Question
A radioactive material of half life $$\ell n2$$ was produced in a nuclear reactor. Consider two different instants $$A$$ and $$B.$$ The number of undecayed nuclei at instant $$B$$ was twice of that of instant $$A.$$ If the activities at instants $$A$$ and $$B$$ are $${A_1}$$ and $${A_2}$$ respectively then the difference in the age of he sample at these instants equals.
A.
$$\left| {\ell n\left( {\frac{{2{A_1}}}{{{A_2}}}} \right)} \right|$$
B.
$$\ell n2\left| {\ell n\left( {\frac{{{A_1}}}{{{A_2}}}} \right)} \right|$$
C.
$$\left| {\ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right)} \right|$$
D.
$$\ell n2\left| {\ell n\left( {\frac{{{A_2}}}{{{A_1}}}} \right)} \right|$$
Answer :
$$\left| {\ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right)} \right|$$
Solution :
$$\eqalign{
& {A_1} = \left( {\lambda {N_0}} \right){e^{ - \lambda {t_1}}}\,......\left( {\text{i}} \right) \cr
& {A_2} = \left( {\lambda 2{N_0}} \right){e^{ - \lambda {t_2}}}\,......\left( {{\text{ii}}} \right) \cr
& {t_1} - {t_2} = \frac{1}{\lambda }\ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right) = \ell n\left( {\frac{{{A_2}}}{{2{A_1}}}} \right) \cr} $$