If the frequency of incident light is tripled, the stopping potential will

Options

A. be tripled
B. become one third
C. become more than triple
D. become less than triple but more than double

Answer

To solve the problem, we need to analyze the relationship between the frequency of incident light and the stopping potential in the context of the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect states that when light of a certain frequency (ν) shines on a metal surface, it can eject electrons from that surface. The kinetic energy (KE) of the emitted electrons is given by the equation:
\[ KE = h\nu - \phi \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Relating Kinetic Energy to Stopping Potential**:
The kinetic energy of the emitted electrons can also be expressed in terms of the stopping potential (Vs):
\[ KE = eV_s \]
where \( e \) is the charge of the electron and \( V_s \) is the stopping potential.

3. **Setting the Equations Equal**:
Since both expressions represent the kinetic energy of the emitted electrons, we can set them equal to each other:
\[ eV_s = h\nu - \phi \]

4. **Expressing Stopping Potential**:
Rearranging the equation gives us the stopping potential in terms of frequency:
\[ V_s = \frac{h\nu - \phi}{e} \]

5. **Effect of Tripling the Frequency**:
If the frequency of the incident light is tripled (i.e., \( \nu \) becomes \( 3\nu \)), we can substitute this into our equation for stopping potential:
\[ V_s' = \frac{h(3\nu) - \phi}{e} \]
Simplifying this gives:
\[ V_s' = \frac{3h\nu - \phi}{e} \]

6. **Comparing the New Stopping Potential to the Original**:
The original stopping potential was:
\[ V_s = \frac{h\nu - \phi}{e} \]
If we compare \( V_s' \) to \( V_s \):
\[ V_s' = 3 \left(\frac{h\nu - \phi}{e}\right) + \text{(additional term due to the work function)} \]
Since the work function \( \phi \) remains constant, the stopping potential will increase proportionally to the increase in frequency.

7. **Conclusion**:
Therefore, if the frequency of the incident light is tripled, the stopping potential will also be tripled:
\[ V_s' = 3V_s \]

### Final Answer:
The stopping potential will be tripled.

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