When a point source of light is at a distance of 50 cm from a photoelectric cell , the cut-off voltage is found to be ` V_(0)` . If the same source is placed at a distance of 1 m from the cell , then the cut-off voltage will be

Options

A. `V_(0)//4`
B. `V_(0) //2 `
C. `V_(0)`
D. ` 2 V_(0)`

Answer

To solve the problem, we need to understand the relationship between the distance of a light source from a photoelectric cell and the cut-off voltage (or stopping potential) in the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect occurs when light (or photons) hits a metal surface and causes the emission of electrons. The energy of the emitted electrons is given by the equation:
\[ K.E. = h\nu - W \]
where \( K.E. \) is the kinetic energy of the emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( W \) is the work function of the material.

2. **Cut-off Voltage Concept**:
The cut-off voltage \( V_0 \) is the voltage required to stop the emitted electrons from reaching the anode. It is related to the maximum kinetic energy of the emitted electrons:
\[ K.E. = eV_0 \]
where \( e \) is the charge of an electron. This means that the cut-off voltage is directly related to the frequency of the incident light, not the intensity.

3. **Intensity and Distance**:
The intensity \( I \) of light from a point source varies with distance \( r \) according to the inverse square law:
\[ I \propto \frac{P}{r^2} \]
where \( P \) is the power of the light source. However, while intensity affects the number of emitted electrons, it does not affect the maximum kinetic energy of the electrons, as long as the frequency of the light remains above the threshold frequency.

4. **Comparing Distances**:
In this problem, the light source is moved from 50 cm to 1 m. While the intensity decreases due to the increased distance, the frequency of the light (which determines the energy of the photons) remains unchanged. Therefore, the cut-off voltage \( V_0 \) will not change.

5. **Conclusion**:
Since the cut-off voltage is dependent on the frequency of the light and not on the intensity or distance from the source, we conclude that:
\[ V = V_0 \]
Hence, the cut-off voltage when the source is at 1 m will still be \( V_0 \).

### Final Answer:
The cut-off voltage when the source is placed at a distance of 1 m from the cell will be \( V_0 \).

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