Question

A sample of radioactive material $A$, that has an activity of $10\text{mCi}$ $(1\text{Ci}=3.7\times {10}^{10})\text{decays/s}$, has twice the number of nuclei as another sample of a different radioactive material $B$ which has an activity of $20\text{mCi}$. The correct choices for half-lives of $A$ and $B$ would then be respectively:

• A. $10\text{days and}40\text{days}$
• B. $5\text{days and}10\text{days}$
• C. $20\text{days and}5\text{days}$
• D. $20\text{days and}10\text{days}$

Answer 1

The correct option is C $20\text{days and}5\text{days}$

$\text{Activity},A=\lambda N$

$\text{For material A,}10=\left(2N_0\right){\lambda }_A$

$\text{For material B,}20=N_0{\lambda }_B$

$\Rightarrow {\lambda }_B=4{\lambda }_A$

$\because T_{\frac{1}{2}}=\frac{0.693}{\lambda }$

$\Rightarrow T_{\frac{1}{2}A}=4T_{\frac{1}{2}B}$

The above condition is satisfied by option $\left(C\right)$

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