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 will use the concepts from the photoelectric effect and Einstein's photoelectric equation.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given two scenarios involving the stopping potential of photoelectrons emitted from a metal surface when illuminated by monochromatic light of different wavelengths. We need to find the threshold wavelength in terms of the initial wavelength.

2. **Using Einstein's Photoelectric Equation**:
The equation is given by:
\[ \frac{hc}{\lambda} = \phi + eV \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the metal,
- \( e \) is the charge of the electron,
- \( V \) is the stopping potential.

3. **First Scenario** (Wavelength = \( \lambda \)):
For the first case, we have:
\[ \frac{hc}{\lambda} = \phi + eV \]
(1)

4. **Second Scenario** (Wavelength = \( \frac{\lambda}{3} \)):
For the second case, the stopping potential becomes \( \frac{V}{4} \):
\[ \frac{hc}{\frac{\lambda}{3}} = \phi + e\left(\frac{V}{4}\right) \]
Simplifying this gives:
\[ \frac{3hc}{\lambda} = \phi + \frac{eV}{4} \]
(2)

5. **Setting Up the Equations**:
Now we have two equations:
- From (1): \( \frac{hc}{\lambda} = \phi + eV \)
- From (2): \( \frac{3hc}{\lambda} = \phi + \frac{eV}{4} \)

6. **Eliminating \( \phi \)**:
Rearranging both equations to express \( \phi \):
- From (1): \( \phi = \frac{hc}{\lambda} - eV \)
- From (2): \( \phi = \frac{3hc}{\lambda} - \frac{eV}{4} \)

Setting these equal to each other:
\[ \frac{hc}{\lambda} - eV = \frac{3hc}{\lambda} - \frac{eV}{4} \]

7. **Solving for \( eV \)**:
Rearranging gives:
\[ -eV + \frac{eV}{4} = \frac{3hc}{\lambda} - \frac{hc}{\lambda} \]
Simplifying:
\[ -\frac{3eV}{4} = \frac{2hc}{\lambda} \]
Thus:
\[ eV = -\frac{8hc}{3\lambda} \]

8. **Substituting Back**:
Substitute \( eV \) back into either equation to find \( \phi \):
Using (1):
\[ \phi = \frac{hc}{\lambda} + \frac{8hc}{3\lambda} \]
Simplifying gives:
\[ \phi = \frac{hc}{\lambda} + \frac{8hc}{3\lambda} = \frac{11hc}{3\lambda} \]

9. **Finding Threshold Wavelength**:
The threshold wavelength \( \lambda_{th} \) is given by:
\[ \phi = \frac{hc}{\lambda_{th}} \]
Setting equal:
\[ \frac{hc}{\lambda_{th}} = \frac{11hc}{3\lambda} \]
This simplifies to:
\[ \lambda_{th} = \frac{3\lambda}{11} \]

10. **Finding \( k \)**:
If the threshold wavelength is \( k \lambda \), we have:
\[ k \lambda = \frac{3\lambda}{11} \implies k = \frac{3}{11} \]

### Final Answer:
Thus, the value of \( k \) is \( \frac{3}{11} \).