Q 12-13-126JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
A sample of radioactive material $A$, that has an activity of $10\ \text{mCi}$ ($1\ \text{Ci} = 3.7\times10^{10}$ 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
Answer: (D) 20 days and 5 days
Activity $R = \lambda N$. With $N_A = 2N_B$:
$$\frac{R_A}{R_B} = \frac{\lambda_A\cdot2N_B}{\lambda_B N_B} = \frac{10}{20} \Rightarrow \frac{\lambda_A}{\lambda_B} = \frac14$$
Half-life is inversely proportional to $\lambda$, so $T_A = 4T_B$. Only "20 days and 5 days" fits.
Solution by Sreeraj P, M.Sc Physics