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A radioactive nucleus $A$ with a half life $T$, decays into a nucleus $B.$ At $t = 0$, there is no nucleus $B$. At sometime $t$, the ratio of the number of $B$ to that of $A$ is $0.3$. Then, $t$ is given by
$t = \frac{T}{2}\;\frac{{\log 2}}{{\log 1.3}}$
$t = T\;\frac{{\log 1.3}}{{\log 2}}$
$t=T \log(1.3)$
$t = \frac{T}{{{{log}}\left( {1.3} \right)}}$
Solution
Let initially there are total $\mathrm{N}_{0}$ number of nuclei
At time $\mathrm{t} \frac{N_{B}}{N_{A}}=0.3(\text { given })$
$\Rightarrow \quad N_{B}=0.3 N_{A}$
$\mathrm{N}_{0}=N_{A}+N_{B}=N_{A}+0.3 N_{A}$
$\therefore \quad N_{A}=\frac{\mathrm{N}_{0}}{1.3}$
As we know $N_{t}=N_{0} e^{-\lambda t}$
or, $\frac{\mathrm{N}_{0}}{1.3}=N_{0} e^{-\lambda t}$
$\frac{1}{1.3}=e^{-\lambda t} \Rightarrow \ln (1.3)=\lambda t$
or, $t=\frac{\ln (1.3)}{\lambda} \Rightarrow t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)} T$
