Modern Physics Ques 139
- A sample of radioactive material $A$, that has an activity of $10$ $mCi\left(1 Ci=3.7 \times 10^{10}\right.$ decays/s) has twice the number of nuclei as another sample of a different radioactive material $B$ which has an activity of $20 $ $mCi$. The correct choices for half-lives of $A$ and $B$ would, then be respectively
(Main 2019, 9 Jan I)
(a) $20 $ days and $ 10 $ days
(b) $5 $ days and $10 $ days
(c) $10 $ days and $ 40 $ days
(d) $20 $ days and $ 5 $ days
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Answer:
Correct Answer: 139.(d)
Solution:
Formula:
- Activity of a radioactive material is given as
$ R=\lambda N $
where, $\lambda$ is the decay constant and $N$ is the number of nuclei in the radioactive material.
For substance $A$,
$ R _A=\lambda _A N _A=10 $ $mCi $
For substance $B$,
$ R _B=\lambda _B N _B=20 $ $mCi $ $\quad$ …….(i)
As given in the question,
$ N _A=2 N _B $
$\Rightarrow \quad R _A=\lambda _A\left(2 N _B\right)=10 mCi$ $\quad$ …….(ii)
$\therefore$ Dividing Eq. (ii) and Eq.(i), we get
$ \frac{R _A}{R _B}=\frac{\lambda _A\left(2 N _B\right)}{\lambda _B\left(N _B\right)}=\frac{10}{20} $
or $\quad \frac{\lambda _A}{\lambda _B}=\frac{1}{4}$ $\quad$ …….(iii)
As, half-life of a radioactive material is given as
$ T _{1 / 2}=\frac{0.693}{\lambda} $
$\therefore$ For material $A$ and $B$, we can write
$ \frac{\left(T _{1 / 2}\right) _A}{\left(T _{1 / 2}\right) _B}=\frac{\frac{0.693}{\lambda _A}}{\frac{0.693}{\lambda _B}}=\frac{\lambda _B}{\lambda _A} $
Using Eq. (iii), we get
$ \frac{\left(T _{1 / 2}\right) _A}{\left(T _{1 / 2}\right) _B}=\frac{4}{1} $
Hence, from the given options, only option (d) satisfies this ratio.
Therefore, $\left(T _{1 / 2}\right) _A=20$ days
and $\quad\left(T _{1 / 2}\right) _B=5$ days
BETA
