When the incident wavelengths are `lamda` and `lamda//2` the kinetic energies of the emitted photo electrons are E and 2E. The work function of the metal is

Options

A. E/4
B. E/2
C. E/3
D. Zero

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the kinetic energy of emitted photoelectrons to the incident wavelength of light and the work function of the metal.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two incident wavelengths: \( \lambda \) and \( \frac{\lambda}{2} \). The corresponding kinetic energies of the emitted photoelectrons are \( E \) and \( 2E \) respectively. We need to find the work function \( \phi \) of the metal.

2. **Write Einstein's Photoelectric Equation**:
The equation is given by:
\[ KE = \frac{hc}{\lambda} - \phi \]
where \( KE \) is the kinetic energy of the emitted electron, \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength of the incident light, and \( \phi \) is the work function.

3. **Set Up the Equations for Each Wavelength**:
- For the wavelength \( \lambda \):
\[ E = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]
- For the wavelength \( \frac{\lambda}{2} \):
\[ 2E = \frac{hc}{\frac{\lambda}{2}} - \phi \quad \text{(2)} \]
This simplifies to:
\[ 2E = \frac{2hc}{\lambda} - \phi \]

4. **Rearranging the Equations**:
From equation (1):
\[ \phi = \frac{hc}{\lambda} - E \quad \text{(3)} \]
From equation (2):
\[ \phi = \frac{2hc}{\lambda} - 2E \quad \text{(4)} \]

5. **Set Equations (3) and (4) Equal to Each Other**:
\[ \frac{hc}{\lambda} - E = \frac{2hc}{\lambda} - 2E \]

6. **Solve for E**:
Rearranging gives:
\[ E = \frac{hc}{\lambda} - \frac{2hc}{\lambda} + 2E \]
Simplifying:
\[ E = -\frac{hc}{\lambda} + 2E \]
\[ E - 2E = -\frac{hc}{\lambda} \]
\[ -E = -\frac{hc}{\lambda} \]
\[ E = \frac{hc}{\lambda} \]

7. **Substituting Back to Find Work Function**:
Substitute \( E \) back into equation (3):
\[ \phi = \frac{hc}{\lambda} - \frac{hc}{\lambda} \]
\[ \phi = 0 \]

### Conclusion:
The work function \( \phi \) of the metal is \( 0 \).