Modern Physics 3 Question 4
4. 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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Solution:
- 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 $B$,
$$ R _B=\lambda _B N _B=20 mCi $$
As given in the question,
$$ N _A=2 N _B $$
$\Rightarrow \quad R _A=\lambda _A\left(2 N _B\right)=10 mCi$
$\therefore$ 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 $\quad \frac{\lambda _A}{\lambda _B}=\frac{1}{4}$
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
