Chapter 30 Assignment Solutions
Page 828 #45-48, 64-67
45) What force inside a nucleus acts to push the nucleus apart? What force inside the nucleus acts to hold the
nucleus together? (30.1)
The repulsive electric force between the positive protons acts to push the nucleus apart. The strong nuclear
force acts to hold the nucleus together.
46) Define the mass defect of a nucleus. To what is it related? (30.1)
The mass defect is the difference between the sum of the masses of the individual particles of the nucleus and
the mass of the nucleus. It is related to the binding energy of the nucleus by 𝐸 = 𝑚𝑐 2 .
47) Which are generally more unstable, small or large nuclei? (30.1)
Large nuclei are generally more unstable. The greater number of protons makes the repulsive electromagnetic
force overcome the attractive strong nuclear force.
48) Which isotope has the greater number of protons, uranium-235 or uranium-238? (30.1)
The both have the same number of protons, 92. However, they have different numbers of neutrons (U-235
has 143, U-238 has 146).
64) Which particles, and how many of each, make up an atom of 109
47Ag?
47 protons, 62 neutrons, 47 electrons
65) What is the isotopic symbol (the one used in nuclear equations) of a zinc atom composed of 30 protons
and 34 neutrons?
64
30Zn
66) The sulfur isotope 32
16S has a nuclear mass of 31.97207 u.
a) What is the mass defect of this isotope?
(Mass defect) = (isotope mass) – (mass of protons) – (mass of neutrons)
= 31.97207 u − 16(1.007276 u) − 16(1.008665 u)
= −0.282986 u
b) What is the binding energy of its nucleus?
(Binding energy) = (mass defect)(binding energy of 1 u)
= (−0.282986 u)(931.49 MeV/u)
= −263.60 MeV
c) What is the binding energy per nucleon?
−263.60 𝑀𝑒𝑉
= −8.237 MeV/nucleon
32 nucleons
67) A nitrogen isotope, 127N, has a nuclear mass of 12.0188 u.
a) What is the binding energy per nucleon?
(Mass defect) = (isotope mass) – (mass of protons) – (mass of neutrons)
= 12.0188 u − 7(1.007276 u) − 5(1.008665 u)
= −0.075457 u
(Binding energy) = (mass defect)(binding energy of 1 u)
= (−0.075457 u)(931.49 MeV/u)
= −70.287 MeV
−70.287 MeV
= −5.8573 MeV/nucleon
12 nucleons
b) Does it require more energy to separate a nucleon from a 147N nucleus or from a 127N nucleus?
a mass of 14.00307 u.
(Mass defect) = (isotope mass) – (mass of protons) – (mass of neutrons)
= 14.00307 u − 7(1.007276 u) − 7(1.008665 u)
= −0.108517 u
(Binding energy) = (mass defect)(binding energy of 1 u)
= (−0.108517 u)(931.49 MeV/u)
= −101.08 MeV
−101.08 MeV
= −7.2202 MeV/nucleon
14 nucleons
Therefore, removing a nucleon from 147N requires more energy.
14
7N
has
Page 828 #50-52, 54, 70-78
50) Radiation: What are the common names for an 𝛼 particle, 𝛽 particle, and 𝛾 radiation? (30.2)
An 𝛼 particle is a high-energy helium nucleus. A 𝛽 particle is a high-energy electron. 𝛾 radiation is a highenergy photon.
51) What two quantities must always be conserved in a nuclear equation? (30.2)
The atomic number (to conserve charge) and the mass number (to conserve the number of nucleons).
52) Nuclear Power: What sequence of events must occur for a chain reaction to take place? (30.2)
Many neutrons must be released by a fissioned nucleus and absorbed by neighboring nuclei, causing them to
undergo fission. Those neutrons are absorbed by more neighboring nuclei, and the process continues.
54) Fission and fusion are opposite processes. How can each release energy? (30.2)
When a large atom, such as uranium, undergoes fission, the mass of the product is less than that of the original
nucleus. An amount of energy equivalent to the difference in mass is released. However, when small nuclei
fuse into a larger nucleus, the mass of the more tightly bound, large nucleus is less, and the extra mass appears
as energy. The most tightly bound nucleus is iron-56. Fission or fusion that gets closer to iron-56 will result in
a loss of mass, and a corresponding release of energy.
70) Write the complete nuclear equation for the alpha decay of 222
86Rn.
222
86Rn
→
218
84Po
+ 42He
89
71) Write the complete nuclear equation for the beta decay of 36
Kr.
89
36Kr
→
89
37Rb
+ −10e + 00𝜈̅
72) Complete each nuclear reaction.
a)
b)
c)
d)
225
4
221
89Ac → 2He + 87Fr
227
0
0
227
88Ra → −1e + 0𝜈̅ + 89Ac
65
1
66
1
65
29Cu + 0n → 29Cu → 1p + 28Ni
137
235
1
96
1
92U + 0n → 40Zr + 3( 0n) + 52Te
73) An isotope has a half-life of 3.0 days. What percent of the original material will be left after
a) 6.0 days?
6.0
1 2.0
9.0
1 3.0
12
1 4.0
6.0 days is 3.0 =2.0 half-lives, so (2)
(100%) = 25% will remain.
b) 9.0 days?
9.0 days is 3.0 =3.0 half-lives, so (2)
(100%) = 12% will remain.
c) 12 days?
9.0 days is 3.0 =4.0 half-lives, so (2)
(100%) = 6.2% will remain.
74) In an accident in a research laboratory, a radioactive isotope with a half-life of three days is spilled. As a
result, the radiation is eight times the maximum permissible amount. How long must workers wait before they
can enter the room?
1
For the activity to fall to 8 its present amount, you must wait three half-lives, or nine days.
75) When a boron isotope, 115B, is bombarded with protons, it absorbs a proton and emits a neutron.
a) What element is formed?
Carbon
b) Write the nuclear equation for this reaction.
11
1
11
1
5B + 1p → 6C + 0n
c) The isotope formed is radioactive and decays by emitting a positron. Write the complete nuclear
equation for this reaction.
11
11
0
0
6C → 5B + 1𝑒̅ + 0𝜈
76) The first atomic bomb released an energy equivalent to 2.0 × 101 kilotons of TNT. One kiloton of TNT is
equivalent to 5.0 × 1012 J. Uranium-235 releases 3.21 × 10−11 J/atom. What was the mass of the uranium235 that underwent fission to produce the energy of the bomb?
𝐸 = (2.0 × 101 kton TNT) (5.0 × 1012
J
) = 1.0 × 1014 J
kton TNT
1 atom
𝑛 = (1.0 × 1014 J) (
) = 3.12 × 1024 atoms
3.21 × 10−11 J
The mass of each atom is approximately 235 u, or 235(1.66 × 10−27 kg) = 3.90 × 10−25 kg
Therefore, the total mass is approximately (3.12 × 1024 atoms)(3.90 × 10−25 kg/atom) = 1.2 kg
77) During a fusion reaction, two deuterons, 21H, combine to form a helium isotope, 32He. What other particle
is produced.
2
2
1H + 1H
→ 32He + 10n
Therefore, a neutron must also be produced.
78) 209
84Po has a half-life of 103 years. How long would it take for a 100-g sample to decay so that only 3.1 g of
Po-209 was left?
1 𝑡/𝜏
𝑚 = 𝑚0 ( )
2
𝑚
1 𝑡/𝜏
=( )
𝑚0
2
𝑚
1 𝑡/𝜏
ln ( ) = ln (( ) )
𝑚0
2
𝑚
𝑡
1
ln ( ) = ln ( )
𝑚0
𝜏
2
𝑡=𝜏∙
𝑚
ln (𝑚 )
0
1
ln (2)
3.1 g
ln (
)
100 g
𝑡 = (103 years) ∙
1
ln ( )
2
𝑡 = 516.19 years
To one significant figure (because of 100 g), it would take about 500 years for the sample to decay.
Page 828 #44, 56-58, 85-88
44) Organize the following terms into the concept map: Standard Model, quarks, gamma rays, force carriers,
protons, neutrons, leptons, W bosons, neutrinos, electrons, gluons.
56) Forces: In which of the four interactions (strong, weak, electromagnetic, and gravitational) do the
following particles take part? (30.3)
a) Electron
Electromagnetic, weak, gravitational
b) Proton
Strong, electromagnetic, gravitational
c) Neutrino
Weak, gravitational (due to bending of space-time of very massive objects)
57) What happens to the atomic number and mass number of a nucleus that emits a positron? (30.3)
The atomic number decreases by 1 (𝑍 → 𝑍 − 1), and there is no change in the mass number (𝐴 → 𝐴).
58) Antimatter: What would happen if a meteorite made of antiprotons, antineutrons, and positrons landed
on Earth? (30.3)
It would annihilate with an equivalent amount of matter, producing an extremely large amount of energy.
85) Each of the following nuclei can absorb an 𝛼 particle. Assume that no secondary particles are emitted by
the nucleus. Complete each equation.
a)
b)
14
4
18
7N → 2He + 9F
31
27
4
13Al → 2He + 15P
86)
211
86Rn
60
has a half-life of 15 h. What fraction of a sample would be left after 60 h?
1 4.0
60 h is 15 =4.0 half-lives, so (2)
1
= 16 will remain.
87) One of the simplest fusion reactions involves the production of deuterium, 21H (2.014102 u), from a
neutron and a proton. Write the complete fusion reaction and find the amount of energy released.
1
1
0n + 1p
→ 21H
(Mass defect) = (isotope mass) – (mass of protons) – (mass of neutrons)
= 2.014102 u − 1(1.007276 u) − 1(1.008665 u)
= −0.001839 u
(Binding energy) = (mass defect)(binding energy of 1 u)
= (−0.001839 u)(931.49 MeV/u)
= −1.7130 MeV
Therefore, 1.7130 MeV is released.
228
88) A 232
92U nucleus, mass = 232.0372 u, decays to 90Th, mass = 228.0287 u, by emitting an 𝛼 particle, mass =
4.0026 u, with a kinetic energy of 5.3 MeV. What must be the kinetic energy of the recoiling thorium nucleus?
The kinetic energy of both the 𝛼 particle and thorium nucleus come from the conversion of mass into energy.
Δ𝑚 = 𝑚𝛼 + 𝑚 𝑇ℎ − 𝑚𝑈
Δ𝑚 = 4.0026 u + 228.0287 u − 232.0372 u
Δ𝑚 = −0.0059 u
The energy released is 𝐸 = (0.0059 u)(931.49 MeV/u) = 5.496 MeV
The kinetic energy of thorium is the released energy minus the kinetic energy of the 𝛼 particle.
𝐾𝑇ℎ = 5.5 MeV − 5.3 MeV
𝐾𝑇ℎ = 0.2 MeV