The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700 per minute. Calculate
(a) decay constant and (b) half-life of the sample

Explanation

To solve the problem, we need to calculate the decay constant (λ) and the half-life (t₁/₂) of the radioactive sample based on the given disintegration rates at two different times.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial disintegration rate (activity) \( R_0 = 4750 \) disintegrations per minute.
- Disintegration rate after 5 minutes \( R = 2700 \) disintegrations per minute.
- Time interval \( t = 5 \) minutes.

2. **Use the Radioactive Decay Formula:**
The activity of a radioactive sample at any time \( t \) is given by the formula:
\[ R = R_0 e^{-\lambda t} \]
where:
- \( R \) is the activity at time \( t \),
- \( R_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

3. **Set Up the Equations:**
For \( t = 0 \):
\[ R_0 = 4750 \]
For \( t = 5 \):
\[ R = 2700 = 4750 e^{-5\lambda} \]

4. **Divide the Two Equations:**
To eliminate \( R_0 \), divide the second equation by the first:
\[ \frac{2700}{4750} = e^{-5\lambda} \]

5. **Calculate the Left Side:**
\[ \frac{2700}{4750} = 0.5684 \]
Thus, we have:
\[ 0.5684 = e^{-5\lambda} \]

6. **Take the Natural Logarithm:**
Taking the natural logarithm of both sides:
\[ \ln(0.5684) = -5\lambda \]

7. **Solve for the Decay Constant (λ):**
\[ \lambda = -\frac{\ln(0.5684)}{5} \]
Calculate \( \ln(0.5684) \):
\[ \ln(0.5684) \approx -0.566 \]
Therefore:
\[ \lambda = -\frac{-0.566}{5} \approx 0.1132 \text{ min}^{-1} \]

8. **Calculate the Half-Life (t₁/₂):**
The half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
Substitute the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{0.1132} \approx 6.12 \text{ minutes} \]

### Final Answers:
- (a) Decay constant \( \lambda \approx 0.1132 \text{ min}^{-1} \)
- (b) Half-life \( t_{1/2} \approx 6.12 \text{ minutes} \)