A radioactive nucleus A with a half life T, d | vedclass

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Question

Let initially there are total $\mathrm{N}_{0}$ number of nuclei

At time $\mathrm{t} \frac{N_{B}}{N_{A}}=0.3(	ext { given })$

$\Rightarrow \quad

Vedclass · Accepted Answer

$t = T\;\frac{{\log 1.3}}{{\log 2}}$

Vedclass · Answer

Let initially there are total $\mathrm{N}_{0}$ number of nuclei

At time $\mathrm{t} \frac{N_{B}}{N_{A}}=0.3(	ext { given })$

$\Rightarrow \quad N_{B}=0.3 N_{A}$

$\mathrm{N}_{0}=N_{A}+N_{B}=N_{A}+0.3 N_{A}$

$	herefore \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, $t=\frac{\ln (1.3)}{\lambda} \Rightarrow t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)} T$

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

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