Question
The activity of a radioactive sample is measured as $${9750\,{\text{counts}}/\min }$$ at $$t = 0$$ and as $${975\,{\text{counts}}/\min }$$ at $$t = 5\,\min .$$ The decay constant is approximately
A.
$$0.922/\min $$
B.
$$0.691/\min $$
C.
$$0.461/\min $$
D.
$$0.230/\min $$
Answer :
$$0.461/\min $$
Solution :
According to law of radioactivity
\[\begin{array}{l}
\frac{N}{{{N_0}}} = {e^{ - \lambda t}}\,\,......\left( {\rm{i}} \right)\\
\Rightarrow \frac{{{N_0}}}{N} = {e^{\lambda t}}\,\,\left[ {\begin{array}{*{20}{c}}
{N = {\rm{final}}\,{\rm{concentration}}}\\
{{N_0} = {\rm{initial\,concentration}}}\\
{\lambda = {\rm{decay\,constant}}}
\end{array}} \right]
\end{array}\]
Taking logarithm on both sides of Eq. (i), we have
$$\eqalign{
& {\log _e}\left( {\frac{{{N_0}}}{N}} \right) = {\log _e}\left( {{e^{\lambda t}}} \right) \cr
& = \lambda t\,{\log _e}e = \lambda t \cr} $$
As we know that, $${\log _e}x = 2.3026\,{\log _{10}}x$$
Making substitution, we get
$$\lambda = \frac{{2.3026\,{{\log }_{10}}\left( {\frac{{9750}}{{975}}} \right)}}{5}\,\,\left[ {\because {N_0} = 9750\,{\text{counts}}/\min \,{\text{and}}\,N = 975\,{\text{counts}}/\min } \right]$$
$$ = \frac{{2.3026}}{5}{\log _{10}}10 = \frac{{2.3026}}{5}{\min ^{ - 1}} = 0.461\,{\min ^{ - 1}}$$