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A radioactive material of half-life $T$ was produced in a nuclear reactor at different instants, the quantity produced second time was twice of that produced first time. If now their present activities are $A_1$ and $A_2$ respectively then their age difference equals :
$\frac{T}{{\ln \,2}}\,\left| {\ln \,\frac{{{A_1}}}{{{A_2}}}} \right|$
$T\,\left| {\ln \,\frac{{{A_1}}}{{{A_2}}}} \right|$
$\frac{T}{{\ln \,2}}\,\,\left| {\ln \,\frac{{{A_2}}}{{2{A_1}}}} \right|$
$T\,\left| {\ln \,\frac{{{A_2}}}{{2{A_1}}}} \right|$
Solution
$N_{1}=N_{o}, N_{2}=2 N_{o}$
$t_{1}=\frac{1}{\lambda} \ln \frac{\lambda N_{o}}{A_{1}}$
$t_{2}=\frac{1}{\lambda} \ln \frac{\lambda 2 N_{o}}{A_{2}}$
hence, $t_{2}-t_{1}=\frac{1}{\lambda} \ln \frac{A_{2}}{2 A_{1}}$
Using $T=\frac{\ln 2}{\lambda}$
$t_{2}-t_{1}=\frac{T}{\ln 2} \ln \frac{A_{2}}{2 A_{1}}$
