homeowork 6

Nuclear Fusion and Radiation

Homework 6 - Solutions

Problem 1: Problem SF 5.1

Chapter 5

Radioactivity

PROBLEMS

1. Consider a stationary nucleus of mass mn in an excited state with energy E!

above the ground state. When this nucleus decays to the ground state bygamma decay, the emitted photon has an energy E! . (a) By considering theconservation of both energy and momentum of the decay reaction explain whyE! < E!. (b) Show that the two energies are related by

E! = mnc2

!"1 +

2E!

mnc2! 1

#" E!

$1 ! E!

2mnc2

%.

(c) Use an explicit example to verify that the di!erence between E! and E! isfor all practical purposes negligible.

Solution:

(a) If the excited nucleus is initially at rest, the products of the decay musthave zero net linear momentum, i.e., the photon and the ground-statedaughter nucleus must travel in opposite directions each with the sameamount of linear momentum. The Q-value of the decay reaction is E!

and must be equal to the sums of the kinetic energies of the photon andground-state nucleus. Hence it follows that E! < E!, the di!erence equalto the kinetic energy of the recoil ground-state nucleus.

(b) Conservation of total energy requires

mnc2 + E! = mnc2 + En + E! , (P5.1)

and conservation of linear momentum (treating the recoil nucleus as aclassical particle) requires

0 = E!/c!&

2mnEn. (P5.2)

Substitute Eq. (P5.1) into Eq. (P5.2) to eliminate En yields

E2! + 2mnc2E! ! 2mnc2E! = 0.

5-11

Chapter 5

Radioactivity

PROBLEMS

1. Consider a stationary nucleus of mass mn in an excited state with energy E!

above the ground state. When this nucleus decays to the ground state bygamma decay, the emitted photon has an energy E! . (a) By considering theconservation of both energy and momentum of the decay reaction explain whyE! < E!. (b) Show that the two energies are related by

E! = mnc2

!"1 +

2E!

mnc2! 1

#" E!

$1 ! E!

2mnc2

%.

(c) Use an explicit example to verify that the di!erence between E! and E! isfor all practical purposes negligible.

Solution:

(a) If the excited nucleus is initially at rest, the products of the decay musthave zero net linear momentum, i.e., the photon and the ground-statedaughter nucleus must travel in opposite directions each with the sameamount of linear momentum. The Q-value of the decay reaction is E!

and must be equal to the sums of the kinetic energies of the photon andground-state nucleus. Hence it follows that E! < E!, the di!erence equalto the kinetic energy of the recoil ground-state nucleus.

(b) Conservation of total energy requires

mnc2 + E! = mnc2 + En + E! , (P5.1)

and conservation of linear momentum (treating the recoil nucleus as aclassical particle) requires

0 = E!/c!&

2mnEn. (P5.2)

Substitute Eq. (P5.1) into Eq. (P5.2) to eliminate En yields

E2! + 2mnc2E! ! 2mnc2E! = 0.

5-1

5-2 Radioactivity Chap. 5

Solving this quadratic equation for E! produces

E! = mnc2

!!1 ±

"1 +

2E!

mnc2

#.

Only the + sign yields a positive real value for E! so that

E! = mnc2

$"1 + 2E!

mnc2 ! 1%

Since E! is typically much less than mnc2 " thousands of MeV, thendefining ! # (2E!)/(mnc2) << 1, we have

$1 + ! = 1+(1/2)!! (1/8)!2+

· · ·, and neglecting terms of order !2 and higher the above result can beapproximated by

E! " E!&1 ! E!

2mnc2

'. (P5.3)

(c) Consider the decay of the first excited state of 60Co (see Fig. 5.12) whichis E! = 1.17 MeV above the ground state. The energy of the emittedgamma ray upon deexcitation is, from Eq. (P5.3),

E! " E!&1 ! E!

2M(60Co)c2

'

= 1.17 MeV&1 ! 1.17 MeV

2 % 60 % 936 MeV

'

= 1.17 [1 ! 0.000010] MeV " 1.17 MeV.

The recoil kinetic energy of the ground-state 6Co nucleus is

E(60Co) = E! ! E! = 1.17 [0.000010] = 0.0117 keV.

2


Solving this quadratic equation for E! produces

E! = mnc2

!!1 ±

"1 +

2E!

mnc2

#.

Only the + sign yields a positive real value for E! so that

E! = mnc2

$"1 + 2E!

mnc2 ! 1%

Since E! is typically much less than mnc2 " thousands of MeV, thendefining ! # (2E!)/(mnc2) << 1, we have

$1 + ! = 1+(1/2)!! (1/8)!2+

· · ·, and neglecting terms of order !2 and higher the above result can beapproximated by

E! " E!&1 ! E!

2mnc2

'. (P5.3)

(c) Consider the decay of the first excited state of 60Co (see Fig. 5.12) whichis E! = 1.17 MeV above the ground state. The energy of the emittedgamma ray upon deexcitation is, from Eq. (P5.3),

E! " E!&1 ! E!

2M(60Co)c2

'

= 1.17 MeV&1 ! 1.17 MeV

2 % 60 % 936 MeV

'

= 1.17 [1 ! 0.000010] MeV " 1.17 MeV.

The recoil kinetic energy of the ground-state 6Co nucleus is

E(60Co) = E! ! E! = 1.17 [0.000010] = 0.0117 keV.

3

Problem 2: Problem SF 5.2Radioactivity Chap. 5 5-3

2. The radioisotope 224Ra decays by ! emission primarily to the ground state of220Rn (94% probability) and to the first excited state 0.241 MeV above theground state (5.5% probability). What are the energies of the two associated! particles?Solution:For the daughter left in an excited state, the Q-value of Eq. (5.7) is modified to increase the restmass of M (22086Rn) by an amount E!/c2, namely

Q!1 =!M (22488Ra) ! [M (42He)

+M (22086Rn) + E!/c2]"

c2.

With the masses in Ap. B we find

Q!1 = {224.0202020! [4.00260325 + 220.0113841]} (u) " 931.5 (MeV/u)!0.241 MeV = 5.548 MeV.

The kinetic energy of !1 is given by Eq. (5.11), namely

E!1 = Q!1

#M (22086Rn)

M (22082Rn) + M (42He)

$# 5.548

#220224

$= 5.449 MeV.

In a similar manner for alpha decay to the ground state (E! = 0), we have

Q!2 =!M (22488Ra) ! [M (42He) + M (22086Rn)]

"c2

= {224.0202020! [4.00260325 + 220.0113841]}931.5 (MeV/u)= 5.789 MeV.

The kinetic energy of this alpha particle is

E!2 = Q!2

#M (22086Rn)

M (22082Rn) + M (42He)

$# 5.789

#220224

$= 5.686 MeV.

3. The radionuclide 41Ar decays by "" emission to an excited level of 41K thatis 1.293 MeV above the ground state. What is the maximum kinetic energy ofthe emitted "" particle?Solution:First find the Q"! for this decay from Eq. (5.15)with the atomic masses in Ap. B. The result is

Q"! =!M (4118Ar) ! [M (4119K) + E!/c2]

"c2

= [40.9645008! 40.96182597]" 931.5!1.293 MeV

= 1.199 MeV.

The maximum kinetic energy the beta particle can have equals this Q-value,i.e., (E")max = 1.199 MeV.

4


Radioactivity Chap. 5 5-5

6. From the energy level diagram of Fig. 5.5, what are the decay constants forelectron capture and positron decay of 22Na? What is the total decay constant?Solution:

The total decay constant is

! =ln 2T1/2

=ln 2

2.602 y= 0.2664 y!1 = !!+ + !EC .

From Fig. 5.5, the probability of "+ decay is 0.8984 + 0.00006 = 0.8990. Thusthe decay constant for positron decay is

!!+ = 0.8990! ! = 0.2395 y!1.

From Fig. 5.5, the probability of electron capture is 0.101. Thus the decayconstant for positron decay is

!!+ = 0.101! ! = 0.0269 y!1.

The total decay constant is ! = !!+ + !EC = 0.2664 y!1 .

7. The activity of a radioisotope is found to decrease by 30% in one week. Whatare the values of its (a) decay constant, (b) half-life, and (c) mean life?Solution:

(a) From the radioactive decay law A(t)/A(0) = exp("!t), upon solving for t,the time required to reach a specified value of A(t)/A(0) is

t = " 1!

ln!

A(t)A(0)

".

Here we are given A(t = 1 wk)/A(0) = 0.7. From the above result we find

! = " 11 wk

ln(0.7) = 0.357 wk!1 = 0.0510 d!1

= 2.12 ! 10!3 h!1 = 5.90 ! 10!7 s!1.

(b) T1/2 = ln 2/! = 1.18 ! 106 s = 326.5 h = 13.6 d = 1.94 wk.

(c) t = 1/! = 1.69 ! 106 s = 471 h = 19.6 d = 2.80 wk.

5



8. The isotope 132I decays by !! emission to 132Xe with a half-life of 2.3 h. (a)How long will it take for 7/8 of the original number of 132I nuclides to decay?(b) How long will it take for a sample of 132I to lose 95% of its activity?Solution:

(a) After 3 half-lives, the activity is 1/23 = 1/8 of the original activity. Thusthe time to obtain 1/8 of the original activity is t = 3 ! T1/2 = 6.9 h.

(b) From the radioactive decay law A(t)/A(0) = exp(""t), upon solving for t,the time required to reach a specified value of A(t)/A(0) is

t = " 1"

ln!

A(t)A(0)

".

In this problem " = ln 2/T1/2 = 0.3014 h!1 and A(t)/A(0) = 0.05 Thusfrom the above equation we have

t = " 10.3014 h!1 ln(0.05) = 9.34 h.

9. How many grams of 32P are there in a 5 mCi source?Solution:

Since the activity A is related to the mass m of a radionuclide source by

A # "N = "mNa

A ,

and, hence, the mass of the radionuclide is

m (g) =AA"Na

. (P5.4)

From Ap. A the half-life of 32P is 14.28 d!1 so that " = ln 2/T1/2 = 5.618 !10!7 s!1. Then from Eq. (P5.4) we find the 32P mass is

m(32P) =(5 ! 10!3 Ci ! 3.7 ! 1010 Bq/Ci)(32 g/mol)(5.618 ! 10!7 s!1)(6.022 ! 1023atoms/mol)

= 1.75 ! 10!8 g = 17.5 pg.

6



10. How many atoms are there in a 1.20 MBq source of (a) 24Na and (b) 238U?Solution:

Because A ! !N we have N(atoms) = A(Bq)/!(s!1).

(a) For 24Na we find from Table A.4 that T1/2 = 14.96 h = 5.385 " 104 s.Then ! = ln 2/T1/2 = 1.287"10!5 s!1 Thus the number of atoms of 24Nafor a 1.20 MBq source is

N =A

!=

1.200" 106

1.287" 10!5= 9.32 ! 1010 atoms.

(b) For 238U we find from Table A.4 that T1/2 = 4.468"109 y = 1.410"1017 s.Then ! = ln 2/T1/2 = 4.916"10!18 s!1 Thus the number of atoms of 238Ufor a 1.20 MBq source is

N =A

!=

1.200" 106

4.916 " 10!18= 2.44 ! 1023 atoms.

11. A very old specimen of wood contained 1012 atoms of 14C in 1986. (a) Howmany 14C atoms did it contain in 9474 B.C.? (b) How many 14C atoms did itcontain in 1986 B.C.?Solution:

For 14C the half-life is 5730 y so that its decay constant is ! = ln 2/T1/2 =1.210 " 10!4 y!1. From the radioactive decay law N(t) = N(0) exp[#!t]. Inthis problem we identify the year 1986 as t = 0 with earlier times being negative.

(a) For t = #1986 # 9474 = #11, 460 y, the number of 14C atoms in thespecimen at 9474 BC is

N(#11, 460 y) = 1012 exp[+1.210"10!4"11, 460] = 4.00 ! 1012 atoms.

(b) For t = #1986#1985 = #3972 y, the number of 14C atoms in the specimenat 1986 BC is

N(#3972 y) = 1012 exp[+1.210 " 10!4 " 3972] = 1.62 ! 1012 atoms.

7



13. A sample contains 1.0 GBq of 90Sr and 0.62 GBq of 90Y. What will be theactivity of each nuclide (a) 10 days later and (b) 29.12 years later?Solution:From data in Ap. D, we find 90Sr has a half-life of 29.12 y, and 90Y 64.0 h.Thus

!Sr ! !1 = ln 2/T1/2 = 6.517 " 10!5 d!1 = 7.543" 10!10 s!1

!Y ! !2 = ln 2/T1/2 = 0.260 d!1 = 3.009 " 10!6 s!1.

(a) The 90Sr activity decays exponentially, and the 90Y after is given byEq. (5.58) (after multiplying by !), i.e.,

A1(t) = A1(0)e!!1t

A2(t) = A2(0)e!!2t +!2A1(0)!2 # !1

[e!!1t # e!!2t].

With A1(0) = 1 GBq and A2(0) = 0.62 GBq, these activities are evaluatedat t = 10 d as

A1(10 d) = 1 " e!!110 = 0.99935 GBq $ 1 GBq

A2(10 d) = 0.62e!!210 +!2A1(0)!2 # !1

[e!!110 # e!!210].

$ 0.62e!!210 + 1 " [1 # e!!210] = 0.0461 + 0.926 = 0.972 GBq.

Thus the total activity = A1(10 d) + A2(10 d) = 1.972 GBq.

(b) After several weeks, 90Y comes into secular equilibrium with 90Sr (assum-ing a sealed source). Thus

A1 = A2 = A1(0)e!!1t

so that the total activity of the source at t = 29.19 y (= one half-life) is

A(29.12 y) = A1 + A2 = 2A1 = 2A1(0)e!!t = 2A1(0)(1/2) = 1 GBq.

14. Consider the following "! decay chain with the half-lives indicated,

210Pb #%22 y

210Bi #%5.0 d

210Po.

A sample contains 30 MBq of 210Pb and 15 MBq of 210Bi at time zero. (a)Calculate the activity of 210Bi at time t = 10 d. (b) If the sample were originallypure 210Pb, how old would it have been at time t = 0?Solution:

From the half-lives we find the decay constants are

!Pb ! !1 = ln 2/T1/2 = 8.63 " 10!5 d!1 = 9.99 " 10!10 s!1

!Bi ! !2 = ln 2/T1/2 = 0.1386 d!1 = 1.605" 10!6 s!1.5-10 Radioactivity Chap. 5

From Eqs. (5.58) and (5.59) we have

A1(t) = A1(0)e!!1t (P5.5)

A2(t) = A2(0)e!!2t +!2A1(0)!2 ! !1

[e!!1t ! e!!2t]. (P5.6)

(a) As shown in the figure to the right,we are given the activities at t = 0and asked to find the activity of thefirst daughter 10 days later. Specifi-cally, we are given A1(0) = 30 MBqand A2(0) = 15 MBq we find fromEq. (P5.6) !

"Ai(t)

t

!! ""0

i = 2

i = 1

##$given

t = 10 d

##$seek

A2(10 d) = 15e!!210 +30!2

!2 ! !1[e!!110 ! e!!210].

= 3.750 + 30.019 [0.99914! 0.25007] = 26.2 MBq.

(b) Here we change our time scale. We are given theparent and daughter activities at some time to(which the problem statement calls t = 0 but wecall to) as shown in the figure to the right. Specif-ically, A1(to) = 30 MBq and A2(to) = 15 Mbq.We then seek, in this new time scale, A1(0).Also we know A2(0) = 0. From Eq. (P5.5),A1(to) = A1(0)e!!1to from which we obtainA1(0) = A1(to)e+!1to . Substitute this result intoEq. (P5.6) with A2(0) = 0 to obtain

!

"Ai(t)

t

!!

"

"

to

i = 1

i = 2

0

##$given

A2(to) =!2A1(to)e!1to

!2 ! !1[e!!1to ! e!!2to ]

=!2A1(to)!2 ! !1

[1 ! e!(!2!!1)to ]. (P5.7)

Finally, solving Eq. (P5.7) for to and substitution of data gives

to = ! 1!2 ! !1

ln!

1 ! A2(to)A1(to)

!2 ! !1

!2

"= 5.00 d.

8


From Eqs. (5.58) and (5.59) we have

A1(t) = A1(0)e!!1t (P5.5)

A2(t) = A2(0)e!!2t +!2A1(0)!2 ! !1

[e!!1t ! e!!2t]. (P5.6)

(a) As shown in the figure to the right,we are given the activities at t = 0and asked to find the activity of thefirst daughter 10 days later. Specifi-cally, we are given A1(0) = 30 MBqand A2(0) = 15 MBq we find fromEq. (P5.6) !

"Ai(t)

t

!! ""0

i = 2

i = 1

##$given

t = 10 d

##$seek

A2(10 d) = 15e!!210 +30!2

!2 ! !1[e!!110 ! e!!210].

= 3.750 + 30.019 [0.99914! 0.25007] = 26.2 MBq.

(b) Here we change our time scale. We are given theparent and daughter activities at some time to(which the problem statement calls t = 0 but wecall to) as shown in the figure to the right. Specif-ically, A1(to) = 30 MBq and A2(to) = 15 Mbq.We then seek, in this new time scale, A1(0).Also we know A2(0) = 0. From Eq. (P5.5),A1(to) = A1(0)e!!1to from which we obtainA1(0) = A1(to)e+!1to . Substitute this result intoEq. (P5.6) with A2(0) = 0 to obtain

!

"Ai(t)

t

!!

"

"

to

i = 1

i = 2

0

##$given

A2(to) =!2A1(to)e!1to

!2 ! !1[e!!1to ! e!!2to ]

=!2A1(to)!2 ! !1

[1 ! e!(!2!!1)to ]. (P5.7)

Finally, solving Eq. (P5.7) for to and substitution of data gives

to = ! 1!2 ! !1

ln!

1 ! A2(to)A1(to)

!2 ! !1

!2

"= 5.00 d.

9



20. Charcoal found in a deep layer of sediment in a cave is found to have an atomic14C/12C ratio only 30% that of a charcoal sample from a higher level with aknown age of 1850 y. What is the age of the deeper layer?Solution:

Let R denote the atomic ratio of 14C to 12C. We assume the same initial ratiowas incorporated into both layers when they were formed, i.e., Rd(0) = Rh(0).Then the present ratio in the deeper layer to that in the higher layer is

Rd(td)Rh(th)

=exp(!!td)exp(!!th)

= e!!(td!th),

where th and td are the ages of the higher and deeper layers, respectively.Solving this equation for td gives

td = th ! 1!

ln!

Rd(td)Rh(th)

"

For 14C, ! = ln 2/T1/2 = 1.21 " 10!4 y!1, and for the given information thatR(td)/R(th) = 0.30 and th = 1850 y, Eq. (P5.8) gives the desired age as

td = 1850 ! 11.21 " 10!4

ln(0.30) = 11,800 y.

21. 238Pu, an alpha-particle emitter, has been used as a thermal power source, asdescribed in Table 12.2. What is the energy recoverable as heat per decay?Solution:238Pu decays by alpha-particle emission to 234U. The Q-value is calculatedusing Eq. 5.7 and Table B.1, namely

Q"/c2 # M (23892Pu) ! [M (23492U) + M (42He)]

Q" = 931.5[238.0495534! 234.0409456! 4.0026032497] = 5.593 MeV.

Depending on the configuration of the power source, all of this energy whichappears as kinetic energy of the products may be dissipated as heat. If onlythe kinetic energy of the alpha particle is useful, the energy dissipated as heatis given by Eq. 5.11, namely,

E" # Q"

!AU

AU + A"

"= 5.50 MeV,

the value given in Table 12.2. 10

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