Physics 12 Assignment
Chapters 20 & 21
KEY
Chapter 20
7. Define the following terms:
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Proton - a positively charged
particle found in the nucleus of all
atoms
Neutron - a particle with zero
charge, found in the
nucleus of all atoms
Nucleon – a particle that resides in
the nucleus of an atom
Chemical symbol - a shorthand
symbol for an element
Atomic number – the number of
protons in the nucleus of an atom
Atomic mass number – see Mass
number
Nucleon number - the total
number of nucleons (protons and
neutrons) in the nucleus; also
called atomic mass number
Strong Nuclear Force – the
fundamental force that holds the
parts of the nucleus together
Nuclide - the nucleus of a
particular atom, as characterized
by its atomic number and atomic
Mass number – the total number
of neutrons and protons in an
atomic nucleus
Isotope - two or more atoms of an
element that have the same
number of protons but a different
number of neutrons in their nuclei
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Mass defect - the difference
between the mass of a nucleus and
the sum of the masses of its
constituent particles; the mass
equivalent of the binding energy of
a nucleus
Atomic mass unit - the value of
mass equal to mass of the most
common carbon isotope (C-12)
divided by 12; 1 u = 1.6605 × 10−27
kg
Alpha particle - one or more
helium nuclei ejected from a
radioactive nucleus
Beta particle - high speed
electrons or positrons ejected from
a radioactive nucleus
Gamma ray - high frequency
electromagnetic wave emitted
from a radioactive nucleus
Radioactive isotope - an isotope of
an element that has an unstable
nucleus and therefore
disintegrates or decays, emitting
alpha, beta, or gamma radiation
Parent nucleus - the initial nucleus
involved in a transmutation
reaction
Daughter nucleus - the nucleus
remaining after a transmutation
reaction
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Transmutation - the conversion of
one element into a different
element, usually as a result of
radioactive decay
Ionizing radiation - radiation of
sufficient energy to liberate the
electrons from the atoms or
molecules
Neutrino - a chargeless, very lowmass particle that accounts for all
the missing momentum and
energy of beta positive decay
Antineutrino - a chargeless, very
low-mass particle that accounts
for all the missing momentum and
energy of beta negative decay
Positron - a particle with the same
mass as the electron, but with a
positive charge; an antielectron
Half-life - the time in which the
amount of a radioactive nuclide
decays to half its original amount
Binding Energy - the amount of
energy that must be supplied to
nuclear particles in order to
separate them; equal to mass
defect
8. Different isotopes of a given element have the same number of protons and electrons. Because they have the
same number of electrons, they have almost identical chemical properties. Each isotope has a different number
of neutrons from other isotopes of the same element. Accordingly, they have different atomic masses.
9. Identify the element based on the atomic number.
(a) Uranium (A = 92) (b) Nitrogen (A = 7) (c) Hydrogen (A = 1) (d) Strontium (A = 38) (e) Berkelium (A = 97)
10. The number of protons is the same as the atomic number, and the number of protons is the mass number minus the
number of protons.
(a) Uranium: 92 protons, 140 neutrons (b) Nitrogen: 7 protons, 11 neutrons (c) Hydrogen: 1 proton, 0 neutrons
(d) Strontium: 38 protons, 44 neutrons (e) Berkelium: 97 protons, 150 neutrons
11. There must be some force holding the nucleus together. For all nuclei with atomic numbers greater than 1, there are
protons packed very close to each other. If the only force present between the protons were the electrostatic force, the
protons would repel each other and no nuclei would be stable. Since there are stable nuclei, there must be some other
force stronger than the electrostatic force, holding the nucleus together – the strong nuclear force.
12. Similarities between the electromagnetic force and the strong nuclear force: They will both act on charged particles.
Differences between the electromagnetic force and the strong nuclear force: The strong force is a short-range force, while
the electromagnetic force is an infinite-range force. The strong force is only attractive, while the electromagnetic force can
be either attractive or repulsive. The strong force acts between all nucleons no matter whether they are charged or neutral,
while the electromagnetic force acts only on charged particles. The strong nuclear force is much stronger than the
electromagnetic force.
13. Radioactivity is found in every case to be unaffected by the strongest physical and chemical treatments, including
strong heating or cooling and the action of strong chemical reagents. Chemical reactions are a result of electron
interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity
is not due to electron interactions. Another piece of evidence is the fact that the α-particle is much heavier than an electron
and has a different charge than the electron, so it can’t be an electron. Therefore it must be from the nucleus. Finally, the
energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron
orbital transitions. All of these observations support radioactivity being a nuclear process.
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12.
15.
16. Neither hydrogen nor deuterium can emit an α particle. Hydrogen has only one nucleon (a proton) in its nucleus, and
deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither one has the necessary four nucleons
(two protons and two neutrons) to emit an α particle.
17. Many artificially produced radioactive isotopes are rare in nature because they have decayed away over time. If the
half-lives of these isotopes are relatively short in comparison with the age of Earth (which is typical for these isotopes),
then there won’t be any significant amount of these isotopes left to be found in nature. Note: the age of the Earth is
approximately 4.6 billion years.
18. After two months the sample will not have completely decayed. After one month half of the sample will remain, and
after two months, one-fourth of the sample will remain. Each month half of the remaining atoms decay.
Note: The mass of an alpha particle is 4.002602 u.
19. N-14 consists of seven protons and seven neutrons. We find the binding energy from the masses:
Mass defect ∆m = [7(m p + ) + 7(m n o )] − m 14 N = [7(1.007825 u) + 7(1.008665 u)] - (14.003074 u) = 0.112356 u
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0.112356 u * (1.6605 x 10
-27
kg/u) = 1.8656714 x 10-28 kg
Binding Energy ∆E = ∆mc2 = (1.8656714 x 10-28 kg)(3.00 x 108 m/s)2 = 1.6791042 x 10-11 J
1.6791042 x 10-11 J * (1 eV/1.602 x 10-19 J) = 104813000 eV ≈ 1.05 MeV
Thus the binding energy per nucleon is
1.04813000 eV
= 7.49 MeV
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20. Note: if the calculated binding energy is negative, a nucleus is unstable; if it is positive, then the nucleus
is stable
(a) We find the binding energy from the masses:
Mass defect ∆m = 2(m 4 He ) − m 8 Be = 2(4.002602 u) - (8.005305 u) = -1.01 x 10-4 u
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-1.01 x 10-4 u * (1.6605 x 10-27 kg/u) = -1.677105 x 10-31 kg
Binding Energy ∆E = ∆mc2 = (-1.677105 x 10-31 kg)(3.00 x 108 m/s)2 = -1.5093945 x 10-14 J
-1.5093945 x 10-14 J * (1 eV/1.602 x 10-19 J) = -94219.38 eV ≈ -0.094 MeV
Because the binding energy is negative, the nucleus is unstable.
(b) We find the binding energy from the masses:
Mass defect ∆m = 3(m 4 He ) − m 12 C = 3(4.002602 u) - (12.000000 u) = +0.007806 u
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+0.007806 u * (1.6605 x 10-27 kg/u) = +1.2961863 x 10-29 kg
Binding Energy ∆E = ∆mc2 = (+1.2961863 x 10-29 kg)(3.00 x 108 m/s)2 = +1.16656767 x 10-12 J
+1.16656767 x 10-12 J * (1 eV/1.602 x 10-19 J) = +7281945.51 eV ≈ +7.3 MeV
Because the binding energy is positive, the nucleus is stable.
21. The decay is 31 H → 23 He+ −01 e + ν . We find the energy released by calculating the mass defect:
∆m = m 3 H − (m 3 He + m 0 e + m ν ) = 3.016049 u – (3.016029 u + 0.0005485697079 u – 0 u)
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−1
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= -5.28569708 x 10 u
(Note: the mass of the antineutrino is extremely small and therefore negligible here.)
5.28569708 x 10-4 u * (1.6605 x 10-27 kg/u) = 8.7769 x 10-31 kg
Thus, the energy released is ∆E = ∆mc2 = (8.7769 x 10-31 kg)(3.00 x 108 m/s)2 = 7.89921 x 10-14 J
7.89921 x 10-14 J * (1 eV/1.602 x 10-19 J) = 493084.2696 eV ≈ 0.493 MeV
22. We find the final nucleus by balancing the mass and charge numbers:
So the final nucleus is Th-234.
If we ignore the recoil of Th-234 isotope, the kinetic energy of the alpha particle is equal to the binding energy:
5.32 MeV = 5.32 x 106 eV = ∆mc2. Thus, ∆m = (5.32 x 106 eV )(1 eV/1.602 x 10-19 J)/(3.00 x 108 m/s)2 =
9.4696 x 10-30 kg = 0.0057028606 u
Hence , ∆m = m 238 U − m 234 Th − m 4 He = 238.050783 u – m 234 Th - 4.002602 u = 0.0057028606 u.
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Thus, Th-234 has an atomic mass of 234.0424781 u
23. If we ignore the recoil of the lead isotope, the kinetic energy of the alpha particle is equal to the binding
energy. We must first find the mass defect:
Mass defect ∆m = m 210 Po − m 206 Pb − m 4 He = 209.982848 u – 205.9744440 u – 4.002603 u = 0.005801 u
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0.005801 u * (1.6605 x 10-27 kg/u) = 9.6325605 x 10-30 kg
Thus, the energy released is ∆E = ∆mc2 = (9.6325605 x 10-30 kg)(3.00 x 108 m/s)2 = 8.66930446 x 10-13 J
8.66930446 x 10-13 J * (1 eV/1.602 x 10-19 J) = 5411550.843 eV ≈ 5.41 MeV
24. Let T0.5 = half-life. Let No = the initial decay rate of 1280 decays/min. Let N = the final decay rate after 6
hours of 320 decays/min. Thus,
N = N o (0.5)
∆t
T0.5
320 decays/min = (1280 decays/min)(0.5)
6h
T0.5
6h
 320 
T
log
 = log(0.5) 0.5
 1280 
 320  6 h
log
log(0.5)
=
 1280  T0.5
∴ T0.5 =
(6 h) ∗ log(0.5)
= 3h
 320 
log

 1280 
25.
(a)
(b)
N = N o (0.5)
∆t
T0.5
∆t
N
= (0.5) T0.5
No
4 T0.5
N
= (0.5) T0.5 = (0.5) 4 = 0.0625
No
N = N o (0.5)
∆t
T0.5
∆t
N
= (0.5) T0.5
No
N
= (0.5)
No
4.5 T0.5
T0.5
= (0.5) 4.5 = 0.044194
Chapter 21
26. Define the following terms:
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Nuclear fission - the splitting of a large nucleus into two lighter nuclei plus two or more neutrons; usually caused by the
impact of a neutron and accompanied by the release of energy
Nuclear fusion - the formation of a larger nucleus from two lighter nuclei, accompanied by the release of energy
Critical - the condition that one neutron from each fission event causes one more nucleus to fission: the reaction will be
sustained at a constant rate
Subcritical - the condition that fewer than one neutron from each fission event causes another fission: the reactions will
eventually cease
Supercritical - the condition that more than one neutron from each fission event causes another fission: the reactions
rate will rise in a cascade of fissions
Thermal neutron - slow neutrons that have an energy of about 0.03 eV that are used to cause fission reactions
Moderator - a substance that will slow down neutrons
Enriched uranium - fissionable isotope of uranium with the percent of uranium-235 increased
Control rods - rods made of cadmium or boron used to keep fission reactions in nuclear reactors at the critical level; the
rods absorb exactly the right amount of neutrons so that exactly one neutron from each fission reaction causes one more
fission reaction
Primary coolant – a fluid that runs through the core of a nuclear reactor, removing the heat from the fuel rods and
carries it away to the boiler
Secondary coolant – the water in the boiler that is converted to steam to run electrical turbines
27. Neutrons are good projectiles for producing nuclear reactions because they are neutral and they are massive.
If you want a particle to hit the nucleus with a lot of energy, a more massive particle is the better choice. A light
electron would not be as effective. Using a positively charged projectile like an alpha or a proton means that the
projectile will have to overcome the large electrical repulsion from the positively charged nucleus. Neutrons can
penetrate directly to the nucleus and cause nuclear reactions.
28.
Power Generation Method: Fossil fuels
Pros
• inexpensive to build
• the technology is already working
Power Generation Method: Nuclear fission
Pros
• the technology is already working
• no air pollution
• no greenhouse gas emissions
Power Generation Method: Nuclear fusion
Pros
• no radioactive waste
• no air pollution
• no greenhouse gas emissions
Cons
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air pollution
greenhouse gas emissions
limited supply of coal and oil
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Cons
expensive to build
thermal pollution
disposal of radioactive waste
accidents are extremely dangerous, i.e. Three
Mile Island & Chernobyl
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Cons
the technology is not yet working in a
sustainable manner
expensive to build
29. The large amount of mass of a star creates an enormous gravitational attraction of all of that mass, which
causes the gas to be compressed to a very high density. This high density creates a very high pressure and high
temperature situation. The high temperatures give the gas particles a large amount of kinetic energy, which
allows them to break the Coulomb barrier and fuse when they collide. Thus, these conditions at the center of the
Sun and other stars make the fusion process possible.
30. The binding energy, Q, gives the energy released in the reaction, assuming the initial kinetic energy of the
neutron is negligible.
235
1
88
136
1
92 U + 0 n → 38 Sr + 54 Xe + 12 0 n
Q = m 1 n + m 235 U − m 88 Sr − m 136 Xe − 12m 1 n = (1.008665 u + 235.043923 u) - 87.90561 u – 135.907220 u 0
92
38
54
0
12(1.008665 u ) = 0.135774 u
0.135774 u * (1.6605 x 10-27 kg/u) = 2.25452727 x 10-28 kg
Thus, the energy released is ∆E = ∆mc2 = (2.25452727 x 10-28 kg)(3.00 x 108 m/s)2 = 2.02907454 x 10-11 J
2.02907454 x 10-11 J * (1 eV/1.602 x 10-19 J) = 126658835.4 eV ≈ 127 MeV
31.
2
1
H + 31 H → 42 He+ 01 n
Q = m 2 H + m 3 H − m 4 He − m 1 n = (2.014102 u + 3.016049 u) – 4.002603 u – 1.008665 u = 0.18883 u
1
1
2
0
0.18883 u * (1.6605 x 10-27 kg/u) = 3.13552215 x 10-29 kg
Thus, the energy released is ∆E = ∆mc2 = (3.13552215 x 10-29 kg)(3.00 x 108 m/s)2 = 2.82196994 x 10-12 J
2.82196994 x 10-12 J * (1 eV/1.602 x 10-19 J) = 17615292.98 eV ≈ 17.6 MeV