The count rate of a Geiger Muller counter for the radiation of a radioactive material of half-life `30` min decreases to `5 s^(-1)` after `2 h`. The initial count rate was

Options

A. `20 s^(-1)`
B. `25 s^(-1)`
C. `80 s^(-1)`
D. `625 s^(-1)`

Answer

To solve the problem, we need to determine the initial count rate of a Geiger Muller counter for a radioactive material, given its half-life and the count rate after a certain time period.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Half-life of the radioactive material, \( t_{1/2} = 30 \) minutes.
- Time elapsed, \( t = 2 \) hours = \( 120 \) minutes.
- Count rate after 2 hours, \( N = 5 \) counts per second.

2. **Calculate Number of Half-Lives**:
- The number of half-lives \( n \) can be calculated using the formula:
\[ n = \frac{t}{t_{1/2}} \]
- Substituting the values:
\[ n = \frac{120 \text{ min}}{30 \text{ min}} = 4 \]

3. **Use the Radioactive Decay Law**:
- The relationship between the initial count rate \( N_0 \) and the count rate after \( n \) half-lives \( N \) is given by:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]
- Rearranging this gives:
\[ N_0 = N \cdot 2^n \]

4. **Substituting Known Values**:
- Now substitute \( N = 5 \) counts per second and \( n = 4 \):
\[ N_0 = 5 \cdot 2^4 \]
- Calculate \( 2^4 = 16 \):
\[ N_0 = 5 \cdot 16 = 80 \]

5. **Final Result**:
- The initial count rate \( N_0 \) is \( 80 \) counts per second.

### Conclusion:
The initial count rate was \( 80 \) counts per second.

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