Modern Physics 3 Question 3
3. In given time $t=0$, Activity of two radioactive substances $A$ and $B$ are equal. After time $t$, the ratio of their activities $\frac{R _B}{R _A}$ decreases according to $e^{-3 t}$. If the half life of $A$ is $\operatorname{In} 2$, the half-life of $B$ will be
(Main 2019, 9 Jan II)
(a) $4 \ln 2$
(b) $\frac{\ln 2}{4}$
(c) $\frac{\ln 2}{2}$
(d) $2 \ln 2$
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Solution:
- Activity of radioactive material is given as
$$ R=\lambda N $$
where, $\lambda$ is the decay constant $N$ is the number of nuclei in the radioactive material.
For substance $A$,
$$ R _A=\lambda _A N _A=\lambda _A N _{0 A}\left(\text { initially } N _A=N _{0 A}\right) $$
For substance $B$,
$$ \left.R _B=\lambda _B N _B=\lambda _B N _{0 B} \text { (initially } N _B=N _{0 B}\right) $$
At $t=0$, activity is equal, therefore
$$ \lambda _A N _{0 A}=\lambda _B N _{0 B} $$
The half-life is given by
$$ T _{1 / 2}=\frac{0.693}{\lambda}=\frac{\ln 2}{\lambda} $$
So, for substance $A$,
$$ \begin{gathered} \left(T _{1 / 2}\right) _A=\frac{\ln 2}{\lambda _A} \Rightarrow \ln 2=\frac{\ln 2}{\lambda _A} \\ \lambda _A=1 \end{gathered} $$
According to the given question,
at time $t$,
$$ \frac{R _B}{R _A}=e^{-3 t} $$
Using Eqs. (i), (ii) and (iii)
$$ \begin{aligned} & \frac{R _B}{R _A}=e^{-3 t}=\frac{\lambda _B N _{0 B} e^{-\lambda _B t}}{\lambda _A N _{0 A} e^{-\lambda _A t}} \\ & \Rightarrow \quad e^{-3 t}=e^{\left(\lambda _A-\lambda _B\right) t} \\ & \Rightarrow \quad \begin{array}{ll} -3 & =\lambda _A-\lambda _B \\ \lambda _B & =\lambda _A+3 \end{array} \\ & \lambda _B=1+3=4 \end{aligned} $$
The half-life of substance $B$ is
$$ \left(T _{1 / 2}\right) _B=\frac{\ln 2}{\lambda _B}=\frac{\ln 2}{4} $$
