If the wavelength of incident light decrease from ` lamda _ 1 ` to ` lamda _ 2 ` in photoelctric cell then corresponding chages in stopping potential will be

Options

A. an increase of ` (hc//e ) ( 1//lamda _ 2 - 1 //lamda _ 1 ) `
B. a decrease of ` (hc//e)(1//lamda _ 2 - 1 //lamda _ 1)`
C. an increase of `(hc) (1//lamda _ 2 - 1//lamda _ 1 )`
D. a decrease of (hc) ` (1//lamda _2 - 1 //lamda _ 1 ) `

Answer

To solve the problem, we will use Einstein's photoelectric equation and analyze the relationship between the wavelength of incident light and the stopping potential in a photoelectric cell.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect describes how light can eject electrons from a material. The energy of the incident photons is given by the equation:
\[ E = h\nu \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(\nu\) is the frequency of the light.

2. **Relate Frequency and Wavelength**:
The frequency \(\nu\) can be expressed in terms of the wavelength \(\lambda\) using the speed of light \(c\):
\[ \nu = \frac{c}{\lambda} \]
Therefore, the energy of the photon can also be written as:
\[ E = \frac{hc}{\lambda} \]

3. **Apply Einstein's Equation**:
According to Einstein's photoelectric equation, the stopping potential \(V\) is related to the energy of the photon and the work function \(\phi_0\) of the material:
\[ eV = E - \phi_0 \]
Substituting the expression for energy, we have:
\[ eV = \frac{hc}{\lambda} - \phi_0 \]

4. **Calculate Stopping Potential for Two Wavelengths**:
For two different wavelengths, \(\lambda_1\) and \(\lambda_2\), we can write the stopping potentials \(V_1\) and \(V_2\) as:
\[ eV_1 = \frac{hc}{\lambda_1} - \phi_0 \]
\[ eV_2 = \frac{hc}{\lambda_2} - \phi_0 \]

5. **Find the Change in Stopping Potential**:
To find the change in stopping potential when the wavelength decreases from \(\lambda_1\) to \(\lambda_2\), we subtract the two equations:
\[ e(V_2 - V_1) = \left(\frac{hc}{\lambda_2} - \phi_0\right) - \left(\frac{hc}{\lambda_1} - \phi_0\right) \]
Simplifying this gives:
\[ e(V_2 - V_1) = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]
\[ e(V_2 - V_1) = hc\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]

6. **Express Change in Stopping Potential**:
Thus, we can express the change in stopping potential as:
\[ V_2 - V_1 = \frac{hc}{e}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]

7. **Conclusion**:
Since \(\lambda_2 < \lambda_1\), it follows that \(\frac{1}{\lambda_2} > \frac{1}{\lambda_1}\), which means \(V_2 > V_1\). Therefore, the stopping potential increases as the wavelength decreases.

### Final Answer:
The change in stopping potential is given by:
\[ \Delta V = V_2 - V_1 = \frac{hc}{e}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \]