`36%` amount of a radioactive sample disintegrates in t time. Calculate how much percentage fraction will decay in `t//2` time.

Explanation

To solve the problem of how much percentage of a radioactive sample will decay in half the time (t/2), given that 36% disintegrates in time t, we will follow these steps:

### Step 1: Understand the decay law
The radioactive decay law states that the number of active nuclei remaining after time t can be expressed as:
\[ N = N_0 e^{-\lambda t} \]
where:
- \( N \) is the number of active nuclei remaining after time t,
- \( N_0 \) is the initial number of active nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 2: Calculate the fraction remaining after time t
Since 36% of the sample disintegrates in time t, this means that 64% remains. Therefore, we can express this as:
\[ \frac{N}{N_0} = 1 - 0.36 = 0.64 \]

### Step 3: Set up the equation for time t
From the decay law, we can write:
\[ \frac{N}{N_0} = e^{-\lambda t} \]
Substituting the fraction we found:
\[ 0.64 = e^{-\lambda t} \]
This is our equation (1).

### Step 4: Calculate the fraction remaining after time t/2
Now, we need to find out how much remains after half the time, t/2. Using the decay law again:
\[ \frac{N}{N_0} = e^{-\lambda (t/2)} \]

### Step 5: Relate the two equations
We can express the fraction remaining after time t/2 in terms of the fraction remaining after time t:
\[ \frac{N}{N_0} = (0.64)^{1/2} \]
This is because the exponential function can be rewritten as:
\[ e^{-\lambda (t/2)} = (e^{-\lambda t})^{1/2} \]

### Step 6: Calculate the value
Now, we compute:
\[ (0.64)^{1/2} = 0.8 \]
Thus, the fraction remaining after time t/2 is 0.8.

### Step 7: Calculate the percentage decayed
To find out how much percentage has decayed in time t/2, we subtract the fraction remaining from 1:
\[ \text{Percentage decayed} = 1 - 0.8 = 0.2 \]
Converting this to percentage:
\[ 0.2 \times 100 = 20\% \]

### Final Answer
Therefore, the percentage fraction that will decay in time \( t/2 \) is **20%**.

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