FYS3500 - Solutions to problem set 7
Spring term 2018
Problem 1 – in class
a) Did you calculate all problems from the previous sets? Any questions?
b) Start with the problem below; after some examples of (a), focus especially (b to d).
Problem 2 γ-decay and Weisskopf units
The reduced transition probabilities are expressed in e2 fm2L for EL multipoles and in µ2N fm2L−12
for ML multipoles, see eg. Krane.
a) For the following transitions between levels, give all permitted γ-ray multipoles and indicate
which multipole might be the most intense in the emitted radiation.
− + − −
a) 29 → 27 : E[1,3,5,7], M[2,4,6,8] b) 12 → 27 : M3, E4 c) 1− → 2+ : E1, (E3, M2) d)
0+ → 0+ : No gamma; internal conversion e) 3+ → 3+ : M[1,3,5], E[2,4,6] e) 4+ → 2+ :
E[2,4,6], M[3,5]
− +
f) 11
2 → 32 : M[4,6], E5
3− 7− 5+ 1−
b) A nucleus has the following sequence of states beginning with the ground state: 2 ,2 ,2 ,2
−
and 23 Draw a level scheme showing the intense γ transitions likely to be emitted and indi-
cate their multipole assignment. Which of the transitions would you expect to be have the
smallest chance to happen?
Figure 1
1
Answer: See discussion in class. If we to first order neglect the Eγ dependence – the energies
are not given in the level scheme – The highest multipolarities & for the same multipolarity
the magnetic transitions have the smallest chance to happen.
c) Which γ would you expect to be emitted in coincidence with each other (your timing res-
olution may be something like 50ns). How would a low-lying excited state with high spin
behave?
Answer: Rule of the thumb: all transition cascades with E1, E2 & M1, maybe also M2. Rather
less likely: The 3/2−, M1 → 1/2−, M2 → 5/2−, M3 → 7/2−, M1 → 3/2− transition.
The gs spins are often rather low, so one needs high multipolarities to decay down from a low-
lying excited state with high spin. That increases their lifetime. They are called metastable,
or isomeric states. An important example of this is given below, 99m 43 Tc56 . – Look up the
spin-parities to confirm this.
d) Assume now, that the daughter nucleus is populated by β-decay with a Q-value of 7 MeV
from the I = 3/2− gs of a mother nucleus. Sketch the electron spectrum. Hint: Is this enough
information? What else do you potentially need? If in class: Ask for the information. If at
home: Make assumptions where necessary.
Answer: In class; Important is the superposition of the spectra, with the ”correct” intensities
of the different transitions and the ”correct” end-point values. Potentially one would also see
discrete lines of internal conversion.
e) For a light nucleus (A = 10), compute the ratio of the emission probabilities for electric quadrupole
(E2) and magnetic dipole (M1) radiation according to the Weisskopf estimates. Consider all
possible choices for the parities of the initial and final states.
Answer: Assuming a gamma-ray energy of 1 MeV, In the cases where both transitions are
possible: 2.8 × 10−5
f) Compare this to the ratio calculated for a heavy nucleus (A = 200).
Answer: Same assumptions; 1.5 × 10−3
Problem 3 Medical isotope production
Technetium-99m (99m 99
43 Tc56 ) is a metastable nuclear isomer of Tc (itself an isotope of Tc) that is
used in tens of millions of medical diagnostic procedures annually, making it the most commonly
used medical radioisotope.1
The technetium 99m Tc (excited state of 99 Tc) is a γ-emitter used in nuclear medicine to for
example detect cervical tumors with a γ-camera.2
99m 99
Tc → Tc + γ
99m Tc is a byproduct of the β− -decay of 99 Mo:
99
Mo → 99m
Tc + e− + ν̄e
A simplified decay scheme is shown in Figure 2
1 Tc-99m: https://en.wikipedia.org/wiki/Technetium-99m
2 Based on Exercise 2.1 in https://lpsc.in2p3.fr/schien/PHY113a/TD_radio_en_2011-2012.pdf
2
99 Mo
42
82.2 %
β−
17.8 % 99m Tc
43
β−
γ
99 Tc
43
100 %
β− 99 Ru
44
Figure 2: Simplified decay scheme of 99 Mo.
99 Mo can be produced for example in a cyclotron. Let’s assume that the initial activity of a
99 Mo source is A(0) = 8.5 Ci and that source does not contain any Tc initially.
a) Plot the number of nuclei (Nx (t) vs t) and activity (A x (t) vs t) for 99 Mo, 99m Tc, 99 Tc and 99 Ru
for the first 300 hours.
Hint: You may want to find informations such as half-lives etc. from http://www.nndc.bnl.
gov/chart/ or https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html. Start by
calculating the number of 99 Mo nuclei present at t = 0. You may either derive the equations
for the radioactive decay series, or look them up. In any case, remember to take into account
the branching in the decay of 99 Mo.
Answer: First we consider a decay chain 1 → 2 → 3, where nuclei of type 3 are stable.
N1 (t) = N0 e−λ1 t (Parent)
λ1 ( e − λ1 t − e − λ2 t )
N2 (t) = N0 (Daughter)
λ2 − λ1
λ (1 − e − λ2 t ) − λ 2 (1 − e − λ1 t )
N3 (t) = N0 1 (Stable)
λ1 − λ2
where Nx (t) is the exponential law of radioactive decay for the different nucleis, N0 is the
number of nuclei present at t = 0 for the parent nucleus, λ x is is the disintegration constant
for the different nucleis and t is the time.
The related activities are (Equation 6.24 on page 169 and Equation 6.32 on page 171 in Krane)
A1 (t) = λ1 N1 (t)
A2 (t) = λ2 N2 (t)
Since nuclei of type 3 are stable, the activity is zero.
For the radioactive decay of Mo you can use N1 (t) and N3 (t) as given above. However, 99 Mo
decays to two isotopes with a certain probability, and therefore we have to take into account
the weights w2A and w2B to decay into the daughter nuclei (99m Tc and 99 Tc). The equations
become
λ2A λ
N2A (t) = w2A · N2 (t) w2A = = 2A
λ2A + λ2B λt
λ2B λ
N2B (t) = w2B · N2 (t) w2B = = 2B
λ2A + λ2B λt
where N2 (t) given above and λ2 = λt = λ2A + λ2B . You can read about weighted decays on
page 164 in Krane.
3
For the activity you can use A1 (t) and A2 (t) (one each for A and B) as in the simple case.
There is no activity for 99 Ru because it is stable.
b) For 99m Tc to be useful for nuclear medical applications, the activity A2 must be larger than 5
Ci at the time of the medical exam. Estimate with the help of the curve in which time interval
the source can be used.
Answer: 5 Ci are 1.85 × 1011 Bq. With the curve I estimate the activity of 99m Tc to be higher
then this for the first 8-60 h.
c) Why don’t we see much activity from 99 Tc in the first 300 hours?
Answer: t1/2 ≈ 2.5 × 105 y, so it decays only quite slowly in comparison to the other isotopes.
Problem 4 Bonus: How constant is the decay constant?
In most of our work in nuclear physics, we regard the decay constant λ as a true constant for a
given nuclear species. However, you have studied two processes in which the nuclear decay rate
could be sensitive to the chemical state of the atom. Discuss these two processes and explain how
the atomic state might influuence the nuclear decay rate.
For a discussion and some examples of cases in which this can occur, see the review by G.T.
Emery, https://doi.org/10.1146/annurev.ns.22.120172.001121
