@SOLUTIONSSTUDY
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 E b 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
3.31662 mal mb 3.31662 mal mb
pal pb
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.
E
hc
Λ
1
, Chapter 1 Solutions
where we substitute h
6.626 1034 J s, c 299 792
458 m s and E 195 keV
2
, Chapter 1 Solutions
1.01869 J–m
0.0636 Angstroms
KeV
■ Problem 1.5. 235 UFission Energy Release.
Using the reaction 235 U 117 Sn 118 Sn, and mass values, we calculate the mass defect of:
M 235 U M 117 Sn M 118 Sn M and an expected
energy release of Mc2.
931.5 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
3
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 E b 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
3.31662 mal mb 3.31662 mal mb
pal pb
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.
E
hc
Λ
1
, Chapter 1 Solutions
where we substitute h
6.626 1034 J s, c 299 792
458 m s and E 195 keV
2
, Chapter 1 Solutions
1.01869 J–m
0.0636 Angstroms
KeV
■ Problem 1.5. 235 UFission Energy Release.
Using the reaction 235 U 117 Sn 118 Sn, and mass values, we calculate the mass defect of:
M 235 U M 117 Sn M 118 Sn M and an expected
energy release of Mc2.
931.5 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
3
