A freshly prepared sample of a radioisotope of half - life `1386 s ` has activity `10^(3) ` disintegrations per second Given that `ln 2 = 0.693` the fraction of the initial number of nuclei (expressed in nearest integer percentage ) that will decay in the first `80 s ` after preparation of the sample is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Half-life of the radioisotope, \( t_{1/2} = 1386 \, \text{s} \)
- Activity of the sample, \( A = 10^3 \, \text{disintegrations per second} \)
- Time interval, \( t = 80 \, \text{s} \)
- Natural logarithm of 2, \( \ln 2 = 0.693 \)

### Step 2: Calculate the decay constant (\( \alpha \))
The decay constant \( \alpha \) is related to the half-life by the formula:
\[ \alpha = \frac{\ln 2}{t_{1/2}} \]
Substituting the values:
\[ \alpha = \frac{0.693}{1386} \approx 5 \times 10^{-4} \, \text{s}^{-1} \]

### Step 3: Calculate the fraction of nuclei that decay in the first 80 seconds
The fraction of the initial number of nuclei that remain after time \( t \) is given by:
\[ \frac{N}{N_0} = e^{-\alpha t} \]
Thus, the fraction that decays is:
\[ \frac{N_0 - N}{N_0} = 1 - e^{-\alpha t} \]
Substituting \( \alpha \) and \( t \):
\[ \frac{N_0 - N}{N_0} = 1 - e^{-5 \times 10^{-4} \times 80} \]

### Step 4: Calculate \( e^{-\alpha t} \)
Calculating the exponent:
\[ -\alpha t = -5 \times 10^{-4} \times 80 = -0.04 \]
Now, we find \( e^{-0.04} \):
Using the approximation \( e^{-x} \approx 1 - x \) for small \( x \):
\[ e^{-0.04} \approx 1 - 0.04 = 0.9608 \]
Thus,
\[ \frac{N}{N_0} \approx 0.9608 \]
So,
\[ \frac{N_0 - N}{N_0} \approx 1 - 0.9608 = 0.0392 \]

### Step 5: Convert to percentage
To express this as a percentage:
\[ \frac{N_0 - N}{N_0} \times 100 \approx 0.0392 \times 100 \approx 3.92\% \]
Rounding to the nearest integer gives:
\[ \approx 4\% \]

### Final Answer
The fraction of the initial number of nuclei that will decay in the first 80 seconds is approximately **4%**.

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