A caesium photocell , with a steady potential difference of `60 V` across , is alluminated by a bright point source of light `50 cm` away. When the same light is placed `1 m` away the photoelectrons emitted from the cell

Options

A. `(n/2)`
B. `2n`
C. `(n/4)`
D. `n`

Answer

To solve the problem involving the caesium photocell and the effect of distance on the number of emitted photoelectrons, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The number of photoelectrons emitted (n) is directly proportional to the intensity (I) of the light. The intensity of light from a point source is inversely proportional to the square of the distance (d) from the source. Therefore, we can express this relationship as:
\[ n \propto \frac{1}{d^2} \]

2. **Setting Up the Ratios**:
Let \( n_1 \) be the number of photoelectrons emitted when the light source is at distance \( d_1 \) (50 cm) and \( n_2 \) be the number of photoelectrons emitted when the light source is at distance \( d_2 \) (1 m = 100 cm). We can set up the ratio:
\[ \frac{n_1}{n_2} = \frac{d_2^2}{d_1^2} \]

3. **Substituting the Distances**:
Substitute \( d_1 = 50 \, \text{cm} \) and \( d_2 = 100 \, \text{cm} \) into the equation:
\[ \frac{n_1}{n_2} = \frac{(100 \, \text{cm})^2}{(50 \, \text{cm})^2} \]

4. **Calculating the Ratio**:
Calculate the squares:
\[ \frac{n_1}{n_2} = \frac{10000}{2500} = 4 \]
This means:
\[ n_1 = 4n_2 \]

5. **Conclusion**:
From the above calculation, we conclude that when the light source is moved from 50 cm to 1 m away, the number of photoelectrons emitted decreases to one-fourth of the original number. Therefore, the answer is:
\[ n_2 = \frac{n_1}{4} \]

### Final Answer:
The number of photoelectrons emitted when the light source is 1 m away is one-fourth of the number emitted when it is 50 cm away.

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