Solutions Manual for Radiation Detection and Measurement, 4th Edition – (Knoll, 2011) | All 20 Chapters Covered

m All 20 Chapters Covered
m m m




SOLUTION MANUAL
m

, Chapter 1 Solutions m m




Radiation Sources m




■ Problem 1.1. Radiation Energy Spectra: Line vs. Continuous
m m m m m m m




Line (or discrete energy): a, c, d, e, f, and i.
m m m m m m m m m m




Continuous energy: b, g, and h.
m m m m m m




■ Problem 1.2. Conversion electron energies compared.
m m m m m




Since the electrons in outer shells are bound less tightly than those in closer shells, conversion electrons from outer shells will
m m m m m m m m m m m m m m m m m m m m




mhave greater emerging energies. Thus, the M shell electron will emerge with greater energy than a K or L shell electron.
m m m m m m m m m m m m m m m m m m m m




■ Problem 1.3. Nuclear decay and predicted energies.
m m m m m m




We write the conservation of energy and momentum equations and solve them for the energy of the alpha particle. Momentum is
m m m m m m m m m m m m m m m m m m m m




given the symbol "p", and energy is "E". For the subscripts, "al" stands for alpha, while "b" denotes the daughter nucleus.
m m m m m m m m m m m m m m m m m m m m m




p al2 p b2
p al p b 0 m m
m
m m Eal Eb
m Eal Eb Qm m
m
m and Q 5.5 MeV
m m m



2 m al
m 2 mb m




Solving our system of equations for Eal, Eb, p al, p b, we get the solutions shown below. Note that we have two possible sets of
m m m m m m m m m m m m m m m m m m m m m m m




solutions (this does not effect the final result).
m m m m m m m m




m al 5.5 m al m


Eb 5.5 m
m m 1 m Eal m


m m mal mb m m




3.31662 m al mb 3.31662 m al mb
p al m m pb m
m




m m




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
m m m m m m m m m m m m m m m m m m m m m m m m




the daughter nucleus, the result is easily found. By substituting our known values of mal 4 and mb 206 into our derived
m m m m m m m m m m m m m m m mm m m mm m m m




Ealequation we get:
m m m




Eal 5.395 MeV m m m




Note : We can obtain solutions for all the variables by substituting mb 206 and mal 4 into the derived equations above :
m m m m m m m m m m m m mm m m mm m m m m m m




Eal 5.395 MeVmm m Eb 0.105 MeVm
m m p al 6.570 m mm amu MeV m pb m
mm 6.570 amu MeV m




■ Problem 1.4. Calculation of Wavelength from Energy.
m m m m m m




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
m m m m m m m m m m m m m m m m m m m m m m




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
m m m m m m m m m m m m m m m m m m m m m m m m




Λ=hc/E. Plugging in our maximum energy value into this equation gives the minimum x-ray wavelength.
m m m m m m m m m m m m m m m




hc mm


Λ m where we substitu te h
m m m m m 6.626 10 34 J s, c 299 792 458 m s and E 195 keV
mm m mm m mm m m m mm m m mm m



E




1

, Chapter 1 Solutions m m




1.01869 J–m m


m m 0.0636 Angstroms m


KeV



235
■ Problem 1.5. m m m m UFission Energy Release. m m




235 117 118
Using the reaction
m m m mU m m m m Sn mm m Sn, and mass values, we calculate the mass defect of:
m m m m m m m m m




M 235 U M 117 Sn M 118 Sn M and an expected energy
m
m
mm m m
m
mm m
m
mm m m m m




2
release of Mc .
m m m




931.5 MeV m


mm m m m m m


AMU

This is one of the most exothermic reactions available to us. This is one reason why, of course, nuclear power from uranium
m m m m m m m m m m m m m m m m m m m m m




fission is so attractive.
m m m m




■ Problem 1.6. Specific Activity of Tritium. m m m m m




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
m m m m m m m m m m m m m m m m m m m m m




isotope, and M is the molecular weight of the sample.
m m m m m m m m m m




ln2 Avogadro ' s Constant
m m m m


Specific Activity m m


T 12 M m




3 grams
We substitute T12 12.26 years and M= to get the specific activity in disintegrations/(gram–year).
m

m m mm m m m m m m m m m

mole


1.13492 10 22 mm


Specific Activity m m



gram –year m




The same result expressed in terms of kCi/g is shown below
m m m m m m m m m m




9.73 kCi m


Specific Activity m m



gram



■ Problem 1.7. Accelerated particle energy. m m m m




The energy of a particle with charge q falling through a potential V is qV. Since V= 3 MV is our maximum potential difference,
m m m m m m m m m m m m m m m m m m m m m m m




the maximum energy of an alpha particle here is q*(3 MV), where q is the charge of the alpha particle (+2). The maximum
m m m m m m m m m m m m m m m m m m m m m m m




alpha particle energy expressed in MeV is thus:
m m m m m m m m




Energy 3 Mega Volts 2 Electron Charges
mm m m m m m m m 6. MeV
m m




2

, Chapter 1 Solutions m m




2 1 1
■ Problem 1.8. Photofission of m m m D Γ
m m m m m n m m p + Q (-2.226 MeV)
m m m m




deuterium.m
1 0 1

The reaction of interest is
m m m m m m
2
D
m m m
0
mΓ m m
1
n
m m m
1
m p+ Q (-2.226 MeV). Thus, the Γ must bring an energy of at least 2.226 MeV
m m m m m m m m m m m m m m m


1 0 0 1
in order for this endothermic reaction to proceed. Interestingly, the opposite reaction will be exothermic, and one can expect to
m m m m m m m m m m m m m m m m m m m




mfind 2.226 MeV gamma rays in the environment from stray neutrons being absorbed by hydrogen nuclei.
m m m m m m m m m m m m m m m




■ Problem 1.9. Neutron energy from D-T reaction by 150 keV deuterons.
m m m m m m m m m m




We write down the conservation of energy and momentum equations, and solve them for the desired energies by eliminating the
m m m m m m m m m m m m m m m m m m m




momenta. In this solution, "a" represents the alpha particle, "n" represents the neutron, and "d" represents the deuteron (and, as
m m m m m m m m m m m m m m m m m m m m




before, "p" represents momentum, "E" represents energy, and "Q" represents the Q-value of the reaction).
m m m m m m m m m m m m m m m




p a2 p n2 pd 2 m


pa pn pdmm mm Ea
m m En m Ed Ea En Ed Q
mm mm
m
m


2 mam 2 mnm 2 mdm




Next we want to solve the above equations for the unknown energies by eliminating the momenta. (Note : Using computer
m m m m m m m m m m m m m m m m m m m




software such as Mathematica is helpful for painlessly solving these equations).
m m m m m m m m m m m




We evaluate the solution by plugging in the values for particle masses (we use approximate values of "ma," "mn,"and "md"
m m m m m m m m m m m m m m m m m m m




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,
m m m m m m m m m m m m m m m m m m m m m m m m m




namely the Q-value, and the given energy of the deuteron (both energy values are in MeV). So we substitute ma = 4, mn = 1,
m m m m m m m m m m m m m m m m m m m m m m m m m m




md
m




= 2, Q = 17.6, Ed = 0.15 into our momenta independent equations. This yields two possible sets of solutions for the energies (in
m m m m m m m m m m m m m m m m m m m m m m m




MeV). One corresponds to the neutron moving in the forward direction, which is of interest.
m m m m m m m m m m m m m m m




En m m 13.340 MeV m Ea m m 4.410 MeV m




En m m 14.988 MeV m Ea m m 2.762 MeV m




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
m m m m m m m m m m m m m m m m m m m m m m m m m m m m




these equations we get the following results.
m m m m m m m




pd 1 1
pn m m 2 3 p d 2 352
m
m
m
m pa m 8 pd 2
m
m
m 2 3 p d 2 352
m
m
m
m


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
m m m m m m m m m m m m m m m m m m m m m




p n and p a by substituting:
m m m m




pd m mm mm




The particle momenta ( in units of amuMeV ) for each set of solutions is thus:
m m m m m m m m m m m m m




p n 5.165
m m p a 5.940 m m




p n 5.475
m m p a 4.700 m m




The largest neutron momentum occurs in the forward (+) direction, so the highest neutron energy of 14.98 MeV corresponds
m m m m m m m m m m m m m m m m m m




mto this direction. m m




3