In a photoelectric experiment the stopping potential for the incident light of wavelength 4000Å is 2 volt. If the wavelength be changed to 3000 Å, the stopping potential will be

Options

A. 2V
B. zero
C. less than 2 V
D. more than 2 V

Explanation

To solve the problem, we need to understand the relationship between the stopping potential, the wavelength of incident light, and the energy of the emitted electrons in the photoelectric effect.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect states that when light of a certain frequency (or wavelength) hits a metal surface, it can eject electrons from that surface. The energy of the incident photons is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.

2. **Calculate the Energy for 4000 Å**:
For the first case, where the wavelength is \(4000 \, \text{Å}\) (or \(4000 \times 10^{-10} \, \text{m}\)):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{hc}{4000 \times 10^{-10}} \]

3. **Relate Stopping Potential to Energy**:
The stopping potential \(V_s\) is related to the maximum kinetic energy of the emitted electrons. The maximum kinetic energy \(K.E.\) of the emitted electrons can be expressed as:
\[ K.E. = eV_s \]
where \(e\) is the charge of an electron. Thus, for the first case, we have:
\[ eV_{s1} = E_1 - \phi \]
where \(\phi\) is the work function of the material.

4. **Calculate the Energy for 3000 Å**:
For the second case, where the wavelength is \(3000 \, \text{Å}\):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{hc}{3000 \times 10^{-10}} \]

5. **Compare Energies**:
Since \(E\) is inversely proportional to \(\lambda\), we can see that:
\[ E_2 > E_1 \]
This means that the energy of the photons at \(3000 \, \text{Å}\) is greater than that at \(4000 \, \text{Å}\).

6. **Determine the New Stopping Potential**:
Since the energy of the incident photons has increased, the maximum kinetic energy of the emitted electrons will also increase. Therefore, the stopping potential will also increase.

Given that the stopping potential for \(4000 \, \text{Å}\) is \(2 \, \text{V}\), we can conclude that the stopping potential for \(3000 \, \text{Å}\) will be greater than \(2 \, \text{V}\).

### Conclusion:
The stopping potential for the incident light of wavelength \(3000 \, \text{Å}\) will be greater than \(2 \, \text{V}\).