40 Chapter Nuclear Physics
Conceptual Problems
1 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: (b) Two other isotopes of 56Fe are: (c) Two other isotopes of 118Sn are:
15
N, 16N Fe, 55Fe Fe, 55Fe
54
54
2 Determine the Concept The parent of that series, 237Np, has a half-life of 2106 y that is much shorter than the age of the earth. There is no naturally occurring Np remaining on earth. 3 Determine the Concept Generally, -decay leaves the daughter nucleus neutron rich, i.e., above the line of stability. The daughter nucleus therefore tends to decay via emission which converts a nuclear neutron to a proton. *4 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 Determine the Concept It would make the dating unreliable because the current concentration of 14C is not equal to that at some earlier time. 6 Determine the Concept An element with such a high Z value would either fission spontaneously or decay almost immediately by emission (see Figure 40-3). 7 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.
1355
1356 Chapter 40
8 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 Determine the Concept Beta decay occurs in nuclei that have too many or too few neutrons for stability. In decay, A remains the same while Z either increases by 1 ( - decay) or decreases by 1 ( + decay). (a) The reaction is:
22 11 0 Na 22 Ne+ +1 + decay 10
(b) The reaction is:
24 11
0 Na 24 Mg + -1 - decay 12
10
Advantages Disadvantages 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.
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.
11 (a) False. The nucleus does not contain electrons. (b) True. (c) False. After two half-lives, three-fourths of the radioactive nuclei in a given sample have decayed. (d) True (given an unlimited supply of 238U).
Nuclear Physics 1357
12 Determine the Concept Pressure and temperature changes have no effect on the internal structure of the nucleus. They do have an effect on the electronic configuration; consequently, they can influence K-capture processes. *13 Determine the Concept Knowing the parent nucleus and one of the decay products, we can use the conservation of charge, the conservation of energy, and the conservation of the number of nucleons to identify the participants in the decay. (a) beta decay of 16N (b) alpha decay of 248Fm (c) positron decay of 12N (d) beta decay of 81Se (e) positron decay of 61Cu (f) alpha decay of 228Th
16 7 0 0 N 16 O + -1 + 0 + Q 8
248 100 12 7 81 34 61 29
Fm244 Cf + 4 He + Q 98 2
0 0 N12 C+ +1 + 0 + Q 6 81 0 0 Se35 Br + -1 + 0 + Q 61 0 0 Cu 28 Ni + +1 + 0 + Q
228 90
Th 224 Ra + 4 He + Q 88 2
*14 Determine the Concept We can use the information regarding the daughter nuclei to write and balance equations for each of the reactions. (a) (b) (c) (d) (e)
240 94 1 Pu 30 n+90 Sr +147 Ba 38 56
72 32
1 75 Ge+ 4 He0 n + 34 Se 2
127 53
1 I+ 2 H0 n +128 Xe 1 54
235 92
1 1 U + 0 n 20 n +113 Ag+121Rh 47 45
55 25
59 Mn + 7 Li3 H + 27 Co 3 1
1 (f) 238 U + 0 n 239 U; 92 92
239 92
0 0 U-1 + 0 + 239 Np; 93
239 93
0 0 Np -1 + 0 +
239 94
Pu
1358 Chapter 40
Estimation and Approximation
15 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 Mass density Half life Nuclear masses 16 Ratio (order of magnitude) 1015 1015 2
Picture the Problem The mass of 235U required is given by m235 =
N M 235 , where NA
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:
m235 =
N M 235 NA
(1)
where M235 is the molecular mass of 235U. Determine N:
N=
Eannual Eper fission
Substitute numerical values and evaluate N:
10 20 J N= 1.60 10 -19 J 200 MeV eV 30 = 3.13 10
Substitute numerical values in equation (1) and evaluate m235:
m235 =
3.13 10 30 (235 g/mol) = 5.20 10 6 kg 6.02 10 23 nuclei/mol
Nuclear Physics 1359
(b) Relate the mass of 2H and 3H required to the number of fusions N required:
m 2 H +3 H =
N M 2 H+3 H NA
(2)
where M 2 H + 3 H is the molecular mass of
2
H + 3H.
Determine N:
N=
Eannual Eper fission
Substitute numerical values and evaluate N:
N=
10 20 J 1.60 10 -19 J 18 MeV eV 31 = 3.47 10
Substitute numerical values in equation (2) and evaluate m 2 H + 3 H :
m235 =
3.47 10 31 (5 g/mol) = 2.88 10 6 kg 23 6.02 10 nuclei/mol
Properties of Nuclei
*17 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. To convert to MeV we multiply this result by 931.5 MeV/u. 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:
6mp + 6mn = 6 1.007825 u + 6 1.008665 u = 12.098940 u
Subtract the mass of 12C from this result:
(6m
p
+ 6mn ) - m12 C = 12.098940 u - 12 u = 0.098940 u
Multiply the mass difference by c2 and convert to MeV:
Eb = (m )c 2 = (0.098940 u ) c 2
931.5 MeV/c 2 = 92.2 MeV 1u
1360 Chapter 40
and the binding energy per nucleon is
Eb 92.2 MeV = = 7.68 MeV A 12
(b) For 56Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the protons:
26mp + 30mn = 26 1.007825 u + 30 1.008665 u = 56.463400 u
Subtract the mass of 56Fe from this result:
(26m
p
+ 30mn ) - m12 C = 56.463400 u - 55.934942 u = 0.528458 u
Multiply the mass difference by c2 and convert to MeV:
931.5 MeV/c 2 = 492 MeV 1u E 492 MeV and the binding energy per nucleon is b = = 8.79 MeV A 56 Eb = (m ) c 2 = (0.528458 u ) c 2
(c) For 238U, Z = 92 and N = 146. Add the mass of the neutrons to that of the protons:
92mp + 146mn = 92 1.007825 u + 146 1.008665 u = 239.984990 u
Subtract the mass of 238U from this result:
(92m
p
+ 146mn ) - m 238 U = 239.984990 u - 238.050783 u = 1.934207 u
Multiply the mass difference by c2 and convert to MeV:
931.5 MeV/c 2 = 1802 MeV 1u E 1802 MeV = 7.57 MeV and the binding energy per nucleon is b = A 238 Eb = (m ) c 2 = (1.934207 u )c 2
Nuclear Physics 1361
18 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. To convert to MeV we multiply this result by 931.5 MeV/u. 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:
3mp + 3mn = 3 1.007825 u + 3 1.008665 u = 6.049470 u
Subtract the mass of 6Li from this result:
(3m
p
+ 3mn ) - m6 Li = 6.049470 u - 6.015122 u = 0.034348 u
Multiply the mass difference by c2 and convert to MeV:
931.5 MeV/c 2 = 32.0 MeV 1u E 32.0 MeV = 5.33 MeV and the binding energy per nucleon is b = A 6 Eb = (m )c 2 = (0.034348 u )c 2
(b) For 39K, Z = 19 and N = 20. Add the mass of the neutrons to that of the protons:
19mp + 20mn = 19 1.007825 u + 20 1.008665 u = 39.321975 u
Subtract the mass of 39K from this result:
(19m
p
+ 20mn ) - m39 K = 39.321975 u - 38.963707 u = 0.358268 u
Multiply the mass difference by c2 and convert to MeV:
931.5 MeV/c 2 Eb = (m )c = (0.358268 u )c = 334 MeV 1u E 334 MeV and the binding energy per nucleon is b = = 8.56 MeV A 39
2 2
(c) For 208Pb, Z = 82 and N = 126. Add the mass of the neutrons to that of the protons:
1362 Chapter 40 82mp + 126mn = 82 1.007825 u + 126 1.008665 u = 209.733440 u
Subtract the mass of 208Pb from this result:
(82m
p
+ 126mn ) - m 208 Pb = 209.733440 u - 207.976636 u = 1.756804 u
Multiply the mass difference by c2 and convert to MeV:
931.5 MeV/c 2 = 1636 MeV 1u E 1636 MeV = 7.87 MeV and the binding energy per nucleon is b = A 208 Eb = (m ) c 2 = (1.756804 u ) c 2
19 Picture the Problem The nuclear radius is given by R = R0 A1 3 where R0 = 1.2 fm. (a) The radius of 16O is: (b) The radius of 56Fe is: (c) The radius of 197Au is:
R16 O = (1.2 fm )(16 ) = 3.02 fm
13
R56 Fe = (1.2 fm )(56 ) = 4.59 fm
13
R197 Au = (1.2 fm )(197 ) = 6.98 fm
13
20 Picture the Problem The nuclear radius is given by R = R0 A1 3 where R0 = 1.2 fm. The radii of the daughter nuclei are given by: Because the ratio of the mass numbers of the daughter nuclei is 3 to 1: Substitute in equation (1) to obtain:
R = R0 A1 3
(1)
1 3 A1 = (239) and A2 = (239) 4 4
3 239 R1 = (1.2 fm ) 4
and
1/3
= 6.77 fm
239 R2 = (1.2 fm ) 4
1/3
= 4.69 fm
*21 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-
Nuclear Physics 1363
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:
Ethermal = kT
= 1.38 10-23 J/K (25 + 273)K = 4.11 10 -21 J = 4.11 10 -21 J = 25.7 meV 1eV 1.60 10 -19 J
(
)
(b) Equate Ethermal and the kinetic energy of the neutron to obtain: Solve for v to obtain:
Ethermal = 1 mn v 2 2
v= 2 Ethermal mn 2 4.11 10 - 21 J = 2.22 km/s 1.67 10- 27 kg
Substitute numerical values and evaluate v:
v=
(
)
(c) Relate the half-life, t1 2 , to the speed of the neutrons in the beam: Substitute numerical values and evaluate t1 2 :
t1 2 =
t1/2 =
x v
1350 km 1min = 608 s 2.22 km/s 60 s
= 10.1min
22 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: The approximate mass is: Express the volume of the nucleus:
=
m V
m = 1.66 10-27 kg A
3 V = 4 R0 A1 3 = 4 R0 A 3 3 3
(
)
(
)
1364 Chapter 40
Substitute for m and V to obtain:
=
=
(1.66 10 (
4 3
-27
kg A
R A
3 0
)
3 1.66 10- 27 kg 3 4R0
3 1.66 10 -27 kg 3 4 (1.2 fm )
)
)
Substitute numerical values and evaluate :
=
(
= 2.29 1017 kg/m 3 = 2.29 1014 g/cm3
23 Picture the Problem The separation of the nuclei when they are just touching is the sum of their radii, which is given by R = R0 A1 3 . The electrostatic potential energy of the system is given by:
U =k
(Z e)(Z La e ) q1q2 = k Mo R RMo + RLa
where R is the distance from the center of the 95Mo nucleus to the center of the 139La nucleus. The radii of the nuclei are: Substitute for RMo and RLa and simplify to obtain:
1/3 1/3 RMo = Ro AMo and RLa = Ro ALa
U = ke 2 =
(Z Mo )(Z La )
1/3 1/3 Ro AMo + Ro ALa
ke 2 (Z Mo )(Z La ) 1/3 1/3 Ro AMo + ALa
(
)
Substitute numerical values and evaluate U:
(8.99 10 Nm /C )(1.60 10 U= (1.2 10 m)
9 2 2
-15
-19
C
)
2
(95
(42)(57 )
1/3
+ 1391/3
)
= 4.71 10-11 J
1eV = 295 MeV 1.60 10-19 J
*24 Picture the Problem The Heisenberg uncertainty principle relates the uncertainty in position, x, to the uncertainty in momentum, p, by xp Solve the Heisenberg equation for p:
1 h. 2
p
h 2x
Nuclear Physics 1365
Substitute numerical values and evaluate p:
p
1.05 10-34 J s 2 10 10-15 m
(
)
= 5.25 10
The kinetic energy of the electron is given by: Substitute numerical values and evaluate K:
- 21
kg m/s
K = pc K = (5.25 10 -21 kg m/s )(3 108 m/s ) 1eV = 1.58 10 -12 J 1.60 10-19 J = 9.88 MeV
This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of 1 to 1000 eV.
Radioactivity
25 Picture the Problem The counting rate, as a function of the number of half-lives n, is n given by R = ( 1 ) R0 . 2 (a) The counting rate after n halflives is: Solve for n to obtain:
R = ( 1 ) R0 2
n
n=
ln (R R0 ) ln ( 1 ) 2
Substitute numerical values and evaluate n:
1000 counts/s ln 4000 counts/s =2 n= ln ( 1 ) 2
t1 2 = 5 min R = ( 1 ) (4000 counts/s ) = 250 Bq 2
4
Because there are two half-lives in 10 min: (b) At the end of 4 half-lives:
26 Picture the Problem The counting rate, as a function of the number of half-lives n, is given by R = ( 1 ) R0 . 2
n
(a) When t = 4 min, two half-lives will have passed and n = 2:
R = ( 1 ) (2000 counts/s ) = 500 Bq 2
2
1366 Chapter 40
(b) When t = 6 min, three half-lives will have passed and n = 3: (c) When t = 8 min, four half-lives will have passed and n = 4:
R = ( 1 ) (2000 counts/s ) = 250 Bq 2
3
R = ( 1 ) (2000 counts/s ) = 125 Bq 2
4
27 Picture the Problem The counting rate, as a function of the number of half-lives n, is n given by R = ( 1 ) R0 and the decay constant is related to the half-life by t1 2 = ln 2 . 2 (a) Relate the counting rate at time t = 10 min to the counting rate at t = 0: Solve for n:
R10 min = ( 1 ) R0 2
n
n=
ln (R10 min R0 ) ln ( 1 ) 2
Substitute numerical values and evaluate n:
1000 counts/s ln 8000 counts/s =3 n= 1 ln ( 2 )
3t1 2 = 10 min t1 2 = 200 s
Therefore, 3 half-lives have passed in 10 min: (b) The decay constant is related to the half-life by: Substitute for t1/2 and evaluate :
t1 2 =
ln 2
=
ln 2 t1 2
=
ln 2 = 3.47 10-3 s -1 200 s
6
(c) Six half-lives will have passed in 20 min:
R20 min = ( 1 ) (8000 Bq ) = 125 Bq 2
28 Picture the Problem We can use R = N 0e - t to show that the disintegration rate is approximately 1 Ci. The decay rate is given by:
R = N 0 e - t
where N0 is the number of nuclei at t = 0.
Nuclear Physics 1367
The decay constant is related to the half-life by: Substitute for t1/2 and evaluate :
t1 2 =
ln 2
=
ln 2 t1 2
=
ln 2 y = 4.28 10-4 y -1 1620 y 31.56 Ms
= 1.356 10-11 s -1
The number of nuclei at t = 0 is given by: N0 = NA/M where M is the atomic mass of radium and NA is Avogadro's number.
Substitute numerical values and evaluate N0:
N0 =
6.02 1023 = 2.664 1021 226
Substitute numerical values for and N0 and evaluate R:
R = 1.356 10 -11 s -1 2.664 10 21 e - (1.35610 3.7 1010 s -1 = 1Ci
(
)(
)
-11 -1
s
)(1200 s ) = 3.61 1010 s -1
29 n Picture the Problem We can use R = ( 1 ) R0 to relate the counting rate R to the number 2 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: When t = 4.8 min, n = 2 and: (b) The number of radioactive nuclei is related to the decay rate R, and the decay constant : The decay constant is related to the half-life:
R2.4 min = ( 1 ) (1000 counts/s ) = 500 Bq 2
1
R4.8 min = ( 1 ) (1000 counts/s ) = 250 Bq 2
2
R = N N =
R
(1)
=
0.693 0.693 0.693 = = t1 2 2.4 min 144 s
= 4.813 10 -3 s -1
Calculate the decay rate at t = 0
R0 = 5 1000 counts/s = 5000 s -1
1368 Chapter 40
from the counting rate: Substitute in equation (1) and evaluate N0 at t = 0: Calculate the decay rate at t = 2.4 min from the counting rate: Substitute in equation (1) and evaluate N2.4 min at t = 0:
N0 =
R0
=
5000 s -1 = 1.04 106 -3 -1 4.813 10 s
R2.4 min = 5 500 counts/s = 2500 s -1
N 2.4 min =
R2.4 min
=
2500 s -1 4.813 10 -3 s -1
= 5.19 105
(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 halflife: The counting rate R after n halflives is related to the counting rate at t = 0 by: Solve for n:
t = nt1 2
(2)
R = ( 1 ) R0 2
n
n=
ln (R R0 ) ln ( 1 ) 2
ln (30 counts/s 1000 counts/s ) ln ( 1 ) 2
Substitute numerical values and evaluate n:
n=
= 5.059
Substitute numerical values for n and t1/2 in equation (2) and evaluate t:
t = (5.059)(2.4 min ) = 12.1 min
30 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:
226
Ra 222 Rn + 4 He
Nuclear Physics 1369
Use Table 40-1 to find E:
E =
931.5 MeV/c 2 (226.025403 u - 222.017571u - 4.002603 u )c 2 = 4.87 MeV 1u
242
(b) Write the reaction: Use Table 40-1 to find E:
Pu 238 U + 4 He
E =
931.5 MeV/c 2 (242.058737 u - 238.050783 u - 4.002603 u )c 2 = 4.98 MeV 1u
*31 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 be 1000 alpha particles per second. (a) The activity of the nuclei present in the ingested 239Pu is given by:
A = N
(1)
Find the constant for the decay of 239 Pu:
=
ln 2 0.693 = t1 / 2 (24360 y )(31.56 Ms/y )
= 9.02 10 -13 s -1
Express the number of nuclei present in the quantity of 239Pu ingested: Substitute numerical values and evaluate N:
N = mPu
NA M Pu
where MPu is the atomic mass of 239Pu.
6.02 1023 nuclei/mol N = (2.0 g ) 239 g/mol 15 = 5.04 10 nuclei A = (9.02 10 -13 s -1 )(5.04 1015 ) = 4.55 103 /s
Substitute numerical values in equation (1) and evaluate A:
(b) The activity varies with time according to:
A = Aoe- t
1370 Chapter 40
Solve for t to obtain:
A ln A t= o - 1 103 / s ln 4.55 103 / s t= 31.56 Ms - 9.02 10-13 s -1 1y
Substitute numerical values and evaluate t:
(
)
= 5.32 104 y
32 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:
2 p K = m v = 2m 1 2 2
(1)
and
2 pY KY = m v = 2mY 1 2 2 Y Y
(2)
2 Solve equations (1) and (2) for p
2 and pD :
2 p = 2m K
and
2 pY = 2mY K Y
From the conservation of linear momentum we have:
r r pi = pf
or, because the parent is initially at rest, 0 = p - pY and p = pY
Because the momenta are equal: Solve for KY:
2mY K Y = 2m K KY = 4 m K = K mY A-4
Because the daughter nucleus and the alpha particle share the Q-value:
Q = K Y + K 4 4 + 1 K K + K = A-4 A-4 A = K A-4 =
Nuclear Physics 1371
Solve for K:
A-4 K = Q A
(b) Substitute for K in the expression for Q to obtain: Solve for KY:
A-4 Q = KY + Q A
4Q A-4 KY = Q - Q = A A
*33 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 32 is given by Q = -(m)c2. Pu undergoes alpha decay according to: The Q value for the decay is given by: Substitute numerical values and evaluate Q:
239
239 94 4 Pu 235 U + 2 + Q 92
931.5 MeV Q = [(mPu ) - (mU + m )] 1u
931.5MeV Q = [(239.052156 u ) - (235.043923 u + 4.002603 u )] = 5.24 MeV 1u
From Problem 32, the kinetic energy of the alpha particle is given by: Substitute numerical values and evaluate K:
A-4 K = Q A
239 - 4 K = (5.24 MeV ) 239 = 5.15 MeV
From Problem 32, the kinetic energy of the 239U is given by: Substitute numerical values and evaluate KU:
KU = KU =
4Q A 4(5.24 MeV ) = 87.7 keV 239
34 n Picture the Problem We can find the age of the sample using Rn = ( 1 ) R0 to find n and 2 then applying t = nt1 2 . Express the age of the bone in terms of the half-life of 14C and the
t = nt1 2
(1)
1372 Chapter 40
number n of half-lives that have elapsed: The decay rate Rn after n half-lives is related to the counting rate R0 at t = 0 by: Solve for n:
Rn = ( 1 ) R0 2
n
n=
ln (R R0 ) ln ( 1 ) 2
Because there are 15.0 decays per minute per gram of carbon in a living organism: Substitute numerical values for R and R0 and evaluate n:
R0 = 15.0
decays 1 min 175 g min g 60 s = 43.75 Bq
8.1 Bq ln 43.75 Bq = 2.433 n= 1 ln ( 2 )
t = (2.433)(5730 y ) = 13,940 y
Substitute numerical values in equation (1) and evaluate t:
35 Picture the Problem We can solve R = R0 e - t for to find the decay constant of the sample and use t1 2 =
ln 2
sample initially can be found from R0 = N 0 . (a) The decay rate is given by: Solve for to obtain:
to find its half-life. The number of radioactive nuclei in the
R = R0e - t R ln R = 0 -t
85.2 Bq ln 115 Bq = 0.133 h -1 = - 2.25 h
Substitute numerical values and evaluate :
Nuclear Physics 1373
The half-life is related to the decay constant: (b) The number N0 of radioactive nuclei in the sample initially is related to the decay constant and the initial decay rate R0: Substitute numerical values and evaluate N0:
t1 2 =
ln 2
=
ln 2 = 5.20 h 0.133 h -1
R0 = N 0 N 0 =
R0
N0 =
115 Bq 1h 0.133 h -1 3600 s
= 3.11 106
*36 Picture the Problem We can use R0 = N to find the initial activity of the sample and
R = Ro e - t 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 for 60Co and the number of atoms N of 60Co initially present in the sample: Express N in terms of the mass m of the sample, the molar mass M of 60 Co, and Avogadro's number NA: Substitute numerical values and evaluate N:
R0 = N
(1)
N=
m NA M
1.00 10 -6 g 23 16 N = 60 g/mol 6.02 10 nuclei/mol = 1.00 10 nuclei
(
)
The decay constant is given by:
=
=
ln2 t1/2
0.693 (5.27 y )(31.56 Ms/y )
Substitute numerical values and evaluate :
= 4.17 10 -9 s -1
Substitute numerical values in equation (1) and evaluate A0:
R0 = 4.17 10 -9 s -1 1.00 1016 nuclei = 4.17 10 7 s -1 = 1.13 mCi 1 Ci 3.7 1010 s -1
(
)(
)
1374 Chapter 40
(b) The activity varies with time according to: Evaluate R at t = 1.75 y:
R = Ro e
- t
= Ro e
0.693t - 5.27y
R = (1.13 mCi ) e = 0.898 mCi
0.6931.75y - 5.27y
37 Picture the Problem The following graph was plotted using a spreadsheet program. Excel's Add Trendline feature was used to determine the equation of the line.
9
8
7
ln(R )
6
5 ln(R ) = -0.0771t + 8.3395 4
3 0 10 20 30 40 50 60
t (min)
The linearity and negative slope of this graph tell us that it represents an exponential decay. The decay rate equation is: Take the natural logarithm of both sides of the equation to obtain: This equation is of the form:
R = R0e - t
ln R = ln e - t + ln R0 = -t + ln R0
y = mx + b
where y = ln R, x = t, m = -, and b = lnR0.
The decay constant is the negative of the slope of the graph:
= 0.0771 min -1
Nuclear Physics 1375
The half-life of the radioisotope is:
t1 2 =
ln 2
=
ln 2 = 8.99 min 0.0771 min -1
38 Picture the Problem We can solve Equation 40-7 for to show that = t1-1ln(R0/R1). (a) Express the half-life as a function of the decay constant : From Equation 40-7 it follows that:
t1 2 =
ln 2
(1)
R0 = e t R1 R ln 0 R R = 1 = t -1 ln 0 R t 1
Solve for :
(b) Substitute numerical values for t, R1, and R0 and evaluate : Use the decay constant to find the half-life:
=
1 1200 Bq = 0.00676 s -1 ln 60 s 800 Bq
t1 2 =
ln 2
=
ln 2 = 103 s 0.00676 s -1
39 Picture the Problem 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 R = ( 1 ) R0 to find the current 2
n
counting rate per gram of 14C. The mass of carbon required is:
M =
5 counts/min R
n n
(1)
Because there were about 15.0 decays per minute per gram of the living wood, the counting rate per gram is: We can find n from the assumed age of the casket and the half-life of 14C:
R = ( 1 ) R0 = ( 1 ) (15 counts/min g ) 2 2
n=
18,000 y = 3.141 5730 y
1376 Chapter 40
Substitute for n and evaluate R:
R = (1 ) 2
3.141
(15 counts/min g )
= 1.70 counts/min g
Substitute for R in equation (1) and evaluate M:
M =
5 counts/min = 2.94 g 1.70 counts/min g
40 Picture the Problem The decay constant 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 and the number of radioactive nuclei N at the moment of interest: 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: Substitute numerical values and evaluate N:
R = N =
R N
(1)
N = NA
m M
N = (6.02 1023 nuclei/mol)
10 -3 g = 1.004 1019 59.934 g / mol
1.131Ci 3.7 1010 Bq Ci 1.004 1019
Substitute numerical values in equation (1) and evaluate :
=
= 4.17 10-9 s -1
Find the half-life from the decay constant:
t1 2 =
ln 2 4.17 10-9 s -1 1y = 1.67 108 s 31.56 Ms = = 5.27 y
ln 2
Nuclear Physics 1377
*41 Picture the Problem The following graph was plotted using a spreadsheet program. Excel's Add Trendline feature was used to determine the equation of the line.
7.0 6.8 6.6 6.4 6.2 ln( R ) 6.0 5.8 5.6 5.4 5.2 5.0 0 1 2 3 t (min) 4 5 6 7 ln(R ) = -0.198t + 6.9076
The linearity and negative slope of this graph tells us that it represents an exponential decay. The decay rate equation is: Take the natural logarithm of both sides of the equation to obtain: This equation is of the form:
R = R0e - t
ln R = ln e - t + ln R0 = -t + ln R0
y = mx + b
where y = ln R, x = t, m = -, and b = lnR0.
The half-life of the radioisotope is:
t1 2 =
ln 2
=
ln 2 = 3.50 min 0.198 min -1
42 Picture the Problem We can use the decay rate equation R = R0e - t 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.
1378 Chapter 40
(a) The half-life of the material is given by: The decay rate is given by: Solve for :
t1 2 =
ln 2
(1)
R = R0e - t
R ln 0 R = t
(2)
Substitute for in equation (1) to obtain:
t1 2 =
ln 2 ln 2 = t R0 R0 ln ln R R t ln 2 (101h ) 115 decays/min ln 73.5 decays/min
Substitute numerical values and evaluate t1/2:
t1 2 =
= 156 h
(b) Solve equation (2) for t:
R ln R t= 0 -
Express in terms of t1/2:
=
ln 2 t1 2
Substitute for in the expression for t to obtain:
R ln R t = 0 t1 2 - ln 2
10 decays/min ln 115 decays/min (156 h ) t= - ln 2 1d = 550 h = 22.9 d 24 h
Substitute numerical values and evaluate t:
43 Picture the Problem We can use the decay rate equation R = R0e - t and the expression relating the half-life of a source to its decay constant to find the age of the fossils.
Nuclear Physics 1379
The decay rate is given by: Solve for t to obtain:
R = R0e - t R ln R t= 0 -
Express in terms of t1/2:
=
ln 2 t1 2
Substitute for in the expression for t to obtain:
R ln R t = 0 t1 2 - ln 2
or, because the activity at any time is proportional to the number of radioactive nuclei present,
N ln Rb N 0,Rb t= t1 2 - ln 2
The number of 87Sr nuclei present in the rocks is given by: We're given that:
(1)
NSr = N 0,Rb - N Rb N 0,Rb = N Sr + N Rb
NSr = 0.01N Rb
N Sr = 0.01 N Rb
Express the ratio of N0,Rb to NRb:
N 0,Rb N Sr + N Rb N Sr = = +1 N Rb N Rb N Rb = 0.01 + 1 = 1.01
Substitute numerical values in equation (1) and evaluate the age of the fossils:
1 ln 1.01 (4.9 1010 y ) t= - ln 2 = 7.03 108 y
44 Picture the Problem We can evaluate this integral by changing variables to obtain a form that we can find in a table of integrals. Change variables by letting:
x = t
1380 Chapter 40
Then:
dx = dt , =
dx
, dt and t =
x
= xe
0
Substitute to obtain:
= te dt = e 0 0
- t
x
-x
dx
1
-x
dx
From integral tables:
xe
0
-x
dx = 1
Substitute in the expression for to obtain:
=
1
Nuclear Reactions
45 Picture the Problem We can use Q = -(m )c 2 to find the Q values for these reactions. (a) Find the mass of each atom from Table 40-1:
m1 H = 1.007825 u m 3 H = 3.016049 u m 3 He = 3.016029 u
mn = 1.008665 u
Calculate the initial mass mi of the incoming particles: Calculate the final mass mf:
mi = 1.007825 u + 3.016049 u = 4.023874 u mf = 3.016029 u + 1.008665 u = 4.024694 u
Calculate the increase in mass:
m = mf - mi = 4.024694 u - 4.023874 u = 0.000820 u
Calculate the Q value:
Q = -(m ) c 2 MeV/c 2 = -(0.000820 u ) c 2 931.5 u = - 0.764 MeV
Nuclear Physics 1381
(b) Proceed as in (a) to obtain:
MeV/c 2 2 c Q = (0.003510 u ) 931.5 u = 3.27 MeV
Remarks: Because Q < 0 for the first reaction, it is endothermic. Because Q > 0 for the second reaction, it is exothermic. 46 Picture the Problem We can use Q = -(m )c 2 to find the Q values for these reactions. (a) Find the mass of each atom from Table 40-1:
m 2 H = 2.014102 u m 3 H = 3.016049 u m1 H = 1.007825 u
Calculate the initial mass mi of the incoming particles: Calculate the final mass mf:
mi = 2(2.014102 u ) = 4.028204 u mf = 3.016049 u + 1.007825 u = 4.023874 u
Calculate the increase in mass:
m = mf - mi = 4.023874 u - 4.028204 u = -0.004330 u
Calculate the Q value:
Q = -(m ) c 2 MeV/c 2 = -(- 0.004330 u ) c 931.5 u
2
= 4.03 MeV
(b) Proceed as in (a) to obtain:
Q = -(m ) c 2 MeV/c 2 2 c = -(- 0.019703 u ) 931.5 u = 18.4 MeV
1382 Chapter 40
(c) Proceed as in (a) to obtain:
Q = -(m ) c 2 MeV/c 2 2 c = -(- 0.005135 u ) 931.5 u = 4.78 MeV
*47 Picture the Problem We can use Q = -(m )c 2 to find the Q values for this reaction. (a) The masses of the atoms are:
m 14 C = 14.003 242 u m14 N = 14.003 074 u
Calculate the increase in mass:
m = mf - mi = 14.003 074 u - 14.003 242 u = -0.000168 u
Calculate the Q value:
Q = -(m )c 2 MeV/c 2 931.5 = -(- 0.000168 u ) c u
2
= 0.156 MeV
(b)
The masses given are for atoms, not nuclei, so for nuclear masses the masses are too large by the atomic number times the mass of an electron. For the given nuclear reaction, the mass of the carbon atom is too large by 6me and the mass of the nitrogen atom is too large by 7 me . Subtracting 6me 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 1me from the right side of the equation.
48 Picture the Problem We can use Q = -(m ) c 2 to find the Q values for this reaction. (a) The masses of the atoms are:
m13 N = 13.005 738 u m13 C = 13.003 354 u
Nuclear Physics 1383
For + decay:
Q = (mi - mf - 2me ) c 2
= (mi - mf )c 2 - 2me c 2
Calculate mi - mf:
mi - mf = 13.005 738 u - 13.003 354 u
= 0.002 384 u
Calculate the Q value:
MeV/c 2 - 2(0.511 MeV ) = 1.20 MeV Q = (0.002 384 u )c 2 931.5 u
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 only has
(b)
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
*49 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: Substitute numerical values and evaluate N:
N=
P Eper fission
N=
500 MW 200 MeV
J 1 eV 5 108 s 1.60 10 -19 J = 200 MeV = 1.56 1019 s -1
1384 Chapter 40
50 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:
(1.1)N
=2 N ln 1.1 = ln 2 ln 2 N= = 7.27 ln 1.1 = 10 N ln 1.1 = ln 10 ln 10 N= = 24.2 ln 1.1 = 100 N ln 1.1 = ln 100 ln 100 N= = 48.3 ln 1.1
(b) Set 1.1N equal to 10 and solve for N:
(1.1)N
(c) Set 1.1N equal to 100 and solve for N:
(1.1)N
(d) Multiply the number of generations by the generation time:
t2 = Nt1 = (7.27 )(1 ms ) = 7.27 ms t10 = Nt1 = (24.2 )(1 ms ) = 24.2 ms t100 = Nt1 = (48.3)(1 ms ) = 48.3 ms
(e) Multiply the number of generations by the generation time:
t2 = Nt1 = (7.27 )(100 ms ) = 0.727 s t10 = Nt1 = (24.2 )(100 ms ) = 2.42 s t100 = Nt1 = (48.3)(100 ms ) = 4.83 s
*51 Picture the Problem We can use Q = -(m)c2, where m = mf - mi, value. The Q value is given by:
to calculate the Q
Q = -(m ) c 2
931.5 MeV/c 2 1u
Calculate the change in mass m:
Nuclear Physics 1385
m = mf - mi = 94.905842 u + 138.906348 u + 2(1.008665 u ) - (235.043923 u + 1.008665 u ) = -0.223068 u
Substitute for m and evaluate Q:
Q = -(- 0.223068 u ) = 208 MeV
931.5 MeV 1u
The ratio of Q to U found in Problem 23 is:
Q 208 MeV = = 88.1% U 236 MeV
52 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 = 1 N 2
where N is the number of reactions per second.
The number of reactions per second is:
J 1eV 4 -19 s 1.60 10 J N = 2 3.27 MeV + 4.03 MeV 12 -1 = 6.85 10 s
Substitute for N and evaluate Nn:
Nn =
1 2
(6.85 10
12
s -1
)
= 3.43 1012 neutrons/s
53 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 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:
Q = 3.27 MeV + 4.03 MeV + 17.6 MeV = 24.9 MeV
1386 Chapter 40
The number of H atoms in 4 L of H2O is: Substitute numerical values and evaluate NH:
mH 2O N H = 2 18 g/mol N A
4 kg 23 26 N H = 2 18 g/mol 6.02 10 atoms/mol = 2.676 10
(
)
The number of D atoms in 4 L of H2O is:
N D = 1.5 10-4 N H
-4
( ) = (1.5 10 )(2.676 10 )
26
= 4.01 10 22
The energy produced is given by:
E=
ND Q 5
Substitute numerical values and evaluate E:
4.01 10 22 (24.9 MeV ) E= 5 1.60 10 -19 J = 1.997 10 23 MeV eV
= 3.20 1010 J
*54 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: Apply conservation of momentum to obtain: Solve equation (2) for vHe:
2 2 18.6 MeV = 1 mHe vHe + 1 mn vn 2 2
= K He + K n
(1) (2)
mHe vHe + mn vn = 0
vHe
m 2 mv 2 = - n n vHe = n vn m mHe He
2
2
2 Substitute for vHe in equation (1):
m 2 2 18.6 MeV = 1 mHe n vn + 1 mn vn 2 2 m He
or
Nuclear Physics 1387
m 2 18.6 MeV = 1 mn vn 1 + n 2 m He m = K n 1 + n m He
Solve for Kn:
Kn =
18.6 MeV m 1+ n mHe 18.6 MeV = 14.86 MeV 1.008665 u 1+ 4.002603 u
Substitute numerical values for mn and mHe and evaluate Kn:
Kn =
Use equation (1) to find KHe:
K He = 18.6 MeV - K n = 18.6 MeV - 14.86 MeV = 3.74 MeV
55 Picture the Problem Adding the three reactions will yield their net effect. We can use (m)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:
1
H + 1H + 1H + 2H + 1H + 3He 2H + + + e + 3He + + 4He +
+
+ e
Simplify to obtain:
41 H 4 He + 2 + + 2 e +
(b) Express the rest energy released in this cycle:
(m )c 2 = (4mp - m - 4me )c 2
Use Table 40-1 to find the masses of the participants in the reaction and evaluate (m)c2:
(m )c 2 = [4(1.007825 u ) - 4.002603 u ]c 2 931.5 MeV/c
u
2
- 4(0.511 MeV )
= 24.7 MeV
1388 Chapter 40
(c) Express the rate R of proton consumption:
R=
P E
(1)
where E is the energy released per proton in fusion. Find N, the number of protons in the sun:
N=
1 2
msun 1 1.99 1030 kg = 2 mp 1.67 10-27 kg
(
)
= 5.96 1056
where we have assumed that protons constitute about half of the total mass of the sun. The energy released per proton in fusion is:
E=
1 4
(26.7 MeV ) = 6.675 MeV
1.60 10 -19 J = 6.675 MeV eV -12 = 1.07 10 J
Substitute numerical values in equation (1) and evaluate R: The time T for the consumption of all protons is:
R=
4 10 26 W = 3.74 1038 s -1 -12 1.07 10 J
T=
N 5.96 1056 = R 3.74 1038 s -1 1y = 1.59 1018 s 31.56 Ms = 5.04 1010 y
General Problems
56 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 MeVfm and hc = 1240 MeVfm (a) Evaluate ke2 to obtain:
ke 2 = 8.99 109 N m 2 / C 2 1.60 10 -19 C = 2.307 10 -28 N m 2 1 eV = 2.307 10 -28 J m = 1.44 10 -9 eV m 1.60 10 -19 J 1 fm 1 MeV = 1.44 10 -9 eV m -15 6 = 1.44 MeV fm 10 m 10 eV
(
)(
)
2
Nuclear Physics 1389
(b) Evaluate hc to obtain:
hc = 6.63 10 -34 J s 3 108 m/s = 1.99 10-25 J m = 1.99 10-25 J m
(
)(
)
1eV = 1240 10 -9 eV m -19 1.60 10 J 1fm 1 MeV 6 = 1240 MeV fm -15 10 m 10 eV
= 1240 10-9 eV m
*57 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: The activity of the source is given by:
=
ln 2 ln 2 = = 0.0693 s -1 t1 2 10 s
-1 R = R0 e - t = (6400 Bq ) e - (0.0693 s )t
-1 The following graph of R = (6400 Bq ) e - (0.0693 s )t was plotted using a spreadsheet
program.
7000 6000 5000 R (Bq) 4000 3000 2000 1000 0 0 10 20 30 t (s) 40 50 60
58 Picture the Problem The energy needed to remove a neutron is given by Q = (m ) c 2 where m is the difference between the sum of the masses of the reaction products and the mass of the target nucleus.
1390 Chapter 40
(a) The reaction is: The masses are (see Table 40-1):
4
He + Q= 3 He + n
m 4 He = 4.002603 u m 3 He = 3.016029 u
mn = 1.008665 u
Calculate the final mass:
mf = 3.016029 u + 1.008665 u = 4.024694 u
Calculate the increase in mass:
m = mf - mi = 4.024694 u - 4.002603 u = 0.022091 u
Calculate the energy Q needed to remove a neutron from 4He:
Q = (m )c 2 MeV/c 2 = (0.022091u )c 2 931.5 u = 20.6 MeV
(b) The reaction is: The masses are (see Table 40-1):
7
Li + Q= 6 Li + n
m 7 Li = 7.016004 u m 6 Li = 6.015122 u
mn = 1.008665 u
Calculate the final mass:
mf = 6.015122 u + 1.008665 u = 7.023787 u
Calculate the increase in mass:
m = mf - mi = 7.023787 u - 7.016004 u = 0.007783 u
Calculate the energy Q needed to remove a neutron from 4He:
Q = (m )c 2 931.5 MeV/c 2 = (0.007783 u )c 1u
2
= 7.25 MeV
Nuclear Physics 1391
59 Picture the Problem The maximum kinetic energy of the electron is given by K max = Q = m14 C - m14 N c 2 .
(
)
The maximum kinetic energy of the electron is the Q value for the reaction: Find the mass of each atom from Table 40-1: Calculate m = m14 C - m14 N :
K max = Q = m14 C - m14 N c 2
(
)
m14 C = 14.003242 u m14 N = 14.003074 u m = 14.003242 u - 14.003074 u = 0.000168 u Q = (m ) c 2 MeV/c 2 = (0.000168 u ) c 2 931.5 u = 156 keV
Calculate the maximum kinetic energy of the electron:
60 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 : Solve for R:
M = V = 4 R 3 3
where R is the radius of the star.
R=3
3M 4
In Problem 20 it was established that: Substitute numerical values and evaluate R:
= 1.174 1017 kg/m 3
R=3
3 1.99 10 30 kg 4 1.174 1017 kg/m 3
(
(
)
)
= 15.9 km
1392 Chapter 40
*61 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:
MeV/c 2 Q = - (mRh + m ) - mAg c 2 931.5 u
[
]
Substitute numerical values and evaluate Q:
Q = -[(4.002603 u + 104.905250 u ) - 108.904756 u ](931.5MeV/u ) = - 2.88 MeV
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, Q = K + K D . 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, m P > m + m D . Alpha decay cannot take place unless the total rest mass decreases. 62 Picture the Problem We can use E threshold = hf threshold = hc threshold , 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: Solve for the threshold wavelength:
Ethreshold = hf threshold =
hc
threshold
(1)
threshold =
hc Ethreshold
The threshold energy equals the binding energy of the deuteron:
Ethreshold = EB = mD - (mp + mn ) c 2
[
]
931.5 MeV/c 2 u
Substitute numerical values and evaluate Eth, the energy that must be added to the deuteron that will cause it to fission:
Ethreshold = [2.014102 u - (1.007825 u + 1.008665 u )](931.5 MeV/u ) = -2.22 MeV 1.60 10 -19 J = -3.55 10 -13 J eV
Nuclear Physics 1393
Substitute numerical values in equation (1) and evaluate threshold:
threshold =
(6.63 10
J s 3.00 108 m/s 3.55 10 -13 J
-34
)(
)
= 5.60 10 -13 m = 0.560 pm
63 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: Find N, the number of K nuclei in the student:
R = N 40 =
N 40 ln 2 t1 2
(1)
N = 0.0036
mN A M
where m is the mass of the student and M is the atomic mass of K. Substitute numerical values and evaluate N:
N = 0.0036
(60 kg )(6.02 1023 nuclei/mol) = 3.326 1024
39.098 g/mol
N 40 = Relative abundance N = 3.991 10 20 = (1.2 10 -4 )(3.326 10 24 )
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: Substitute numerical values in equation (1) and evaluate R:
R=
(3.99110 )ln 2
20
1.3 109 y
31.56 Ms y
= 6.74 103 Bq
64 Picture the Problem We can find the energy released in the reaction + + - Q 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: Substitute numerical values and evaluate Q:
Q = E = 2me c 2
1394 Chapter 40
Q = 2 9.11 10-31 kg 3 108 m/s = 1.64 10 -13 J
(
)(
)
2
1eV = 1.02 MeV 1.60 10-19 J
65 Picture the Problem We can use the fact that, after n half-lives, the decay rate of the 24Na n isotope is R = ( 1 ) R0 , where R0 is its decay rate at t = 0. 2 The counting rate after n half-lives is related to the initial counting rate: Divide both sides of the equation by the volume V of blood in the patient: 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: Solve for and evaluate V:
R = ( 1 ) R0 2
n
R 1 n R0 = (2 ) V V
60 Bq/mL = ( 1 ) 2
23
600 kBq V
V = (1 ) 2
23
600 kBq = 6.30 103 mL 60 Bq/mL
= 6.30 L
*66 Picture the Problem We can solve this problem in the center of mass reference frame for the general case of an particle in a head-on collision with a nucleus of atomic mass M u and then substitute data for a nucleus of 197Au and a nucleus of 10B. In the CM frame, the kinetic energy is:
K CM =
K lab K lab = 4u m 1+ 1+ M M kq1q2 k (2e )(Ze ) ke 2 2Z = = Rmin Rmin Rmin
At the point of closest approach:
K CM =
or, because ke2 = 1.44 MeVfm,
K CM =
(1.44 MeV fm )(2Z )
Rmin
Solve for Rmin to obtain:
Rmin =
(1.44 MeV fm )(2Z )
K CM
(1)
Nuclear Physics 1395
(a) Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab. Evaluate equation (1) for 197Au: Evaluate equation (1) for 10B:
Rmin =
(1.44 MeV fm )(2 79)
8 MeV
= 28.4 fm
Rmin =
(1.44 MeV fm )(2 5)
8 MeV
= 1.80 fm
(b) Find KCM for the 197Au nucleus:
K CM =
8 MeV = 7.841 MeV 4u 1+ 197 u
Substitute numerical values in equation (1) and evaluate Rmin:
Rmin =
(1.44 MeV fm )(2 79)
7.841 MeV
= 29.0 fm
Note that this result is about 2% greater that Rmin calculated ignoring recoil. Find KCM for the 10B nucleus:
K CM =
8 MeV = 5.714 MeV 4u 1+ 10 u
Substitute numerical values in equation (1) and evaluate Rmin:
Rmin =
(1.44 MeV fm )(2 5)
5.714 MeV
= 2.52 fm
Note that this result is about 40% greater that Rmin calculated ignoring recoil. 67 Picture the Problem The allowed energy levels in a one-dimensional infinite square well are given by Equation 35-13: En = n 2
h2 . 2 8mL
1396 Chapter 40
(a) The lowest energy of a nucleon of mass 1 u in the well corresponds to n = 1:
J s E1 = 2 -27 8(1 u ) 1.66 10 kg/u (3 fm ) 1eV = 3.678 10 -12 J 1.60 10 -19 J
(
(6.63 10
-34
) )
2
= 23.0 MeV
(b) Because neutrons are fermions, there can be only two per state:
E = 2(E1 + E2 + E3 + E4 + E5 + E6 ) = 2 E1 + 22 E1 + 32 E1 + 4 2 E1 + 52 E1 + 6 2 E1 = 182 E1 = 182(23.0 MeV ) = 4.19 GeV
(c) Find E for 4 protons and 4 neutrons:
(
) )
E = 4(E1 + E2 + E3 ) = 4 E1 + 22 E1 + 32 E1 = 56 E1 = 56(23.0 MeV ) = 1.29 GeV
(
68 Picture the Problem We can apply BE = (m)c2 to the model to find the binding energies and the binding energies/bond. (a) Find the binding energy BE for this model:
BE = (4m - m16 O )c 2
= [4(4.002603 u ) - 15.994915 u ]c 2 = (0.015497 u ) c 2
There are 6 bonds for the regular tetrahedron:
BE = 1 BE = 6 bond =
1 6
1 6
(0.015497 u )c 2
c
(0.015497 u )c 2 931.5 MeV/u 2
= 2.406 MeV
(b) 12C has 3 pairwise particle bonds. Find the total BE for 12C with this model: Calculate BE(4He):
BE
( C) = 3 BE( He)+ 3(2.406 MeV )
12 4
Nuclear Physics 1397 BE 4 He = 2(mp + mn ) - m 4 He c 2 = [2(1.007825 u + 1.008665 u ) - 4.002603 u ]c 2 = 28.30 MeV
Substitute numerical values and evaluate BE
(
) [
]
931.5 MeV/c 2 u
( C):
12
BE
( C) = 3(28.30 MeV ) + 3(2.406 MeV ) =
12
92.1 MeV
Use Table 40-1 to find BE
( C):
12
BE(12 C ) = 6(mp + mn ) - m12 C c 2 = [6(1.007825 u + 1.008665 u ) - 12.000000 u ]c 2 = 92.2 MeV
Note that this result is good agreement with the model. 69 Picture the Problem We can separate the variables in the differential equation dN/dt = Rp N and integrate to express N as a function of t. When dN/dt 0, Rp N = 0, a condition we can use to find N. (a) Separate the variables in the differential equation to obtain: Integrate the left side of the equation from 0 to N and the right side from 0 to t to obtain: Let u = Rp - N. Then:
[
]
931.5 MeV/c 2 u
dN = dt Rp - N
N
dN ' Rp - N ' = dt ' 0 0
t
du = -dN '
and
N
2 dN' 1 2 du 1 =- = - ln u Rp - N' l u l 0 1 1
l
l
=-
1
ln (Rp - N ')
N
0
1 1 = - ln (Rp - N ) + ln (Rp )
=
Rp ln Rp - N 1
1398 Chapter 40
Because
dt ' = t :
0
t
Rp =t ln Rp - N 1
N= Rp
Solve for N to obtain:
(1 - e )
- t
The following graph of N(t) = (Rp/)(1 -e-t) was plotted using a spreadsheet program. Note that N(t) approaches Rp/ in the same manner that the charge on a capacitor approaches the value CV.
1.2
1.0
0.8
0.6
N
0.4
0.2
0.0 0 1 2 3 4
t
(b) When dN/dt = 0:
Rp - N = 0 N =
Rp
The decay constant is:
=
ln 2 t1 2
Substitute for to obtain:
N =
Rp ln 2
t1 2
Substitute numerical values and evaluate N:
N =
100 s -1 60 s 10 min ln 2 min
= 8.66 104
Nuclear Physics 1399
*70 Picture the Problem The mass of 235U required is given by m235 =
N M 235 , where NA
M235 is the molecular mass of 235U and N is the number of fissions required to produce 7.01019 J. Relate the mass of 235U required to the number of fissions N required:
m235 =
N M 235 NA
(1)
where M235 is the molecular mass of 235U. Determine N:
N=
Eannual Eper fission
Substitute numerical values and evaluate N:
N=
7.0 1019 J 1.60 10-19 J 200 MeV eV 30 = 2.18 10
Substitute numerical values in equation (1) and evaluate m235:
m235
2.18 1030 (235 g/mol) = 8.51105 kg = 23 6.02 10 nuclei/mol
71 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 2 the particle and the relationship E 2 = E0 + p 2 c 2 with E0 << pc to show that E pc. (a) In the ground state of a onedimensional infinite square well of length L: (b) Use de Broglie's relation to obtain: Substitute numerical values and evaluate p: (c) Relate the total energy of the electron to its rest energy and
= 2 L = 2(2 fm ) = 4.00 fm
p=
h
=
hc c
p=
1240 eV nm = 310 MeV/c (4 fm )c
E2 E 2 = E02 + p 2c 2 = p 2c 2 1 + 2 0 2 pc
1400 Chapter 40
momentum: Because E0 << pc:
E 2 p 2c 2 E pc K = E - E0 E = pc = (310 MeV/c )c = 310 MeV
(d) The kinetic energy of an electron in the ground state of this well is given by:
72 Picture the Problem When a single proton is removed from a 12C nucleus, a 11B nucleus remains and we can use Q = mc 2 to determine the minimum energy required to remove a proton. The nuclear reaction is: The minimum energy Q required is: Substitute numerical values and evaluate Q:
12 6
C + Q11 B+1 H 5 1
Q = m11 B + m1 H - m12 C c 2
(
)
931.5 MeV Q = [(11.009306 u + 1.007825 u ) - 12.000000 u ] = 16.0 MeV u
*73 Picture the Problem The momentum of the electron is related to its total energy through E 2 = p 2 c 2 + E02 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: The total relativistic energy E of the electron is the sum of its kinetic energy and its rest energy: Substitute for E in equation (1) to obtain: Solve for p:
E 2 = p 2 c 2 + E02
(1)
E = K + E0
(K + E0 )2 = p 2c 2 + E02
p= K ( K + 2 E0 ) c
Substitute numerical values and evaluate p:
Nuclear Physics 1401 p=
(0.782 MeV )(0.782 MeV + 2 0.511MeV )
c
Kp =
2 pp
= 1.188 MeV/c
(b) Because pp = -pe:
2mp
Substitute numerical values (see Table 7-1 for the rest energy of a proton) and evaluate Kp: (c) The percent correction is:
Kp =
(1.188 MeV/c )2
2 938.28 MeV/c 2
(
)=
752 eV
Kp K
=
752 eV = 0.0962% 0.782 MeV
74 Picture the Problem Conservation of momentum and conservation of energy allow us to find the final velocities. Because the initial kinetic energy of the nucleus is zero, its final kinetic energy equals the energy lost by the neutron. (a) Apply conservation of momentum to the collision to obtain: Solve for V:
(m + M )V
= mvL
V =
mvL m+M
(b) In the CM frame, VMi = V and so: In the CM frame, Vf = -Vi and so: (c) Use conservation of momentum to obtain one relation for the final velocities: 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:
VMi = V VMf = - V mvL = mvf + MVMf
(1)
VMf - vf = -(VMi - vL ) = 0 + vL
and so
vf = VMf - vL
1402 Chapter 40
To eliminate vf, substitute in equation (1) : Solve for VMf:
mvL = m(VMf - vL ) + MVMf
V Mf =
2m vL M +m
(d) The kinetic energy of the nucleus after the collision in the laboratory frame is: Substitute for VMf and simplify to obtain:
2 K M = 1 MVMf 2
K Mf
2m = M vL M +m 4mM 1 2 mv = (M + m )2 2 L
1 2
2
(
)
(e) The fraction of the energy lost by the neutron in the elastic collision is given by:
4mM 1 2 2 mvL 4mM E - K Mf (M + m )2 = 1 2 = = = 2 1 E (M + m )2 2 mvL 2 mvL
(
)
4m M = 2 2 m m 1+ M 2 1 + M M 4mM
75 Picture the Problem We can use the result of Problem 74, part (e), to find the fraction f = Ef E0 of its initial energy lost per collision and then use this result to show that, after N collisions, E = (0.714)NE0. (a) Determine f = Ef E0 per collision: From Problem 74, part (e):
f =
E0 - E E = 1- E0 E0
4m
E = E0
m M 1 + M
4m
2
Substitute for E/E0 in the expression for f to obtain:
f = 1-
m M 1 + M
2
Nuclear Physics 1403
Substitute numerical values and evaluate f:
f = 1-
4(1.008665 u )
2
1.008665 u 12.000000 u 1 + 12.000000 u = 0.714
After N collisions: (b) Solve equation (1) for N:
E fN = f N E0 =
(0.714)N E0
(1)
E fN ln E 0 N= ln(0.714)
0.02 eV ln 2 MeV = 54.7 N= ln (0.714 )
Substitute numerical values and evaluate N:
55 head - on collisions are required to reduce the energy of the neutron from 2 MeV to 0.02 eV.
76 Picture the Problem We can use the result of Problem 74, part (e), to find the fraction f = Ef E0 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 f = Ef E0 per collision: In a collision with a hydrogen atom:
f =
E E0 - E = 1- E0 E0
E = 0.63 E0
and so
f = 0.37
N
After N collisions: Solve equation (1) for N:
E fN = f N E0 = (0.37 ) E0 E ln fN E N= 0 ln (0.37 )
(1)
(2)
1404 Chapter 40
Substitute numerical values and evaluate N:
0.02 eV ln 2 MeV = 18.5 N= ln (0.37 )
19 head - on collisions with an atom of hydrogen are required to reduce the energy of the neutron from 2 MeV to 0.02 eV.
(b) In a collision with a carbon atom:
E = 0.11 E0
and so
f = 0.89
Equation (2) becomes:
E ln fN E N= 0 ln (0.89 )
0.02 eV ln 2 MeV = 158 N= ln (0.89 )
Substitute numerical values and evaluate N:
158 head - on collisions with an atom of carbon are required to reduce the energy of the neutron from 2 MeV to 0.02 eV.
*77
Picture the Problem We can differentiate N B (t ) =
N A0A -At -Bt e -e with respect B - A
(
)
to t to show that it is the solution to the differential equation dNB/dt = ANA - BNB. (a) The rate of change of NB is the rate of generation of B nuclei minus the rate of decay of B nuclei. The generation rate is equal to the decay rate of A nuclei, which equals ANA. The decay rate of B nuclei is BNB. (b) We're given that:
dN B = A N A - B N B dt N N B (t ) = A0 A e - A t - e - B t B - A
(1)
(
)
(2) (3)
N A = N A0e - At
Nuclear Physics 1405
Differentiate equation (2) with respect to t to obtain:
d [N B (t )] = N A0A d e-At - e-Bt = N A0A - A e-At + Be-Bt B - A B - A dt dt
Substitute this derivative in equation (1) to get:
[(
)]
[
]
N N A0A - A e -At + Be -Bt = A N A0e -At - B A0 A e -At - e -Bt B - A B - A
[
]
(
)
Multiply both sides by
B - A and simplify to obtain: BA
+ Be -Bt =
N A0
B
[- e
A
-At
]
B - A N A0e - t - N A0 (e - t - e - t ) B N = N A0e - t - A0 A e - t - N A0e - t + N A0e - t B N = - A0 A e - t + N A0e - t B
A A B A A A B A B
=
N A0
B
[- e
A
-At
+ Be -Bt
]
which is an identity and confirms that equation (2) is the solution to equation (1).
If A > B the denominator and the expression in the parentheses are both
(c) negative for t > 0. If A < B the denominator and the expression in the
parentheses are both positive for t > 0.
1406 Chapter 40
(d) The following graph was plotted using a spreadsheet program.
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 5 10 Time , arbitrary units 15 20
Number of parent (A) nuclei Number of daughter ( B) nuclei
78 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 77 we have:
dN B = A N A - B N B = 0 for extrema dt
and
Replace ANA by A N A0e
- At
N A0A -At -Bt NB by e -e : B - A
(
)
A N A0e - t - B
A
N A0A -At -Bt e -e =0 B - A
(
)
Simplify to obtain:
e -At -
B (e - t - e - t ) = 0 B - A (B - A )e - t - B (e - t - e - t ) = 0
A B A A B
Remove the parentheses and combine like terms to obtain: Solve for t:
A e - t = Be - t
A B
t=
ln (B A ) B - A
Nuclear Physics 1407
Remarks: Note that all we've shown is that an extreme value exists at
t=
ln( B A ) . To show that this value for t maximizes NB, we need to either B - A
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) . 79 Picture the Problem We can show that, provided A >> B , e - At - e - Bt 1 and
A A and, hence, that NB = (A/B)NA. B - A B
We have, from Problem 77 (b):
N B (t ) =
N A A -At -Bt e -e B - A
(
)
(1)
Because A >> B : When several years have passed, because A t << 1 : Also, when A << B :
A << B
e - At - e - Bt 1
(2)
A A B - A B
N B (t ) =
(3)
Substitute (2) and (3) in (1) to obtain:
A N B A
1408 Chapter 40