Q94 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
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.
- A$\lambda_A=\lambda_B$
- B$\lambda_A=2 \lambda_B$
- Correct answerC$\lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ln 2$
- D$\lambda_{\mathrm{A}} \ln 2=\lambda_{\mathrm{B}}$
Solution
It is given that
$\begin{aligned}<br />& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\<br />& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\<br />& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2<br />\end{aligned}$
$\begin{aligned}<br />& \left(\frac{\mathrm{T}_1}{2}\right)_{\mathrm{A}}=\left(\mathrm{T}_{\mathrm{av}}\right)_{\mathrm{B}} \\<br />& \because \mathrm{T}_{1 / 2}=\frac{\ell \mathrm{n} 2}{\lambda} \text { and } \mathrm{T}_{\mathrm{av}}=\frac{1}{\lambda} \\<br />& \therefore \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{A}}}=\frac{1}{\lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{A}}=\lambda_{\mathrm{B}} \ell n 2<br />\end{aligned}$
Hence, the correct answer is option: 'C'.