The half-lives of radioactive sample are 30 years and 60 years for two decay processes. If the sample decays by both the processes simultaneously. The time after which, only one-fourth of the sample will remain is

Options

A. (a) 10 years
B. (b) 20 years
C. (c) 40 years
D. (d) 60 years

Explanation

To solve the problem, we need to find out the time after which only one-fourth of a radioactive sample remains when it decays through two processes with half-lives of 30 years and 60 years.

### Step-by-Step Solution:

1. **Identify the Half-Lives**:
- The half-life for the first decay process, \( t_1 = 30 \) years.
- The half-life for the second decay process, \( t_2 = 60 \) years.

2. **Calculate the Decay Constants**:
- The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
- For the first process:
\[ \lambda_1 = \frac{\ln(2)}{30} \]
- For the second process:
\[ \lambda_2 = \frac{\ln(2)}{60} \]

3. **Total Decay Constant**:
- When both decay processes occur simultaneously, the total decay constant \( \lambda_{total} \) is the sum of the individual decay constants:
\[ \lambda_{total} = \lambda_1 + \lambda_2 = \frac{\ln(2)}{30} + \frac{\ln(2)}{60} \]
- To combine these, find a common denominator (which is 60):
\[ \lambda_{total} = \frac{2\ln(2)}{60} + \frac{\ln(2)}{60} = \frac{3\ln(2)}{60} = \frac{\ln(2)}{20} \]

4. **Calculate the Effective Half-Life**:
- The effective half-life \( t_{1/2, effective} \) can be calculated using:
\[ t_{1/2, effective} = \frac{\ln(2)}{\lambda_{total}} = \frac{\ln(2)}{\frac{\ln(2)}{20}} = 20 \text{ years} \]

5. **Determine the Time for One-Fourth of the Sample to Remain**:
- To find the time \( t \) when only one-fourth of the sample remains, we note that:
- After one half-life, \( N = \frac{N_0}{2} \)
- After two half-lives, \( N = \frac{N_0}{4} \)
- Therefore, it takes two half-lives to reduce the sample to one-fourth:
\[ t = 2 \times t_{1/2, effective} = 2 \times 20 = 40 \text{ years} \]

### Final Answer:
The time after which only one-fourth of the sample will remain is **40 years**.