The fraction of a radioactive material which reamins active after time t is `9//16`. The fraction which remains active after time `t//2` will be .

Options

A. `(4)/(5)`
B. `(7)/(8)`
C. `(3)/(5)`
D. `(3)/(4)`

Explanation

To solve the problem, we need to determine the fraction of a radioactive material that remains active after a time of \( \frac{t}{2} \), given that the fraction remaining after time \( t \) is \( \frac{9}{16} \).

### Step-by-step Solution:

1. **Understanding the decay formula**:
The fraction of radioactive material remaining after time \( t \) can be described by the formula:
\[ \frac{N(t)}{N_0} = e^{-\lambda t} \]
where \( N(t) \) is the amount remaining at time \( t \), \( N_0 \) is the initial amount, and \( \lambda \) is the decay constant.

2. **Using the given information**:
From the problem, we know:
\[ \frac{N(t)}{N_0} = \frac{9}{16} \]
This implies:
\[ e^{-\lambda t} = \frac{9}{16} \]

3. **Finding the fraction remaining after time \( \frac{t}{2} \)**:
We need to find the fraction remaining after time \( \frac{t}{2} \):
\[ \frac{N\left(\frac{t}{2}\right)}{N_0} = e^{-\lambda \left(\frac{t}{2}\right)} \]

4. **Relating the two fractions**:
We can express \( e^{-\lambda t} \) in terms of \( e^{-\lambda \left(\frac{t}{2}\right)} \):
\[ e^{-\lambda t} = \left(e^{-\lambda \left(\frac{t}{2}\right)}\right)^2 \]
Thus, we can write:
\[ e^{-\lambda \left(\frac{t}{2}\right)} = \sqrt{e^{-\lambda t}} = \sqrt{\frac{9}{16}} \]

5. **Calculating the square root**:
Now, we calculate the square root:
\[ \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4} \]

6. **Conclusion**:
Therefore, the fraction of the radioactive material that remains active after time \( \frac{t}{2} \) is:
\[ \frac{N\left(\frac{t}{2}\right)}{N_0} = \frac{3}{4} \]

### Final Answer:
The fraction which remains active after time \( \frac{t}{2} \) is \( \frac{3}{4} \).

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