Modern Physics Ques 129
- 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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Answer:
Correct Answer: 129.(b)
Solution:
Formula:
- 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} \quad …….(i)$
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 \quad …….(ii) \end{gathered} $
According to the given question,
at time $t$,
$ \frac{R _B}{R _A}=e^{-3 t} \quad …….(iii) $
Using Eqs. (i), (ii) and (iii)
$\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 -3 =\lambda _A-\lambda _B $
$\lambda _B =\lambda _A+3$
$\lambda _B=1+3=4 \quad …….(iv)$
The half-life of substance $B$ is
$ \left(T _{1 / 2}\right) _B=\frac{\ln 2}{\lambda _B}=\frac{\ln 2}{4} $
BETA
