A If the wavelength of the incident radiation changes from `lamda_(1)` to `lamda_(2)` then the maximum kinetic energy of the emitted photo electrons changes from K to K,, then the work function of the emitter surface is

Options

A. `(lamda_(1)K_(1)-lamda_(2)K_(2))/(lamda_(2)-lamda_(1))`
B. `(lamda_(1)K_(2)-lamda_(2)K_(1))/(lamda_(1)-lamda_(2))`
C. `(K_(2)-K_(1))/(lamda_(1)K_(1)-lamda_(2)K_(2))`
D. `(lamda_(2)-lamda_(1))/(lamda_(2)K_(1)-lamda_(1)K_(2))`

Answer

To solve the problem, we need to apply the principles of the photoelectric effect. The photoelectric effect can be described by the equation:

\[ E = \phi + K.E. \]

Where:
- \( E \) is the energy of the incident photon,
- \( \phi \) is the work function of the emitter surface,
- \( K.E. \) is the maximum kinetic energy of the emitted photoelectrons.

The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light.

### Step-by-Step Solution:

1. **Write the energy equations for both wavelengths:**
For the first wavelength \( \lambda_1 \):
\[ \frac{hc}{\lambda_1} = \phi + K_1 \]
For the second wavelength \( \lambda_2 \):
\[ \frac{hc}{\lambda_2} = \phi + K_2 \]

2. **Rearrange both equations to isolate the work function \( \phi \):**
From the first equation:
\[ \phi = \frac{hc}{\lambda_1} - K_1 \]
From the second equation:
\[ \phi = \frac{hc}{\lambda_2} - K_2 \]

3. **Set the two expressions for \( \phi \) equal to each other:**
\[ \frac{hc}{\lambda_1} - K_1 = \frac{hc}{\lambda_2} - K_2 \]

4. **Rearrange the equation to express \( \phi \):**
\[ K_2 - K_1 = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]

5. **Factor out \( hc \):**
\[ K_2 - K_1 = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

6. **Multiply both sides by \( \lambda_1 \lambda_2 \) to eliminate the fractions:**
\[ (K_2 - K_1) \lambda_1 \lambda_2 = hc (\lambda_1 - \lambda_2) \]

7. **Solve for the work function \( \phi \):**
Rearranging gives:
\[ \phi = \frac{hc}{\lambda_1} - K_1 \]
or
\[ \phi = \frac{hc}{\lambda_2} - K_2 \]

### Final Expression for Work Function:
From the rearranged equations, we can express the work function as:
\[ \phi = \frac{K_2 \lambda_2 - K_1 \lambda_1}{\lambda_2 - \lambda_1} \]