Two radioactive samples `1` and `2` have equal number of nuclei initially. They have halg-lives of `10` seconds and `20` seconds. The ratio of number of nuclei of `1` and `2` at `t=60` seconds is :

Options

A. `8:1`
B. `1:8`
C. `3:8`
D. `1:9`

Answer

To solve the problem, we will use the formula for radioactive decay, which describes how the number of nuclei of a radioactive substance decreases over time. The formula is given by:

\[ n(t) = n_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

where:
- \( n(t) \) is the number of nuclei at time \( t \),
- \( n_0 \) is the initial number of nuclei,
- \( T_{1/2} \) is the half-life of the substance,
- \( t \) is the time elapsed.

### Step-by-Step Solution

1. **Identify the half-lives and initial conditions**:
- Let \( n_{0} \) be the initial number of nuclei for both samples.
- For sample 1, \( T_{1/2} = 10 \) seconds.
- For sample 2, \( T_{1/2} = 20 \) seconds.

2. **Calculate the number of nuclei for sample 1 at \( t = 60 \) seconds**:
\[ n_1(60) = n_0 \left(\frac{1}{2}\right)^{\frac{60}{10}} = n_0 \left(\frac{1}{2}\right)^{6} = n_0 \cdot \frac{1}{64} \]

3. **Calculate the number of nuclei for sample 2 at \( t = 60 \) seconds**:
\[ n_2(60) = n_0 \left(\frac{1}{2}\right)^{\frac{60}{20}} = n_0 \left(\frac{1}{2}\right)^{3} = n_0 \cdot \frac{1}{8} \]

4. **Find the ratio of the number of nuclei of sample 1 to sample 2 at \( t = 60 \) seconds**:
\[ \text{Ratio} = \frac{n_1(60)}{n_2(60)} = \frac{n_0 \cdot \frac{1}{64}}{n_0 \cdot \frac{1}{8}} = \frac{\frac{1}{64}}{\frac{1}{8}} = \frac{1}{64} \cdot \frac{8}{1} = \frac{8}{64} = \frac{1}{8} \]

### Final Answer
The ratio of the number of nuclei of sample 1 to sample 2 at \( t = 60 \) seconds is \( \frac{1}{8} \).