV is this correction? Picture the Problem The momentum of the electron is related to its total energy through 2

0222 EcpE += and its total relativistic energy E is the sum of its kinetic

and rest energies. (a) Relate the total energy of the electron to its momentum and rest energy:

20

222 EcpE += (1)

The total relativistic energy E of the electron is the sum of its kinetic energy and its rest energy:

0EKE +=

Substitute for E in equation (1) to obtain:

( ) 20

2220 EcpEK +=+

Solving for p gives:

( )c

EKKp 02+=

Nuclear Physics

353

Substitute numerical values and evaluate p:

( )( )cc

cp MeV/19.1MeV/188.1

MeV511.02MeV782.0MeV782.0==

×+=

(b) Because pp = −pe:

p

2p

p 2mp

K =

Substitute numerical values (see Table 7-1 for the rest energy of a proton) and evaluate Kp:

( )( ) eV752

MeV/28.9382MeV/188.1

2

2

p ==c

cK

(c) The percent correction is: %0962.0MeV0.782eV752p ==

KK

70 ••• In the laboratory reference frame, a neutron of mass m moving with speed vL makes an elastic head-on collision with a nucleus of mass M that is at rest. (a) Show that the speed of the center of mass in the lab frame is V = mvL/(m + M). (b) What is the speed of the nucleus in the center-of-mass frame before the collision and after the collision? (c) What is the speed of the nucleus in the laboratory frame after the collision? (d) Show that the energy of the nucleus after the collision in the laboratory frame is

12

M 2V( )2 = 4mMm + M( )2

12

mvL2⎛

⎝⎜⎞⎠⎟

(e) Show that the fraction of the energy lost by the neutron in this elastic collision is

( )

( )( )22 1

44MmMm

MmmM

EE

+=

+=

Δ−

Picture the Problem Conservation of momentum and conservation of energy allow us to find the final speeds of the neutron and nucleus. (a) Apply conservation of momentum to the collision to obtain: ( ) LmvVMm =+ ⇒ Lv

MmmV ⎟

⎠⎞

⎜⎝⎛

+=

(b) In the CM frame, VMi = V and so: VVM =i

Chapter 40

354

In the CM frame, Vf = −Vi and so:

VVM −=f

(c) Use conservation of momentum to obtain one relation for the final velocities:

ffL MMVmvmv += (1)

The equality of the initial and final kinetic energies provides a second equation relating the two final velocities. This is implemented by equating the speeds of recession and approach:

( ) LLiff 0 vvVvV MM +=−−=− and so

Lff vVv M −=

To eliminate vf, substitute for mvL in equation (1) :

( ) fLfL MM MVvVmmv +−=

Solving for VMf yields: Lf

2 vMm

mVM ⎟⎠⎞

⎜⎝⎛

+=

(d) The kinetic energy of the nucleus after the collision in the laboratory frame is:

2f2

1MM MVK =

Substitute for VMf and simplify to obtain:

( )( )2

L21

2

2

L21

f

4

2

mvMm

mM

vMm

mMKM

+=

⎟⎠⎞

⎜⎝⎛

+=

(e) The fraction of the energy lost by the neutron in the elastic collision is given by:

( )

( )

( )2L2

1

2L

2f2

1

2L2

1

2L2

12f2

1

2L2

1if

mvvvm

mvmvmv

mvKK

EE mm

−−=

−−=

−−=

Δ−

Because Lff vVv M −= :

( )[ ]2L2

1

2L

2Lf2

1

mvvvVm

EE M −−−=

Δ−

Nuclear Physics

355

Substituting for VMf from Part (c) gives:

12112112

1122

222

2L2

1

22L2

1

2L2

1

2L

2

LL21

−⎥⎦⎤

⎢⎣⎡

+−−

−=−⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

+−=−⎥

⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛

+−

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛

+−

=Δ−

MmMmm

Mmm

Mmm

mv

Mmmmv

mv

vvvMm

mm

EE

Further simplification yields:

( ) ( )222

2 14

1

44MmMm

MmM

mMmM

mME

E+

=

⎟⎠⎞

⎜⎝⎛ +

=+

=Δ−

71 ••• (a) Use the result from Part (e) of Problem 72 to show that after N head-on collisions of a neutron with carbon nuclei at rest, the energy of the neutron is approximately (0.714)NE0, where E0 is its original energy. (b) How many head-on collisions are required to reduce the energy of the neutron from 2.0 MeV to 0.020 eV, assuming stationary carbon nuclei? Picture the Problem We can use the result of Problem 70, Part (e), to find the fraction 0f EEf = of its initial energy lost per collision and then use this result to show that, after N collisions, E = (0.714)NE0. (a) Determine 0f EEf = per collision:

00

0 1EE

EEEf Δ

−=Δ−

=

From Problem 70, Part (e): 2

0 1

4

⎟⎠⎞

⎜⎝⎛ +

=Δ

MmM

mEE

Substitute for ΔE/E0 in the expression for f to obtain:

2

1

41⎟⎠⎞

⎜⎝⎛ +

−=

MmM

mf

Chapter 40

356

Substitute numerical values and evaluate f:

( )

714.0u000000.12

u008665.11u000000.12

u008665.141 2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=f

After N collisions:

( ) 00 714.0 EEfE NNfN == (1)

(b) Solve equation (1) for N:

( )714.0ln

ln0⎟⎟⎠

⎞⎜⎜⎝

⎛

=EE

N

fN

Substitute numerical values and evaluate N: ( ) 55

714.0lnMeV0.2

eV .0200ln≈

⎟⎟⎠

⎞⎜⎜⎝

⎛

=N

55 head-on collisions are required to reduce the energy of the neutron from 2.0 MeV to 0.020 eV. 72 ••• On the average, a neutron loses 63 percent of its energy in a collision with a hydrogen atom and 11 percent of its energy in a collision with a carbon atom. Calculate the number of collisions needed to reduce the energy of a neutron from 2.0 MeV to 0.020 eV if the neutron collides with (a) hydrogen atoms and (b) carbon atoms. Picture the Problem We can use the result of Problem 70, Part (e), to find the fraction 0f EEf = of its initial energy lost per collision. Note the difference between the energy loss per collision specified here and that of the preceding problem. In the preceding problem it was assumed that all collisions are head-on collisions. (a) Determine 0f EEf = per collision:

00

0 1EE

EEEf Δ

−=Δ−

=

In a collision with a hydrogen atom:

63.00

=ΔEE ⇒ 37.0=f

After N collisions:

( ) 00 37.0 EEfE NNfN == (1)

Nuclear Physics

357

Solve equation (1) for N:

( )37.0ln

ln0⎟⎟⎠

⎞⎜⎜⎝

⎛

=EE

N

fN

(2)


37.0lnMeV0.2

eV020.0ln≈

⎟⎟⎠

⎞⎜⎜⎝

⎛

=N

19 head-on collisions with an atom of hydrogen are required to reduce the energy of the neutron from 2.0 MeV to 0.020 eV.

(b) In a collision with a carbon atom:

11.00

=ΔEE

⇒ 89.0=f

Equation (2) becomes:

( )89.0ln

ln0⎟⎟⎠

⎞⎜⎜⎝

⎛

=EE

N

fN


0.89lnMeV 2.0

eV 020.0ln≈

⎟⎠⎞

⎜⎝⎛

=N

160 head-on collisions with an atom of carbon are required to reduce the energy of the neutron from 2.0 MeV to 0.020 eV. 73 ••• [SSM] Frequently, the daughter nucleus of a radioactive parent nucleus is itself radioactive. Suppose the parent nucleus, designated by P, has a decay constant λP; while the daughter nucleus, designated by D, has a decay constant λD. The number of daughter nuclei ND are then given by the solution to the differential equation dND/dt = λPNP – λDND where NP is the number of parent nuclei. (a) Justify this differential equation. (b) Show that the solution for this equation is

N D t( )= N P0λP

λD − λP

e−λP t − e−λD t( ) where NP0 is the number of parent nuclei present at t = 0 when there are no daughter nuclei. (c) Show that the expression for ND in Part (b) gives ND (t) > 0

Chapter 40

358

whether λP > λD or λD > λP. (d) Make a plot of NP (t) and ND (t) as a function of time when τD = 3τP, where τD and τP are the mean lifetimes of the daughter and parent nuclei, respectively.

Picture the Problem We can differentiate ( ) ( )tt eeNtN DP

PD

PP0D

λλ

λλλ −− −−

= with

respect to t to show that it is the solution to the differential equation dND/dt = λPNP − λDND. (a) The rate of change of ND is the rate of generation of D nuclei minus the rate of decay of D nuclei. The generation rate is equal to the decay rate of P nuclei, which equals λPNP. The decay rate of D nuclei is λDND. (b) We’re given that: DDPP

D NNdt

dNλλ −= (1)

( ) ( )tt eeNtN DP

PD

PP0D

λλ

λλλ −− −−

= (2)

teNN PP0P

λ−= (3)

Differentiate equation (2) with respect to t to obtain:

( )[ ] ( )[ ] [ ]tttt eeNeedtdNtN

dtd

DPDPDP

PD

PP0

PD

PP0D

λλλλ λλλλλ

λλλ −−−− +−

−=−

−=

Substitute this derivative in equation (1) to obtain:

[ ] ( )⎥⎦

⎤⎢⎣

⎡−

−−=+−

−−−−−− ttttt eeNeNeeN

DPPDP

PD

PP0DP0PDP

PD

PP0 λλλλλ

λλλλλλλ

λλλ

Multiply both sides byPD

PD

λλλλ − and simplify to obtain:

[ ] ( )

[ ]tt

tt

tttt

ttttt

eeN

eNeN

eNeNeNeN

eeNeNeeN

DP

DP

DPPP

DPPDP

DPD

P0

P0D

PP0

P0P0D

PP0P0

P0P0D

PDDP

D

P0

λλ

λλ

λλλλ

λλλλλ

λλλ

λλ

λλ

λλλλλ

λ

−−

−−

−−−−

−−−−−

+−=

+−=

+−−=

−−−

=+−

Nuclear Physics

359

Because this is an identity, it confirms that equation (2) is the solution to equation (1). (c) If λP > λD the denominator and expression in parentheses are both negative for t > 0. If λP < λD the denominator and expression in parentheses are both positive for t > 0. (d) A spreadsheet program to calculate NP and ND as functions of time is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form A4 0 t A5 A4+1 t + 1 B4 EXP(−A4) teN P

P0λ−

C4 −1.5*(EXP(−A4) −EXP(−(1/3)*A4)) ( )tt eeN

DP

PD

PP0 λλ

λλλ −− −−

A B C 1 2 3 Time NP ND 4 0 1 0 5 1 0.367879 0.5229786 2 0.135335 0.5671237 3 0.049787 0.477139

22 18 1.52E−08 0.00371823 19 5.6E−09 0.00266424 20 2.06E−09 0.001909

Chapter 40

360

A graph of NP and ND as functions of time follows:

0.000 5 10 15 20

Number of parent (P) nucleiNumber of daughter ( D) nuclei

τ τ τ τp p p p

NPO

N

N

N

NPO

PO

PO

PO

0.20

0.40

0.60

0.80

1.00

Time 74 ••• Suppose isotope A decays to isotope B and has a decay constant λA, and isotope B in turn decays and has a decay constant λB. Suppose a sample contains, at t = 0, only isotope A nuclei. Derive an expression for the time at which the number of isotope B nuclei will be a maximum. (See Problem 75.) Picture the Problem We can express the time at which the number of isotope B nuclei will be a maximum by setting dNB/dt equal to zero and solving for t. From Problem 75 we have:

0BBAAB =−= NN

dtdN λλ for extrema

Replace λANA by teN A

A0Aλλ − and

NB by ( )tt eeNBA

AB

AA0 λλ

λλλ −− −−

:

( ) 0BAA

AB

AA0BA0A =−

−− −−− ttt eeNeN λλλ

λλλλλ

Simplify to obtain: ( ) 0BAA

AB

B =−−

− −−− ttt eee λλλ

λλλ

( ) ( ) 0BAABAB =−−− −−− ttt eee λλλ λλλ

Remove the parentheses and combine like terms to obtain:

tt ee BABA

λλ λλ −− = ⇒( )

AB

ABlnλλλλ

−=t

Nuclear Physics

361

Remarks: Note that all we’ve shown is that an extreme value exists at ( )

AB

AB

λλλλt

−=

ln . To show that this value for t maximizes NB, we need to either

1) examine the second derivative at this value for t, or 2) plot a graph of NB as a function of time (see Problem 77) . 75 ••• An example of the situation discussed in Problem 75 is the radioactive isotope 229Th, an α emitter that has a half-life of 7300 y. Its daughter, 225Ra, is a β emitter that has a half-life of 14.8 d. In this instance, as in many instances, the half-life of the parent is much longer than the half-life of the daughter. Using the expression given in Problem 75, Part (b), starting with a sample of pure 229Th containing NP0 nuclei, show that the number, ND, of 225Ra nuclei will, after several years, be given by

P0D

PD NN

λλ

=

where NP is number of 229Th nuclei. The number of daughter nuclei are said to be in secular equilibrium. Picture the Problem We can show that, provided ,DP ττ >> 1DP ≈− −− tt ee λλ and

D

P

PD

P

λλ

λλλ

≈−

and, hence, that P0D

PD NN

λλ

= .

We have, from Problem 75 (b): ( ) ( )tt eeNtN DP

PD

PP0D

λλ

λλλ −− −−

= (1)

Because :DP ττ >> DP λλ <<

When several years have passed, because 1P <<tλ :

1DP ≈− −− tt ee λλ (2)

Also, when DP λλ << : D

P

PD

P

λλ

λλλ

≈−

(3)

Substitute (2) and (3) in (1) to obtain: ( ) P0

D

PD NtN

λλ

=

Chapter 40

362

Chap 40 SM

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