A photocell is illuminated by a small bright source places `1` m away when the same source of light is placed `(1)/(2)` m away. The number of electron emitted by photocathode would be

Options

A. (a)decrease by a factor of 2
B. (b)increase by a factor of 2
C. (c)decrease by a factor of 4
D. (d)increase by a factor of 4

Answer

To solve the problem regarding the photocell and the number of electrons emitted when the distance of the light source is changed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between intensity and distance**:
The intensity \( I \) of light from a point source is inversely proportional to the square of the distance \( r \) from the source. This can be expressed mathematically as:
\[ I \propto \frac{1}{r^2} \]

2. **Define the distances**:
Let \( r_1 = 1 \, \text{m} \) (the initial distance) and \( r_2 = \frac{1}{2} \, \text{m} \) (the new distance).

3. **Calculate the intensities**:
Using the relationship from step 1, we can express the intensities at these distances:
\[ I_1 \propto \frac{1}{r_1^2} = \frac{1}{1^2} = 1 \]
\[ I_2 \propto \frac{1}{r_2^2} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4 \]

4. **Determine the ratio of intensities**:
Now, we can find the ratio of the intensities:
\[ \frac{I_2}{I_1} = \frac{4}{1} = 4 \]
This means that the intensity at \( r_2 \) is 4 times that at \( r_1 \).

5. **Relate intensity to the number of emitted electrons**:
The number of electrons emitted from the photocathode is directly proportional to the intensity of light incident on it. Therefore, if the intensity increases by a factor of 4, the number of emitted electrons also increases by the same factor.

6. **Conclusion**:
Thus, the number of electrons emitted when the source is placed at \( \frac{1}{2} \, \text{m} \) is 4 times the number emitted when the source is at \( 1 \, \text{m} \).

### Final Answer:
The number of electrons emitted by the photocathode would be increased by a factor of 4.