Question

At a given instant, say $t=0$, two radioactive substances $A$ and $B$ have equal activities. The ratio $\frac{R_B}{R_A}$ of their activities after time $t$ itself decays with time $t$ as $e^{-3t}$. If the half-life of $A$ is $\ln 2$, the half-life of $B$ is:

• A. $\frac{\ln 2}{2}$
• B. $4\ln 2$
• C. $2\ln 2$
• D. $\frac{\ln 2}{4}$

Answer 1

The correct option is D $\frac{\ln 2}{4}$

Given,
Half life of $A,(t_{1/2}{)}_A=\ln 2$

As we know that,

$(t_{1/2}{)}_A=\frac{\ln 2}{{\lambda }_A}$

$\therefore {\lambda }_A=1$

$\text{at}t=0,R_A=R_B$

$N_A{e}^{-{\lambda }_{A}T}=N_B{e}^{-{\lambda }_{B}T}$
$N_A=N_B$ at $t=0$

$\text{At}t=t,\frac{R_B}{R_A}=\frac{N_B{e}^{-{\lambda }_{B}t}}{N_A{e}^{-{\lambda }_{A}t}}$

$\Rightarrow e^{-3t}=e^{-({\lambda }_B-{\lambda }_A)t}(\because N_A=N_B)$

$\Rightarrow {\lambda }_B-{\lambda }_A=3$

$\Rightarrow {\lambda }_B=3+{\lambda }_A=3+1=4$

$\therefore (t_{1/2}{)}_B=\frac{\ln 2}{{\lambda }_B}=\frac{\ln 2}{4}$

Hence, option $\left(\text{C}\right)$ is correct.