Which sample, $A$ or $B$ shown in figure has shorter mean-life?

Here, as compared to $A$, activity of $B$ decreases rapidly. Hence $\lambda_{B}>\lambda_{A} \Rightarrow \tau_{B}<\tau_{A}$

$\left(\because\right.$ Average life $\left.\tau=\frac{1}{\lambda}\right)$

Proof: $(i)$ For element $A$,

$\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{0 \mathrm{~A}} e^{-\lambda_{\mathrm{A}} t_{0}}\left(\text { Where } t=t_{0}\right)$

$(ii)$ For element $B$,

$\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{0 \mathrm{~B}} e^{-\lambda_{\mathrm{B}} t_{0}}\left(\text { Where } t=t_{0}\right)$

Taking ratio,

$\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{e^{-\lambda_{\mathrm{A}} t_{0}}}{e^{-\lambda_{\mathrm{B}} t_{0}}} \quad\left(\because \mathrm{I}_{0 \mathrm{~A}}=\mathrm{I}_{0 \mathrm{~B}}\right)$

$\therefore \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{e^{\lambda_{\mathrm{B}} t_{0}}}{e^{\lambda_{\mathrm{A}} t_{0}}}$

From the figure,

$\mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}} \Rightarrow e^{\lambda_{\mathrm{B}} t_{0}}>e^{\lambda_{\mathrm{A}} t_{0}}$

$\therefore \lambda_{\mathrm{B}} t_{0}>\lambda_{\mathrm{A}} t_{0}$

$\therefore \lambda_{\mathrm{B}}>\lambda_{\mathrm{A}}$

$\therefore \tau_{\mathrm{B}}<\tau_{\mathrm{A}} \quad\left(\because \tau=\frac{1}{\lambda}\right)$