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A radioactive nucleus $A$ has a single decay mode with half-life $\tau_A$. Another radioactive nucleus $B$ has two decay modes $1$ and $2$. If decay mode $2$ were absent, the half-life of $B$ would have been $\tau_A / 2$. If decay mode $1$ were absent, the half-life of $B$ would have been $3 \tau_A$. If the actual half life of $B$ is $\tau_B$, then the ratio $\tau_B / \tau_A$ is
$3 / 7$
$7 / 2$
$7 / 3$
$1$
Solution
(a)
For material $B$, total disintegration constant.
$\lambda=\lambda_1+\lambda_2$
$\Rightarrow \quad \frac{1}{\tau_B}=\frac{1}{T_1}+\frac{1}{T_2}$
where, $\lambda=$ decay constant
and $\tau_B=$ half-life time of sample.
$\Rightarrow \quad \frac{1}{\tau_B}=\frac{1}{\left(\tau_A / 2\right)}+\frac{1}{3 \tau_A}$
$\Rightarrow \quad \tau_B=\frac{\left(\frac{\tau_A}{2}\right) 3 \tau_A}{\left(\frac{\tau_A}{2}\right)+3 \tau_A}=\left(\frac{\frac{3}{2}}{\frac{7}{2}}\right) \tau_A$
$\text { or } \quad \tau_B=\frac{3}{7} \tau_A \Rightarrow \frac{\tau_B}{\tau_A}=\frac{3}{7}$
