A `5 W` source emits monochromatic light of wavelength `5000 Å`. When placed `0.5 m` away , it liberates photoelectrons from a photosensitive metallic surface . When the source is moved to a distance of `1.0 m`, the number of photoelectrons liberated will be reduced by a factor of

Options

A. `4`
B. `8`
C. `16`
D. `2`

Explanation

To solve the problem, we need to determine how the number of photoelectrons liberated from a photosensitive surface changes when the distance from a light source is altered.

### Step-by-Step Solution:

1. **Understand the relationship between intensity and distance**:
The intensity \( I \) of light from a point source is given by the formula:
\[ I \propto \frac{P}{d^2} \]
where \( P \) is the power of the source and \( d \) is the distance from the source.

2. **Calculate the intensity at the initial distance (0.5 m)**:
Given that the power \( P = 5 \, W \) and the initial distance \( d_1 = 0.5 \, m \):
\[ I_1 \propto \frac{5}{(0.5)^2} = \frac{5}{0.25} = 20 \, W/m^2 \]

3. **Calculate the intensity at the new distance (1.0 m)**:
Now, for the new distance \( d_2 = 1.0 \, m \):
\[ I_2 \propto \frac{5}{(1.0)^2} = \frac{5}{1} = 5 \, W/m^2 \]

4. **Determine the reduction in intensity**:
The ratio of the intensities at the two distances is:
\[ \frac{I_1}{I_2} = \frac{20}{5} = 4 \]
This means that the intensity (and therefore the number of photons and photoelectrons) at 1.0 m is \( \frac{1}{4} \) of that at 0.5 m.

5. **Conclusion**:
Since the number of photoelectrons liberated is directly proportional to the intensity of the incident light, the number of photoelectrons will be reduced by a factor of 4 when the source is moved from 0.5 m to 1.0 m.

### Final Answer:
The number of photoelectrons liberated will be reduced by a factor of **4**.