If light of wavelength `lambda_(1)` is allowed to fall on a metal , then kinetic energy of photoelectrons emitted is `E_(1)`. If wavelength of light changes to `lambda_(2)` then kinetic energy of electrons changes to `E_(2)`. Then work function of the metal is

Options

A. `(E_(1)E_(2)(lambda_(1)-lambda_(2)))/(lambda_(1)lambda_(2))`
B. `(E_(1)lambda_(1)-E_(2)lambda_(2))/(lambda_(1)-lambda_(2))`
C. `(E_(1)lambda_(1)-E_(2)lambda_(2))/(lambda_(2)-lambda_(1))`
D. `(lambda_(1)lambda_(2)E_(1)E_(2))/(lambda_(2)-lambda_(1))`

Answer

To find the work function of the metal based on the given wavelengths and kinetic energies of emitted photoelectrons, we will use Einstein's photoelectric equation. Here’s a step-by-step solution:

### Step 1: Write down Einstein's photoelectric equation
Einstein's photoelectric equation is given by:
\[ E = \frac{hc}{\lambda} = \phi + KE \]
where:
- \( E \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal,
- \( KE \) is the kinetic energy of the emitted photoelectron,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light.

### Step 2: Apply the equation for the first wavelength \( \lambda_1 \)
For the first case where the wavelength is \( \lambda_1 \) and the kinetic energy is \( E_1 \):
\[ \frac{hc}{\lambda_1} = \phi + E_1 \]
This can be rearranged to:
\[ \phi = \frac{hc}{\lambda_1} - E_1 \quad \text{(Equation 1)} \]

### Step 3: Apply the equation for the second wavelength \( \lambda_2 \)
For the second case where the wavelength is \( \lambda_2 \) and the kinetic energy is \( E_2 \):
\[ \frac{hc}{\lambda_2} = \phi + E_2 \]
This can be rearranged to:
\[ \phi = \frac{hc}{\lambda_2} - E_2 \quad \text{(Equation 2)} \]

### Step 4: Set the two expressions for work function equal to each other
Since both equations equal \( \phi \), we can set them equal to each other:
\[ \frac{hc}{\lambda_1} - E_1 = \frac{hc}{\lambda_2} - E_2 \]

### Step 5: Rearrange the equation to isolate \( hc \)
Rearranging gives:
\[ \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = E_1 - E_2 \]
Factoring out \( hc \):
\[ hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = E_1 - E_2 \]

### Step 6: Solve for \( hc \)
Now, we can solve for \( hc \):
\[ hc = \frac{(E_1 - E_2)}{\left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)} \]

### Step 7: Substitute \( hc \) back into one of the equations for \( \phi \)
We can substitute \( hc \) back into either Equation 1 or Equation 2. Let’s use Equation 1:
\[ \phi = \frac{(E_1 - E_2)}{\left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \cdot \frac{1}{\lambda_1}} - E_1 \]

### Step 8: Final expression for work function \( \phi \)
After simplifying, we can express the work function \( \phi \) as:
\[ \phi = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_1 - \lambda_2} \]

### Conclusion
Thus, the work function of the metal is given by:
\[ \phi = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_1 - \lambda_2} \]