when a monochromatic point source of light is at a distance 0.2 m from a photoelectric cell, the saturation current and cut-off voltage are 12.0 mA and 0.5 V. If the same source is placed 0.4 m away from the photoelectric cell, then the saturation current and the stopping potential respectively are

Options

A. 4 mA and 1 V
B. 3 mA and 1 V
C. 12 mA nad 0.5 V
D. 3 mA nad 0.5 V

Answer

To solve the problem step by step, we will analyze the relationship between the distance of the light source from the photoelectric cell, the saturation current, and the stopping potential.

### Step 1: Understand the relationship between distance and intensity
The intensity \( I \) of light from a point source is inversely proportional to the square of the distance \( R \) from the source:
\[ I \propto \frac{1}{R^2} \]
This means that as the distance from the light source increases, the intensity decreases.

### Step 2: Determine the initial conditions
Given:
- Initial distance \( R_1 = 0.2 \, m \)
- Initial saturation current \( I_{s1} = 12.0 \, mA \)
- Stopping potential \( V_s = 0.5 \, V \)

### Step 3: Calculate the new distance
The new distance from the light source is:
\[ R_2 = 0.4 \, m \]

### Step 4: Use the intensity relationship to find the new saturation current
Using the relationship of intensity with distance:
\[ \frac{I_{s2}}{I_{s1}} = \frac{R_1^2}{R_2^2} \]
Substituting the values:
\[ \frac{I_{s2}}{12.0 \, mA} = \frac{(0.2)^2}{(0.4)^2} \]
Calculating the right side:
\[ \frac{(0.2)^2}{(0.4)^2} = \frac{0.04}{0.16} = \frac{1}{4} \]
Thus:
\[ I_{s2} = 12.0 \, mA \times \frac{1}{4} = 3.0 \, mA \]

### Step 5: Determine the stopping potential
The stopping potential \( V_s \) is dependent on the frequency of the light and does not change with distance. Therefore, the stopping potential remains the same:
\[ V_{s2} = V_s = 0.5 \, V \]

### Final Answer
The saturation current and stopping potential when the source is placed 0.4 m away from the photoelectric cell are:
\[ \text{Saturation Current} = 3.0 \, mA, \quad \text{Stopping Potential} = 0.5 \, V \]

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