Radioactive nucleus A decays into B with half-life `10sec`. A also convert into C with half-life `100 sec`. Find half-life of A for both emission.

Options

A. `6sec`
B. `9sec`
C. `3sec`
D. `2sec`

Answer

To solve the problem of finding the half-life of radioactive nucleus A when it decays into both B and C, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Decay Process**:
- Nucleus A decays into B with a half-life \( T_{1/2}^{(B)} = 10 \) seconds.
- Nucleus A decays into C with a half-life \( T_{1/2}^{(C)} = 100 \) seconds.

2. **Define Decay Constants**:
- The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
- For decay into B:
\[ \lambda_1 = \frac{\ln(2)}{10} \]
- For decay into C:
\[ \lambda_2 = \frac{\ln(2)}{100} \]

3. **Combine Decay Constants**:
- Since A decays into both B and C, the total decay constant \( \lambda_{eq} \) is the sum of the individual decay constants:
\[ \lambda_{eq} = \lambda_1 + \lambda_2 \]
- Substituting the values:
\[ \lambda_{eq} = \frac{\ln(2)}{10} + \frac{\ln(2)}{100} \]

4. **Finding a Common Denominator**:
- The common denominator for \( 10 \) and \( 100 \) is \( 100 \):
\[ \lambda_{eq} = \frac{10 \ln(2)}{100} + \frac{\ln(2)}{100} = \frac{11 \ln(2)}{100} \]

5. **Calculate the Equivalent Half-Life**:
- Using the relationship between decay constant and half-life:
\[ T_{1/2}^{(eq)} = \frac{\ln(2)}{\lambda_{eq}} = \frac{\ln(2)}{\frac{11 \ln(2)}{100}} = \frac{100}{11} \]
- Calculating \( \frac{100}{11} \) gives approximately \( 9.09 \) seconds.

### Final Answer:
The half-life of nucleus A for both emissions is approximately \( 9 \) seconds.

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