Half lives for `alpha` and `beta` emission of a radioacative materila are 16 years and 48 years respectively. When material decays giving `alpha` and `beta` emission simultaneously, time in which `3//4^(th)` material decays is .

Options

A. 29 years
B. 24 years
C. 64 years
D. 12 years

Answer

To solve the problem, we need to determine the time it takes for three-fourths of a radioactive material to decay when it undergoes both alpha and beta emissions simultaneously. The half-lives for alpha and beta emissions are given as 16 years and 48 years, respectively.

### Step-by-Step Solution:

1. **Identify the Initial Amount**:
Let the initial amount of the radioactive material be \( N_0 \).

2. **Determine the Remaining Amount**:
If three-fourths of the material has decayed, then the remaining amount is:
\[ N = N_0 - \frac{3}{4}N_0 = \frac{1}{4}N_0 \]

3. **Calculate the Decay Constants**:
The decay constant (\( \lambda \)) is related to the half-life (\( t_{1/2} \)) by the formula:
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
- For alpha decay (\( t_{1/2} = 16 \) years):
\[ \lambda_{\alpha} = \frac{\ln 2}{16} \]
- For beta decay (\( t_{1/2} = 48 \) years):
\[ \lambda_{\beta} = \frac{\ln 2}{48} \]

4. **Calculate the Effective Decay Constant**:
When both decays occur simultaneously, the effective decay constant (\( \lambda_{\text{effective}} \)) is given by:
\[ \lambda_{\text{effective}} = \lambda_{\alpha} + \lambda_{\beta} = \frac{\ln 2}{16} + \frac{\ln 2}{48} \]
To combine these, find a common denominator (which is 48):
\[ \lambda_{\text{effective}} = \frac{3\ln 2}{48} + \frac{\ln 2}{48} = \frac{4\ln 2}{48} = \frac{\ln 2}{12} \]

5. **Determine the Effective Half-Life**:
The effective half-life (\( t_{1/2, \text{effective}} \)) can be calculated from the effective decay constant:
\[ t_{1/2, \text{effective}} = \frac{\ln 2}{\lambda_{\text{effective}}} = \frac{\ln 2}{\frac{\ln 2}{12}} = 12 \text{ years} \]

6. **Calculate the Time for Remaining Amount**:
The time taken to reduce the amount from \( N_0 \) to \( \frac{1}{4}N_0 \) corresponds to 2 half-lives, since:
- After 1 half-life: \( N = \frac{1}{2}N_0 \)
- After 2 half-lives: \( N = \frac{1}{4}N_0 \)

Therefore, the time taken is:
\[ t = 2 \times t_{1/2, \text{effective}} = 2 \times 12 \text{ years} = 24 \text{ years} \]

### Final Answer:
The time in which three-fourths of the material decays is **24 years**.