A Geigger Muller counter is used to study the radioactive process. In the absence of radioactive substances `A`, it counts `3` disintegration per second (dps). At the start in the presence of `A`, it records `23` dps, and after `10 m` in `13 dps`,
i. What does it count after `20 m` in ?

Options

A. `8 dps, 10 mi n`
B. `5 dps, 10 mi n`
C. `5 dps, 20 mi n`
D. `5 dps, 5 mi n`

Explanation

To solve the problem step by step, we will analyze the data provided and apply the principles of radioactive decay.

### Step-by-Step Solution:

1. **Understanding Zero Error**:
- The Geiger Muller counter shows a count of 3 disintegrations per second (dps) in the absence of radioactive substance A. This is considered the zero error of the counter.

2. **Initial Count with Substance A**:
- When substance A is present, the counter records 23 dps. To find the actual count of disintegrations from substance A, we subtract the zero error:
\[ \text{Count from A} = 23 \, \text{dps} - 3 \, \text{dps} = 20 \, \text{dps} \]

3. **Count After 10 Minutes**:
- After 10 minutes, the counter records 13 dps. Again, we subtract the zero error to find the actual count:
\[ \text{Count after 10 minutes} = 13 \, \text{dps} - 3 \, \text{dps} = 10 \, \text{dps} \]

4. **Decay Calculation**:
- We can see that the count has decreased from 20 dps to 10 dps in 10 minutes. This indicates a decay.
- The count halves from 20 dps to 10 dps in 10 minutes, which implies the half-life (t½) of the substance is 10 minutes.

5. **Count After 20 Minutes**:
- Since the half-life is 10 minutes, after another 10 minutes (total of 20 minutes), the count will halve again:
\[ \text{Count after 20 minutes} = \frac{10 \, \text{dps}}{2} = 5 \, \text{dps} \]
- Now, we add back the zero error:
\[ \text{Final count after 20 minutes} = 5 \, \text{dps} + 3 \, \text{dps} = 8 \, \text{dps} \]

### Final Answer:
The Geiger Muller counter will count **8 disintegrations per second (dps)** after 20 minutes.

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