NPRE 441 Spring 2021HW1 Solutions

Q27 (Page 81-82)A parent nuclide decays by beta-particle emission into a stable daughter. The major radiations, energies(MeV), and frequencies are, in the notation of Appendix D:

β−: 3.92 max (7%), 3.10 max (5%), 1.60 max (88%)γ: 2.32 (34%), 1.50 (54%), 0.820 (88%)

e−: 0.818 0.805

(a) Draw the decay scheme

Solution:

(b) What is the maximum energy that the antineutrino can receive in this decay?

Solution:Emax,ν = Emax,β−

= 3.92 MeV

(c) What is the value of the internal-conversion coefficient?

Solution:NeNγ

10%

49%α = = = 0.204 [Equation3.31]

Note: The reciprocal solution is also accepted.

(d) Estimate the L-shell electron binding energy of the daughter nuclide.

Solution:

where:

So:

Ee = E∗ − EB

E∗ = 0.82 MeV

Ee = 0.818, 0.805 MeV

EB = 0.82− 0.818 MeV = 0.002 MeV

EB = 0.82− 0.805 MeV = 0.015 MeV

The larger EB is for K-shell, so the binding energy for L-shell electron is 0.002 MeV

(e) Would daughter X rays be expected also? Why or why not?

Solution: Yes. The daughter atom would be missing a K-shell or L-shell electron after internalconversion. When an outer shell electron moves in to fill this hole, the energy difference couldbe released as X rays (or Auger electrons).

Q33 (Page 81-82)The nuclide 6530Zn decays by electron capture (98.5%) and by positron emission (1.5%).

(a) Calculate the Q value for both modes of decay.

Solution: Electron capture:

6530Zn+0

−1 e− −→65

29m Cu+0

0 γ

Q = ∆P −∆D − EB= (−65.911)− (−67.263)− 0.009 MeV

= 1.343 MeV

Positron emission:65 Cu+0

1 β+ +0

0 γ30Zn −→6529

Q = ∆P −∆D − 2mc2

= (−65.911)− (−67.263)− 1.022

= 0.33 MeV

(b) Draw the decay scheme for 65Zn.

Solution:

(c) What are the physical processes responsible for each of the major radiations listed inAppendix D?

Solution: Positrons are produced by the positron decay of the parent atoms. When thesepositrons annihilate with electrons in the surrounding material, two annihilation photons of∼511 keV are emitted per annihilation event.At least 51% of the decay is via Electron Capture that populates the 1.116 MeV energy level.From this energy level, the excited daughter atoms either decays to the ground state via the1.116 MeV gamma emissions (51%) or via Internal Conversion that ejects (most probably) theK-shell electron with an energy of 1.107 MeV.The remaining Electron Capture events towards the nuclear ground-state of the daughter atomsas well as the above Internal Conversion events contribute to the population of excited daughteratoms (missing K-shell or L-shell electrons). These excited atoms de-excite when the outer shellelectrons move in to fill the hole, producing daughter X-rays, or when the atoms eject augerelectrons.

(d) (3 points) Estimate the binding energy of a K-shell electron in copper.

Solution:EB = E∗ − Ee

= 1.116− 1.107 MeV

= 0.009 MeV

Q35 (Page 81-82)Show that 5526F e, which decays by electron capture, cannot decay by positron decay.

Solution: Q value of decay of 5526Fe : 55

26Fe+0−1 e

− −→55m25 Mn+0

0 γ

Q = ∆P −∆D − EB= −57.474− (−57.705)− EB MeV

= 0.231 MeV − EB < (2×me−)

Since the Q value of the decay of 5526Fe is less than two times of the mass of electrons required for

positron decay, the decay cannot proceed via positron decay.

Note: If you compare the decay equations between Electron Capture and Positron Decay, youwill find that there will be two extra leptons (an electron and an position) in the positron decaymode. The creation of these two leptons require energy and this energy is derived from the Q valueof the reaction. If the Q value is less than two lepton masses, the decay cannot proceed via positrondecay.

Q36 (Page 81-82)The isotope 53126I can decay by EC, β−, and β+ transitions.

(a) Calculate the Q values for the three modes of decay to the ground states of the daughternuclei.

Solution: EC:12653 I +0

−1 e− −→126m

52 Te+00 γ

QEC = ∆P + ∆D − EB= −87.9− (−90.05)− 0.033 MeV

= 2.12 MeV

β+ : 12653 I −→126

52 Te+01 β

+ +00 ν

Qβ+ = ∆P + ∆D − 2mc2

= −87.9− (−90.05)− 1.022 MeV

= 1.13 MeV

β− : 12653 I −→126

54 Xe+0−1 β

− +00 ν

Qβ− = ∆P + ∆D

= −87.9− (−89.15) MeV

= 1.25 MeV

(b) Draw the decay scheme

Solution:

(c) What kinds of radiation can one expect from a 126I source?

Solution: X-ray, Gamma-rays, β−, β+ as well as 511 keV annihilation photons can be ex-pected.

Q17 (Page 104-105): Consider the following β− nuclide decay chain with the half-lives indicated:

21082 Pb

β−−−→22 y

210

83

Biβ−−−−→5.0 d

210

84Po

A sample contains 30 MBq of 210Pb and 15 MBq of 210Bi at time t = 0.

(a) (10 pts): Calculate the activity of 210Bi at time t = 10 d.

Secular Equilibrium (T1 � T2)Eq. (4.37):

A2 = A1(1− e−λ2t) + A20e−λ2t,

where λ2 =ln(2)

T2

= 0.1386 d−1.

Solution:

A2 = (30 MBq)(1 − e−(0.1386 d−1)(10 d)) + (15 MBq)e−(0.1386 d−1)(10 d)

⇒ A2 = 26.25 MBq

(b) If the sample was originally pure 210Pb, then how old is it at time t = 0?

For initial pure sample, A20 = 0. Using Eq. (4.37):

Solution:15 MBq = (30 MBq)(1− e−(0.1386 d−1)∆t)

⇒ ∆t = 5 d

Q24 (Page 104-105) A 40-mg sample of pure 226Ra is encapsulated.

(a) How long will it take for the activity of 222Rn to build up to 10 mCi?

Specific Activity of Radium-226 Eq. (4.24):

SA =6.02× 1023λ

M=

4.17× 1023

MT.

From Appendix D, T = 1600 y and M = A = 226. Converting T to seconds, we have Eq. (4.25):

SA =4.17× 1023

226× 1600× 365× 24× 3600

Eq. (4.26):= 3.66× 1010 s−1g−1 = 3.7× 1010 Bq g−1.

This, by definition, is an activity of 1 Ci. For an arbitrary nuclide of half-life T in years:Eq. (4.27):

SA =1600

T× 226

ACi g−1.

Secular Equilibrium (T1 � T2)For initial pure sample, A20 = 0. Using Eq. (4.37):

A2 = A1(1− e−λ2t)

etT2

ln( 12

)= 1− A2

A1

t = T2

ln(1− A2

A1)

ln(12)

.

Solution:A1 = 40 mg × 1 Ci g−1 = 40 mCi

t = (3.82 d)ln(1− 10 mCi

40 mCi)

ln(12)

⇒ t = 1.6 d

(b) What will be the activity of 222Rn after 2 years?

After about seven daughter half-lives (t & 7T2), e−λ2t � 1, secular equilibrium is met. Since 7 × 3.82 d = 26.74 d, t = 2 y � 26.74 d. So A2 = A1. However, t = 2 y � T1 = 1600 y. Therefore,

Solution:

A2 = A1 = A10

⇒ A2 = 40 mCi

(c) What will be the activity of 222Rn after 1000 years?

Now t = 1000 y . T1 = 1600 y. Following Eq. (4.13):Eq. (4.13):

A

A0

=

(1

2

)t/T.

Solution:A2 = A1 = (40 mCi)(0.5)

1000 y1600 y

⇒ A2 = 26 mCi

(d) What is the ratio of the specific activity of 222Rn to that of 226Ra?

The ratio is simply just the specific activity of 222Rn, since the unit of curie represents the activity of 1-gram of 226Ra. Using Eq. (4.27):

Solution:SA2 =

16003.82 d365 d

× 226

222Ci g−1.

⇒ SA2

SA1

= 156, 000

A1 + A2 = λ1N1 + λ2N2

Q26 P

(a) Verify Eq. (4.44).

Solution:Total activity:

Maximum total activity:

d(A1 + A2)

dt= λ1

dN1

dt+ λ2

dN2

dt= 0

wheredN1

dt= −λ1N1,

dN2

dt= λ1N1 − λ2N2.

⇒ (λ1λ2 − λ21)N1 = λ2

2N2

Eq. (4.40):

N2 =λ1N10

λ2 − λ1

(e−λ1t − e−λ2t)

Substitution:

λ1(λ2 − λ1)e−λ1t =λ1λ

22

λ2 − λ1

(e−λ1t − e−λ2t)

(λ2 − λ1)2

λ22

= 1− e−(λ2−λ1)t

−(λ2 − λ1)t = ln

(1− (λ2 − λ1)2

λ22

)= ln

(2λ1λ2 − λ2

1

λ22

)Eq. (4.44)

⇒ t =1

λ2 − λ1

lnλ2

2

2λ1λ2 − λ21

(b) Show that the time of maximum total activity occurs earlier thanthe time of maximum daughter activity in Fig. 4.5.

The inequality of maximum A2 (Eq. 4.43) > maxiumum A1 + A2 (Eq. 4.44) must satisfy the condition that λ1 < λ2 for Transient Equilibrium (T1 & T2).

Solution:

1

λ2 − λ1

lnλ2

λ1

>1

λ2 − λ1

lnλ2

2

2λ1λ2 − λ21

λ2

λ1

>λ2

2

2λ1λ2 − λ21

2λ1λ22 − λ2

1λ2 > λ1λ22

2λ2 − λ1 > λ2

⇒ λ2 > λ1

(c) Does Eq. (4.43) apply to A2 when there is no equilibrium (Fig. 4.6)?

Solution:

When there is No Equilibrium (T1 < T2), then λ1 > λ2. Since Eq. (4.43) was derived from the General Case, then it should always apply to A2 at the point where A1 = A2, so long that you can show that t > 0 and is real. A more proper form for the ”No Equilibrium” case is

1

λ1 − λ2

lnλ1

λ2

,

which is always positive and real. This is not the case for Eq. (4.44), since the quantityinside the natural logarithm is negative for λ1 > λ2.

Q28 (page 104-105): The average mass of potassium in the human body is about 140 g. From the abundance and half-life given in Appendix D, estimate the average

activity of 40K in the body.

The natural abundance of 40K is 0.0118% and the half-life is 1.28 × 109 years. The specific activity is needed, given by Eq. (4.27):

Solution:

SA =1600

1.28× 109× 226

40= 7.06 µCi g−1.

A = 0.000118× 140 g × 7.06 µCi g−1

⇒ A = 0.117 µCi