In the previous question, if the intensity of light is made `4I_0`, then the stopping potential will become

Options

A. `1.36xx1V`
B. `1.36xx2V`
C. `1.36xx3V`
D. `1.36xx4V`

Explanation

To solve the problem, we need to understand the relationship between the intensity of light, the stopping potential, and the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding Stopping Potential**:
The stopping potential (V₀) is the potential required to stop the most energetic photoelectrons emitted from the surface of a material when light is incident on it.

2. **Effect of Intensity on Photoelectric Effect**:
Intensity (I) of light is related to the number of photons hitting the surface per unit time. When the intensity is increased to `4I₀`, it means that the number of incident photons is increased.

3. **Current and Intensity Relationship**:
The increase in intensity leads to an increase in the number of emitted photoelectrons, which increases the current (I). However, the stopping potential itself is not directly affected by the intensity of light.

4. **Kinetic Energy of Photoelectrons**:
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed as:
\[ K.E. = eV₀ = hf - \phi \]
where:
- \( e \) is the charge of the electron,
- \( V₀ \) is the stopping potential,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light,
- \( \phi \) is the work function of the material.

5. **Effect of Frequency on Stopping Potential**:
The stopping potential can only be increased if the frequency of the incident light is increased. Since the intensity increase does not change the frequency, the stopping potential remains the same.

6. **Conclusion**:
Therefore, if the intensity of light is increased to `4I₀`, the stopping potential will remain unchanged.

Thus, the stopping potential will still be \( V₀ = 13.6 \, \text{volts} \).

### Final Answer:
The stopping potential will remain the same at \( 13.6 \, \text{volts} \).

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