Question

The specific activity of a preparation consisting of radioactive $$Co^{58}$$ and nonradioactive $$Co^{59}$$ is equal to $$2.2 \cdot 10^{12} $$ dis/(s $$\cdot$$ g). The half life of $$Co^{58}$$ is $$71/3$$ days. Find the ratio of the mass of radioactive cobalt in that preparation to the total mass of the preparation (in per cent).

Answer 1

We see that
Specific activity of the sample
$$ = \dfrac{1}{M + M'} $$ [ Activity of $$M\,gm$$ of $$Co^{58} $$ in the sample]
Here $$M$$ and $$M'$$ are the masses of $$Co^{58} $$ and $$Co^{59}$$ in the sample. Now activity of $$M\,gm$$ of $$Co^{58} $$
$$ = \dfrac{M}{58} \times 6.023 \times 10^{23} \times \dfrac{ln\ 2}{71.3 \times 86400} $$ dis/sec
$$ = 1.168 \times 10^{15} M $$
Thus from the problem
$$ 1.168 \times 10^{15} \dfrac{M}{M + M'} = 2.2 \times 10^{12} $$
or $$ \dfrac{M}{M + M'} = 1.88 \times 10^{-3} $$ i.e., $$ 0.188\%$$