All 20 Chapters Covered
SOLUTION MANUAL
, Chapter 1 Solutions
Radiation Sources
■ Problem 1.1. Radiation Energy Spectra: Line vs. Continuous
Line (or discrete energy): a, c, d, e, f, and i.
Continuous energy: b, g, and h.
■ Problem 1.2. Conversion electron energies compared.
Since the electrons in outer shells are bound less tightly than those in closer shells, conversion electrons from outer shells will
have greater emerging energies. Thus, the M shell electron will emerge with greater energy than a K or L shell electron.
■ Problem 1.3. Nuclear decay and predicted energies.
We write the conservation of energy and momentum equations and solve them for the energy of the alpha particle. Momentum is
given the symbol "p", and energy is "E". For the subscripts, "al" stands for alpha, while "b" denotes the daughter nucleus.
pal2 pb2
pal pb 0 Eal Eb Eal Eb Q and Q 5.5 MeV
2 mal 2 mb
Solving our system of equations for Eal, Eb, pal, pb, we get the solutions shown below. Note that we have two possible sets of
solutions (this does not effect the final result).
mal 5.5 mal
Eb 5.5 1 Eal
mal mb mal mb
3.31662 mal mb 3.31662 mal mb
pal pb
mal mb mal mb
We are interested in finding the energy of the alpha particle in this problem, and since we know the mass of the alpha particle and
the daughter nucleus, the result is easily found. By substituting our known values of mal 4 and mb 206 into our derived
Ealequation we get:
Eal 5.395 MeV
Note : We can obtain solutions for all the variables by substituting mb 206 and mal 4 into the derived equations above :
Eal 5.395 MeV Eb 0.105 MeV pal 6.570 amu MeV pb 6.570 amu MeV
■ Problem 1.4. Calculation of Wavelength from Energy.
Since an x-ray must essentially be created by the de-excitation of a single electron, the maximum energy of an x-ray emitted in a
tube operating at a potential of 195 kV must be 195 keV. Therefore, we can use the equation E=h, which is also E=hc/Λ, or
Λ=hc/E. Plugging in our maximum energy value into this equation gives the minimum x-ray wavelength.
hc
Λ where we substitute h 6.626 1034 J s, c 299 792 458 m s and E 195 keV
E
1
, Chapter 1 Solutions
1.01869 J–m
Λ 0.0636 Angstroms
KeV
235
■ Problem 1.5. UFission Energy Release.
235 117 118
Using the reaction U Sn Sn, and mass values, we calculate the mass defect of:
M 235 U M 117 Sn M 118 Sn Mand an expected
energy release of Mc2.
931.5 MeV
223 MeV
AMU
This is one of the most exothermic reactions available to us. This is one reason why, of course, nuclear power from uranium
fission is so attractive.
■ Problem 1.6. Specific Activity of Tritium.
Here, we use the text equation Specific Activity = (ln(2)*Av)/ T12*M), where Av is Avogadro's number, T12 is the half-life of the
isotope, and M is the molecular weight of the sample.
ln2 Avogadro ' s Constant
Specific Activity
T12 M
3 grams
We substitute T12 12.26 years and M= to get the specific activity in disintegrations/(gram–year).
mole
1.13492 1022
Specific Activity
gram –year
The same result expressed in terms of kCi/g is shown below
9.73 kCi
Specific Activity
gram
■ Problem 1.7. Accelerated particle energy.
The energy of a particle with charge q falling through a potential V is qV. Since V= 3 MV is our maximum potential difference, the
maximum energy of an alpha particle here is q*(3 MV), where q is the charge of the alpha particle (+2). The maximum alpha
particle energy expressed in MeV is thus:
Energy 3 Mega Volts 2 Electron Charges 6. MeV
2
, Chapter 1 Solutions
2 1 1
■ Problem 1.8. Photofission of deuterium. 1D Γ 0 n 1 p + Q (-2.226 MeV)
2 0 1 1
The reaction of interest is D Γ n p+ Q (-2.226 MeV). Thus, the Γ must bring an energy of at least 2.226 MeV
1 0 0 1
in order for this endothermic reaction to proceed. Interestingly, the opposite reaction will be exothermic, and one can expect to
find 2.226 MeV gamma rays in the environment from stray neutrons being absorbed by hydrogen nuclei.
■ Problem 1.9. Neutron energy from D-T reaction by 150 keV deuterons.
We write down the conservation of energy and momentum equations, and solve them for the desired energies by eliminating the
momenta. In this solution, "a" represents the alpha particle, "n" represents the neutron, and "d" represents the deuteron (and, as
before, "p" represents momentum, "E" represents energy, and "Q" represents the Q-value of the reaction).
pa2 pn2 pd 2
pa pn pd Ea En Ed Ea En Ed Q
2 ma 2 mn 2 md
Next we want to solve the above equations for the unknown energies by eliminating the momenta. (Note : Using computer
software such as Mathematica is helpful for painlessly solving these equations).
We evaluate the solution by plugging in the values for particle masses (we use approximate values of "ma," "mn,"and "md" in
AMU, which is okay because we are interested in obtaining an energy value at the end). We define all energies in units of MeV,
namely the Q-value, and the given energy of the deuteron (both energy values are in MeV). So we substitute ma = 4, mn = 1, md
= 2, Q = 17.6, Ed = 0.15 into our momenta independent equations. This yields two possible sets of solutions for the energies (in
MeV). One corresponds to the neutron moving in the forward direction, which is of interest.
En 13.340 MeV Ea 4.410 MeV
En 14.988 MeV Ea 2.762 MeV
Next we solve for the momenta by eliminating the energies. When we substitute ma = 4, mn = 1, md = 2, Q = 17.6, Ed = 0.15 into
these equations we get the following results.
pd 1 1
pn 2 3 pd 2 352 pa 8 pd 2 2 3 pd 2 352
5 5 10
We do know the initial momentum of the deuteron, however, since we know its energy. We can further evaluate our solutions for
pn and pa by substituting:
pd
The particle momenta ( in units of amuMeV ) for each set of solutions is thus:
pn 5.165 pa 5.940
pn 5.475 pa 4.700
The largest neutron momentum occurs in the forward (+) direction, so the highest neutron energy of 14.98 MeV corresponds
to this direction.
3
SOLUTION MANUAL
, Chapter 1 Solutions
Radiation Sources
■ Problem 1.1. Radiation Energy Spectra: Line vs. Continuous
Line (or discrete energy): a, c, d, e, f, and i.
Continuous energy: b, g, and h.
■ Problem 1.2. Conversion electron energies compared.
Since the electrons in outer shells are bound less tightly than those in closer shells, conversion electrons from outer shells will
have greater emerging energies. Thus, the M shell electron will emerge with greater energy than a K or L shell electron.
■ Problem 1.3. Nuclear decay and predicted energies.
We write the conservation of energy and momentum equations and solve them for the energy of the alpha particle. Momentum is
given the symbol "p", and energy is "E". For the subscripts, "al" stands for alpha, while "b" denotes the daughter nucleus.
pal2 pb2
pal pb 0 Eal Eb Eal Eb Q and Q 5.5 MeV
2 mal 2 mb
Solving our system of equations for Eal, Eb, pal, pb, we get the solutions shown below. Note that we have two possible sets of
solutions (this does not effect the final result).
mal 5.5 mal
Eb 5.5 1 Eal
mal mb mal mb
3.31662 mal mb 3.31662 mal mb
pal pb
mal mb mal mb
We are interested in finding the energy of the alpha particle in this problem, and since we know the mass of the alpha particle and
the daughter nucleus, the result is easily found. By substituting our known values of mal 4 and mb 206 into our derived
Ealequation we get:
Eal 5.395 MeV
Note : We can obtain solutions for all the variables by substituting mb 206 and mal 4 into the derived equations above :
Eal 5.395 MeV Eb 0.105 MeV pal 6.570 amu MeV pb 6.570 amu MeV
■ Problem 1.4. Calculation of Wavelength from Energy.
Since an x-ray must essentially be created by the de-excitation of a single electron, the maximum energy of an x-ray emitted in a
tube operating at a potential of 195 kV must be 195 keV. Therefore, we can use the equation E=h, which is also E=hc/Λ, or
Λ=hc/E. Plugging in our maximum energy value into this equation gives the minimum x-ray wavelength.
hc
Λ where we substitute h 6.626 1034 J s, c 299 792 458 m s and E 195 keV
E
1
, Chapter 1 Solutions
1.01869 J–m
Λ 0.0636 Angstroms
KeV
235
■ Problem 1.5. UFission Energy Release.
235 117 118
Using the reaction U Sn Sn, and mass values, we calculate the mass defect of:
M 235 U M 117 Sn M 118 Sn Mand an expected
energy release of Mc2.
931.5 MeV
223 MeV
AMU
This is one of the most exothermic reactions available to us. This is one reason why, of course, nuclear power from uranium
fission is so attractive.
■ Problem 1.6. Specific Activity of Tritium.
Here, we use the text equation Specific Activity = (ln(2)*Av)/ T12*M), where Av is Avogadro's number, T12 is the half-life of the
isotope, and M is the molecular weight of the sample.
ln2 Avogadro ' s Constant
Specific Activity
T12 M
3 grams
We substitute T12 12.26 years and M= to get the specific activity in disintegrations/(gram–year).
mole
1.13492 1022
Specific Activity
gram –year
The same result expressed in terms of kCi/g is shown below
9.73 kCi
Specific Activity
gram
■ Problem 1.7. Accelerated particle energy.
The energy of a particle with charge q falling through a potential V is qV. Since V= 3 MV is our maximum potential difference, the
maximum energy of an alpha particle here is q*(3 MV), where q is the charge of the alpha particle (+2). The maximum alpha
particle energy expressed in MeV is thus:
Energy 3 Mega Volts 2 Electron Charges 6. MeV
2
, Chapter 1 Solutions
2 1 1
■ Problem 1.8. Photofission of deuterium. 1D Γ 0 n 1 p + Q (-2.226 MeV)
2 0 1 1
The reaction of interest is D Γ n p+ Q (-2.226 MeV). Thus, the Γ must bring an energy of at least 2.226 MeV
1 0 0 1
in order for this endothermic reaction to proceed. Interestingly, the opposite reaction will be exothermic, and one can expect to
find 2.226 MeV gamma rays in the environment from stray neutrons being absorbed by hydrogen nuclei.
■ Problem 1.9. Neutron energy from D-T reaction by 150 keV deuterons.
We write down the conservation of energy and momentum equations, and solve them for the desired energies by eliminating the
momenta. In this solution, "a" represents the alpha particle, "n" represents the neutron, and "d" represents the deuteron (and, as
before, "p" represents momentum, "E" represents energy, and "Q" represents the Q-value of the reaction).
pa2 pn2 pd 2
pa pn pd Ea En Ed Ea En Ed Q
2 ma 2 mn 2 md
Next we want to solve the above equations for the unknown energies by eliminating the momenta. (Note : Using computer
software such as Mathematica is helpful for painlessly solving these equations).
We evaluate the solution by plugging in the values for particle masses (we use approximate values of "ma," "mn,"and "md" in
AMU, which is okay because we are interested in obtaining an energy value at the end). We define all energies in units of MeV,
namely the Q-value, and the given energy of the deuteron (both energy values are in MeV). So we substitute ma = 4, mn = 1, md
= 2, Q = 17.6, Ed = 0.15 into our momenta independent equations. This yields two possible sets of solutions for the energies (in
MeV). One corresponds to the neutron moving in the forward direction, which is of interest.
En 13.340 MeV Ea 4.410 MeV
En 14.988 MeV Ea 2.762 MeV
Next we solve for the momenta by eliminating the energies. When we substitute ma = 4, mn = 1, md = 2, Q = 17.6, Ed = 0.15 into
these equations we get the following results.
pd 1 1
pn 2 3 pd 2 352 pa 8 pd 2 2 3 pd 2 352
5 5 10
We do know the initial momentum of the deuteron, however, since we know its energy. We can further evaluate our solutions for
pn and pa by substituting:
pd
The particle momenta ( in units of amuMeV ) for each set of solutions is thus:
pn 5.165 pa 5.940
pn 5.475 pa 4.700
The largest neutron momentum occurs in the forward (+) direction, so the highest neutron energy of 14.98 MeV corresponds
to this direction.
3
