Three fourth of the active decays in a radioactive sample in `3//4 sec`. The half-life of the sample is

Options

A. `(1)/(2) sec`
B. `1 sec`
C. `(3)/(8) sec`
D. `(3)/(4)sec`

Explanation

To solve the problem, we need to determine the half-life of a radioactive sample given that three-fourths of it decays in \( \frac{3}{4} \) seconds.

### Step-by-Step Solution:

1. **Understanding the Decay**:
- Let the initial amount of the radioactive sample be \( A_0 \).
- According to the problem, three-fourths of the sample decays in \( \frac{3}{4} \) seconds. This means that the amount that remains after this time is:
\[ A = A_0 - \frac{3}{4} A_0 = \frac{1}{4} A_0 \]

2. **Using the Decay Formula**:
- The relationship between the remaining quantity of a radioactive substance and time is given by:
\[ A = A_0 e^{-\lambda t} \]
- Here, \( \lambda \) is the decay constant, and \( t \) is the time elapsed.
- Substituting the values we have:
\[ \frac{1}{4} A_0 = A_0 e^{-\lambda \left(\frac{3}{4}\right)} \]

3. **Simplifying the Equation**:
- We can cancel \( A_0 \) from both sides (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{4} = e^{-\lambda \left(\frac{3}{4}\right)} \]

4. **Taking the Natural Logarithm**:
- Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{4}\right) = -\lambda \left(\frac{3}{4}\right) \]
- We know that \( \ln\left(\frac{1}{4}\right) = \ln(4^{-1}) = -\ln(4) \), so:
\[ -\ln(4) = -\lambda \left(\frac{3}{4}\right) \]
- This simplifies to:
\[ \lambda = \frac{4 \ln(4)}{3} \]

5. **Finding the Half-Life**:
- The half-life \( T_{1/2} \) is given by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
- Substituting the value of \( \lambda \):
\[ T_{1/2} = \frac{\ln(2)}{\frac{4 \ln(4)}{3}} = \frac{3 \ln(2)}{4 \ln(4)} \]
- Since \( \ln(4) = 2 \ln(2) \):
\[ T_{1/2} = \frac{3 \ln(2)}{4 \cdot 2 \ln(2)} = \frac{3}{8} \text{ seconds} \]

### Final Answer:
The half-life of the radioactive sample is \( \frac{3}{8} \) seconds.