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The half life of a radioactive nucleus is $50$ days. The time interval $\left( t _2-t_1\right)$ between the time $t _2$ when $\frac{2}{3}$ ot it has decayed and the time $t_1$, when $\frac{1}{3}$ of it had decayed is ......days
$30 $
$50$
$15 $
$60$
Solution
According to radioactive decay law
$N=N_{e} e^{-\lambda t}$
where $N_{0}=$ Number of radioactive nuclei at time $t=0$
$N=$ Number of radioactive nuclei left undecayed at any time $t$
$\lambda=$ decay constant
At time $t_{2}, \frac{2}{3}$ of the sample had decayed
$\therefore N=\frac{1}{3} N_{0}$
$\therefore \frac{1}{3} N_{0}=N_{0} e^{-\lambda t_{2}}$ …… $(i)$
At time $t_{1}, \frac{1}{3}$ of the sample had decayed,
$\therefore N=\frac{2}{3} N_{0}$
$\therefore \frac{2}{3} N_{0}=N_{0} e^{-\lambda t_{1}}$ …… $(ii)$
Divide $(i)$ by $(ii)$, we get
$\frac{1}{2} = \frac{{{e^{ – \lambda {t_2}}}}}{{{e^{ – \lambda {t_1}}}}}$ $ \Rightarrow \frac{1}{2} = {e^{ – \lambda \left( {{t_2} – {t_1}} \right)}}$
${\lambda\left(t_{2}-t_{1}\right)=\ln 2} $
${t_2} – {t_1} = \frac{{\ln 2}}{\lambda } = \frac{{\ln 2}}{{\left( {\frac{{\ln 2}}{{{T_{1/2}}}}} \right)}}$ $\left( {\because \lambda = \frac{{\ln 2}}{{{T_{1/2}}}}} \right)$
${=T_{10}=50 \text { days }}$
