Stopping potential of emitted photo electron is V when monochromatic light of wavelength 'lamda' incident on a metal surface. If wavelength of light incident becomes 'lambda/3' stopping potential of photoelectrons becomes V/4 then the threshold wavelength of metal is k'lambda' then k will be

Options

A. 3
B. 6
C. 9
D. 2

Answer

To solve the problem, we need to analyze the relationship between the stopping potential, the wavelength of the incident light, and the threshold wavelength of the metal.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The stopping potential (V) is related to the energy of the emitted photoelectrons. The energy of the incident photons can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the incident light.

2. **Energy and Stopping Potential Relationship**: The energy of the emitted photoelectrons can be related to the stopping potential by the equation:
\[ eV = E - \phi \]
where \(e\) is the charge of the electron, \(V\) is the stopping potential, and \(\phi\) is the work function of the metal.

3. **First Condition**: When the wavelength of the incident light is \(\lambda\), the stopping potential is \(V\). Therefore, we can write:
\[ eV = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]

4. **Second Condition**: When the wavelength of the incident light is \(\frac{\lambda}{3}\), the stopping potential becomes \(\frac{V}{4}\). We can write:
\[ e\left(\frac{V}{4}\right) = \frac{hc}{\frac{\lambda}{3}} - \phi \quad \text{(2)} \]

5. **Substituting and Simplifying**: From equation (2), we can express it as:
\[ \frac{eV}{4} = \frac{3hc}{\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{3hc}{\lambda} - \frac{eV}{4} \quad \text{(3)} \]

6. **Equating the Two Expressions for Work Function**: From equations (1) and (3), we can set the expressions for \(\phi\) equal to each other:
\[ \frac{hc}{\lambda} - eV = \frac{3hc}{\lambda} - \frac{eV}{4} \]

7. **Clearing the Equation**: To eliminate terms, multiply the entire equation by 4:
\[ 4\left(\frac{hc}{\lambda} - eV\right) = 4\left(\frac{3hc}{\lambda} - \frac{eV}{4}\right) \]
Simplifying gives:
\[ \frac{4hc}{\lambda} - 4eV = 3\frac{4hc}{\lambda} - eV \]
Rearranging leads to:
\[ 4eV - eV = 3\frac{4hc}{\lambda} - \frac{4hc}{\lambda} \]
This simplifies to:
\[ 3eV = 3\frac{hc}{\lambda} \]
Thus:
\[ eV = \frac{hc}{\lambda} \]

8. **Finding the Threshold Wavelength**: The work function can also be expressed in terms of the threshold wavelength (\(\lambda_0\)):
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for \(\phi\) equal gives:
\[ \frac{hc}{\lambda_0} = \frac{hc}{9\lambda} \]
From this, we can deduce:
\[ \lambda_0 = 9\lambda \]

9. **Final Result**: The threshold wavelength is \(k\lambda\) where \(k = 9\).

### Conclusion:
The value of \(k\) is \(9\).