Question

Two radioactive substances A and B have same number of initial nuclei. If the half-lives of A and B are 1.5 days and 4.5 days respectively, then the ratio of the number of nuclei remaining in A and B after 9 days is

• A. 1:16
• B. 1:1
• C. 1:4
• D. 1:8

Hint:

Ratio of remaining nuclei $N_A/N_B = (1/2)^{n_A - n_B}$. Here $n_A - n_B = 6 - 2 = 4$. Ratio is $(1/2)^4 = 1/16$.

Answer 1

The Correct Option is A

Step 1: Formula for Remaining Nuclei:
$N = N_0 \left(\frac{1}{2}\right)^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Step 2: Calculate for Substance A:
Half-life $T_A = 1.5$ days. Time $t = 9$ days. Number of half-lives $n_A = \frac{9}{1.5} = 6$. Remaining nuclei $N_A = N_0 \left(\frac{1}{2}\right)^6 = \frac{N_0}{64}$.
Step 3: Calculate for Substance B:
Half-life $T_B = 4.5$ days. Time $t = 9$ days. Number of half-lives $n_B = \frac{9}{4.5} = 2$. Remaining nuclei $N_B = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}$.
Step 4: Calculate Ratio:
\[ \frac{N_A}{N_B} = \frac{N_0/64}{N_0/4} = \frac{4}{64} = \frac{1}{16} \]