EAS 4561/6560 Isotope Geochemistry
Problem Set 1 Solutions
Due Feb. 9, 2017
1. Using the equation and values for the liquid drop model given in the book, predict the binding energy
per nucleon for 4He, 56Fe, and 238U (ignore even-odd effects).
14A-13A^2/3-18.1I^2/4A-.58Z^2/A^0.33
A
Z
N
I
4
He
4
2
2
0
56
Fe
56
26
30
4
238
U
238
92
146
54
Etotal
Eb/nucleon
21.78054
5.44513
489.9410
8.748948
1985.1101
8.340799
2. How many stable nuclides are there with N = 82? List them. How many with N = 83? List them. Why
the difference? Use the Chart of the Nuclides maintained by Brookhaven National Laboratory Site <
http://www.nndc.bnl.gov/chart/> or another online version to answer this question.
N=82: 144Sm, 142Nd, 141Pr, 140Ce, 139La, 138Ba, 136Xe
N-83: 143Nd
82 is a magic number (also even).
3. A certain radionuclide emits radiation at the rate of 15.0µW at one instant of time and at 1.0 µW
(microwatts) one hour latter. What is its half-life?
The activity, i.e., decays per unit time is -dN/dt. From the basic equation of
radioactive decay, -dN/dt = N. Therefore:
Also
dN / dt0
l N0
N0
=
=
dN / dt1hour l N1hour N1hour
æ Nö
- lt = ln ç ÷
è N0 ø
Substituting 1.0/15.0 for N/N0, 1 hour for t, we can solve for  and convert that
to half-life, which works out to be: 0.256 hr (15.4 min) ( = 2.708 hr-1)
4. What is the total energy released when 87Rb (mass = 86.909183) decays to 87Sr (mass = 86.908879u)?
The maximum  – energy is the case where the beta particle has all the energy
involved in the decay. This can be calculated from the mass difference of the 2
nuclides. 4.54  10-14 J = 0.283 MeV
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EAS 4561/6560 Isotope Geochemistry
Problem Set 1 Solutions
Due Feb. 9, 2017
5. Radioactive decay obeys the laws of mass-energy conservation and also of momentum. Consequently,
emission of particles from the nucleus results in recoil of the daughter nucleus. What is the recoil energy
of 208Tl (mass = 207.9820187u) in the 5.601 MeV alpha decay illustrated in Figure 1.7?
To find the recoil energy, we make use of the fact that the decay must obey
conservation of momentum: PTl = P. Since we know the alpha particles mass and
energy, it is straightforward to calculate its momentum. Momentum, p, is related
to energy as p= sqrt(e * 2m). We find the alpha momentum to be 1.098x10-19 kgm/s. Using the mass above, we calculate the energy of recoil to be 0.1082 MeV or
1.73 x 10-14 J
6. The accompanying page shows a section of the chart of the nuclides. Mass numbers of stable isotopes
are shown; unstable nuclides are shown as blank squares and can be assumed to be short-lived.
a. Show the s-process path beginning with 131Ta.
b. Identify all nuclides created, in part or in whole, by the r-process.
c. Identify all nuclides created only by the p-process.
d. Which of the stable nuclides shown should be least abundant and why?
e. Which of the osmium (Os) isotopes shown would you expect to be most abundant and why?
(Your answer may include more than one nuclide in d and e).
d. the least abundant nuclides will be the p-only ones (which includes 180Ta, although
we can’t deduce that without seeing more of the chart). Most abundant Os nuclides
should be 188 and 190, since they are both even and both created by both s- and
r-processes.
Some useful facts:
Bear in mind that radioactive decay obeys all the conversation laws of modern and classical physics,
namely, mass-energy, momentum, spin, charge, and baryon number.
Avegadro's number is 6.02252 ×1023 (mole)-1; Speed of light: c = 2.997925 ×108 m/sec.
1 eV = 1.60218 ×10-19 J; 1 J = 1 kg m2/sec2; 1 J = 1 VC; electron charge: q = 1.6021 ×10-19 C (coloumb)
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EAS 4561/6560 Isotope Geochemistry
Problem Set 1 Solutions
Due Feb. 9, 2017
1 u (dalton, amu) = 1.660541 ×10-27 kg. mass of proton: 1.007276467 u = 1.6726231 × 10-27 kg; mass of
neutron: 1.008664716 u mass of electron = mass of positron: 0.0005485799 u = 9.100938188 × 10-31 kg
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