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

Options

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

Explanation

To solve the problem step by step, we will use the properties of radioactive decay and the relationship between the fraction of material remaining and time.

### Step 1: Understand the given information
We know that the fraction of a radioactive material that remains active after time \( t \) is given as:
\[ \frac{n}{n_0} = \frac{9}{16} \]
where \( n \) is the number of active nuclei at time \( t \) and \( n_0 \) is the initial number of active nuclei.

### Step 2: Relate the fraction to the decay formula
The relationship between the number of active nuclei and time is given by the exponential decay formula:
\[ \frac{n}{n_0} = e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 3: Set up the equation
From the information provided, we can set up the equation:
\[ e^{-\lambda t} = \frac{9}{16} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm on both sides gives:
\[ -\lambda t = \ln\left(\frac{9}{16}\right) \]
This can be rewritten as:
\[ \lambda t = -\ln\left(\frac{9}{16}\right) \]

### Step 5: Find the fraction remaining after time \( \frac{t}{2} \)
Now, we need to find the fraction that remains after time \( \frac{t}{2} \). Using the decay formula again:
\[ \frac{n'}{n_0} = e^{-\lambda \frac{t}{2}} \]

### Step 6: Substitute for \( \lambda t \)
We can express \( \lambda \frac{t}{2} \) in terms of \( \lambda t \):
\[ \lambda \frac{t}{2} = \frac{1}{2} \cdot (-\ln\left(\frac{9}{16}\right)) = -\frac{1}{2} \ln\left(\frac{9}{16}\right) \]

### Step 7: Rewrite the fraction
Thus, we have:
\[ \frac{n'}{n_0} = e^{-\frac{1}{2} \ln\left(\frac{9}{16}\right)} \]
Using the property of exponents:
\[ e^{-\frac{1}{2} \ln\left(\frac{9}{16}\right)} = \left(\frac{9}{16}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{\frac{9}{16}}} = \frac{4}{3} \]

### Step 8: Final result
So, the fraction that remains active after time \( \frac{t}{2} \) is:
\[ \frac{n'}{n_0} = \frac{3}{4} \]

### Conclusion
The fraction of the radioactive material that remains active after time \( \frac{t}{2} \) is \( \frac{3}{4} \).

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