Question
A $$^7Li$$ target is bombarded with a proton beam current of $${10^{ - 4}}A$$ for $$1$$ hour to produce $$^7Be$$ of activity $$1.8 \times {10^8}$$ disintegrations per second. Assuming that one $$^7Be$$ radioactive nucleus is produced by bombarding $$1000$$ protons, determine its half life.
A.
$$8.6 \times {10^6}s$$
B.
$$4.2 \times {10^5}s$$
C.
$$3.1 \times {10^5}s$$
D.
$$1.1 \times {10^6}s$$
Answer :
$$8.6 \times {10^6}s$$
Solution :
The total number of protons bombarded
$$ = \frac{{it}}{\lambda } = \frac{{{{10}^{ - 4}} \times 3600}}{{1.6 \times {{10}^{ - 19}}}} = 22.5 \times {10^{17}}$$
Number of $$^7Be$$ produced
$$N = \frac{{22.5 \times {{10}^{17}}}}{{1000}} = 22.5 \times {10^{14}}$$
We know that activity
$$\eqalign{
& A = \lambda N\,\,{\text{or}}\,\,A = \left( {\frac{{0.693}}{{{t_{\frac{1}{2}}}}}} \right)N \cr
& \therefore {t_{\frac{1}{2}}} = 0.693\frac{N}{A} = 0.693 \times \frac{{22.5 \times {{10}^{14}}}}{{1.8 \times {{10}^8}}} = 8.63 \times {10^6}s \cr} $$