Question
The half-life of $$_6{C^{14}},\left( {\lambda = 2.31 \times {{10}^{ - 4}}\,{\text{per}}\,{\text{year}}} \right)$$ is
A.
$$2 \times {10^2}\,yr$$
B.
$$3 \times {10^3}\,yr$$
C.
$$3.3 \times {10^4}\,yr$$
D.
$$4 \times {10^3}\,yr$$
Answer :
$$3 \times {10^3}\,yr$$
Solution :
$$\eqalign{
& {\text{Radioactive decay is first order reaction}}{\text{.}} \cr
& {\text{So,}}\,\,\,\lambda {\text{ = }}\frac{{0.693}}{{{t_{\frac{1}{2}}}}} \cr
& \therefore \,\,\,{t_{\frac{1}{2}}} = \frac{{0.693}}{{2.31 \times {{10}^{ - 4}}}}yr \cr
& = 0.3 \times {10^4} \cr
& = 3 \times {10^3}\,yr \cr} $$