Page 174 - Essentials of physical chemistry

136 Essentials of Physical Chemistry

Let us consider the decay of Pm by two mechanisms: (1) electron capture where a 1s electron is
pulled into the nucleus to make a neutron from one of the protons and (2) beta emission where an
electron is ejected and a neutron decays into a proton. Neutrons are not stable and when they are out
of a nucleus they have a half-life of 10.3 min but what determines whether a proton becomes a
neutron or a neutron becomes a proton depends on the other protons and neutrons in a given
nucleus.

n ! p þ e :

þ

145 145
1s
61 Pm þ e ! 60 Nd; t 1=2 ¼ 17:7 years:
146 146
" #
1s
61 Pm þ e ! 60 Nd
; t 1=2 ¼ 5:52 years:
146 146 Sm þ e
61 Pm ! 62 b
147 147 t 1=2 ¼ 2:623 years:
b
61 Pm ! 62 Sm þ e ;
With that very brief exposure to the complexity of nuclear chemistry and perhaps a better
appreciation of the significance of the research of the Curies, let us consider the simplest case:
147 147 Sm with t 1=2 ¼ 2:623 years:
61 Pm ! 62
How much of a 10 g sample will remain after 5 years (Figure 7.1)? We simply insert the data into
the equation,

0:693
k 1 (5 years)
a ¼ (10 g)e ¼ (10 g) exp (5 years) ¼ 2:669 g:
2:623 years
Thus, we see that the first-order kinetic equation is much simpler than the consideration of the decay
mechanism. From a geological point of view, this shows that any primal amount of Pm would have
rapidly decayed in the lifetime of the Earth of over 4 billion years. In fact the age of the Earth has
been estimated by the presence or absence of radioactive isotopes in the crust of the Earth related to
their isotopes, although the original amounts have to be estimated. Nuclear decay is a prime example
of first-order kinetics and as with the case of any kinetic problem, the decay is totally dependent on

Pm(147) Sm(147)
12
10

8
Grams 6

4

2
0
0 5 10 15 20 25 30 35
Years
FIGURE 7.1 The radioactive decay of 147 Pm ! 147 Sm based on an initial 10 g sample.
62
61