A photo-cell is illuminated by a source of light which is placed at a distance d from the cell . If the distance become d/2 . Then number of electrons emitted per seconed will be

Options

A. Remain same
B. Four times
C. Two times
D. One-fourth

Answer

To solve the problem, we will analyze how the change in distance from the light source affects the number of electrons emitted from the photocell.

### Step-by-Step Solution:

1. **Understanding the relationship between intensity and distance**:
- The intensity \( I \) of light from a point source is inversely proportional to the square of the distance \( d \) from the source. Mathematically, this can be expressed as:
\[ I \propto \frac{1}{d^2} \]

2. **Initial and final distances**:
- Let the initial distance from the light source to the photocell be \( d \).
- When the distance is changed to \( \frac{d}{2} \), we need to find the new intensity \( I_2 \) at this new distance.

3. **Calculating the intensities**:
- The initial intensity \( I_1 \) at distance \( d \) can be represented as:
\[ I_1 \propto \frac{1}{d^2} \]
- The new intensity \( I_2 \) at distance \( \frac{d}{2} \) is:
\[ I_2 \propto \frac{1}{\left(\frac{d}{2}\right)^2} = \frac{1}{\frac{d^2}{4}} = \frac{4}{d^2} \]

4. **Finding the ratio of intensities**:
- Now, we can find the ratio of the two intensities:
\[ \frac{I_1}{I_2} = \frac{\frac{1}{d^2}}{\frac{4}{d^2}} = \frac{1}{4} \]
- This implies:
\[ I_2 = 4 I_1 \]

5. **Relating intensity to the number of emitted electrons**:
- The number of electrons emitted per second from the photocell is directly proportional to the intensity of the light incident on it:
\[ N \propto I \]
- Therefore, if the intensity increases to \( 4 I_1 \), the number of electrons emitted \( N_2 \) at the new intensity will be:
\[ N_2 = 4 N_1 \]
- Here, \( N_1 \) is the number of electrons emitted at the initial intensity \( I_1 \).

6. **Conclusion**:
- Thus, when the distance is reduced from \( d \) to \( \frac{d}{2} \), the number of electrons emitted per second becomes four times the initial number of electrons emitted.

### Final Answer:
The number of electrons emitted per second will be \( 4N_1 \), where \( N_1 \) is the initial number of electrons emitted.

---