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Two radioactive elements $A$ and $B$ initially have same number of atoms. The half life of $A$ is same as the average life of $B$. If $\lambda_A$ and $\lambda_B$ are decay constants of $A$ and $B$ respectively, then choose the correct relation from the given options.
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Solution:
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Verified Answer
The correct answer is:
$\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$
It is given that
$\begin{aligned}
& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\
& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\
& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2
\end{aligned}$
$\begin{aligned}
& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\
& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\
& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2
\end{aligned}$
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