When a monochromatic point source of light is at a distance `0.2m` from a photoelectric cell, the saturation current and cut-off voltage are `12.0 mA` and `0.5 V`, respec tively. If the same source is placed `0.4 m` away from the photoelectric cell, then the saturation current is `x` milli-ampere and the stopping. potential is `y` volt. Find `(x)/(y)`.

Answer

To solve the problem, we need to find the values of the saturation current \( x \) and the stopping potential \( y \) when the distance of the monochromatic point source from the photoelectric cell is changed from \( 0.2 \, m \) to \( 0.4 \, m \).

### Step-by-Step Solution:

1. **Understanding the Given Values:**
- Initial distance \( d_1 = 0.2 \, m \)
- Saturation current at \( d_1 \) is \( I_s = 12.0 \, mA \)
- Cut-off voltage (stopping potential) at \( d_1 \) is \( V_0 = 0.5 \, V \)
- New distance \( d_2 = 0.4 \, m \)

2. **Using the Inverse Square Law:**
The intensity of light (and thus the number of photoelectrons emitted) is inversely proportional to the square of the distance from the source. Therefore, we can express the relationship between the currents at the two distances as:
\[ I_{s2} = I_{s1} \left( \frac{d_1}{d_2} \right)^2 \]
where \( I_{s1} = 12.0 \, mA \) and \( d_1 = 0.2 \, m \), \( d_2 = 0.4 \, m \).

3. **Calculating the New Saturation Current \( x \):**
\[ I_{s2} = 12.0 \, mA \left( \frac{0.2}{0.4} \right)^2 \]
\[ I_{s2} = 12.0 \, mA \left( \frac{1}{2} \right)^2 = 12.0 \, mA \cdot \frac{1}{4} = 3.0 \, mA \]
Thus, \( x = 3.0 \, mA \).

4. **Determining the Stopping Potential \( y \):**
The stopping potential does not depend on the distance from the source. Therefore, the stopping potential remains the same:
\[ y = V_0 = 0.5 \, V \]

5. **Finding the Ratio \( \frac{x}{y} \):**
\[ \frac{x}{y} = \frac{3.0 \, mA}{0.5 \, V} = 6 \]

### Final Answer:
\[ \frac{x}{y} = 6 \]