The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from `lamda_1 " to " lamda_2` The work function of the metal is

Options

A. `(hc)/(lamda_1lamda_2)(2lamda_2-lamda_1)`
B. `(hc)/(lamda_1lamda_2)(2lamda_1-lamda_2)`
C. `(hc)/(lamda_1lamda_2)(lamda_1+lamda_2)`
D. `(hc)/(lamda_1lamda_2)(lamda_1-lamda_2)`

Answer

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

### Step-by-Step Solution:

1. **Understand the Problem**:
We are given that the kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation changes from \( \lambda_1 \) to \( \lambda_2 \).

2. **Use Einstein's Photoelectric Equation**:
The kinetic energy \( KE \) of the emitted photoelectrons can be expressed as:
\[ KE = \frac{hc}{\lambda} - \phi_0 \]
where \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength of the incident light, and \( \phi_0 \) is the work function of the metal.

3. **Set Up the Equations**:
- For wavelength \( \lambda_1 \):
\[ KE_1 = \frac{hc}{\lambda_1} - \phi_0 \]
- For wavelength \( \lambda_2 \) (where the kinetic energy is doubled):
\[ KE_2 = \frac{hc}{\lambda_2} - \phi_0 \]
Given that \( KE_2 = 2 KE_1 \), we can write:
\[ \frac{hc}{\lambda_2} - \phi_0 = 2\left(\frac{hc}{\lambda_1} - \phi_0\right) \]

4. **Expand and Rearrange the Equation**:
Expanding the equation gives:
\[ \frac{hc}{\lambda_2} - \phi_0 = \frac{2hc}{\lambda_1} - 2\phi_0 \]
Rearranging this leads to:
\[ \frac{hc}{\lambda_2} + \phi_0 = \frac{2hc}{\lambda_1} \]

5. **Isolate the Work Function \( \phi_0 \)**:
Rearranging further gives:
\[ \phi_0 = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
Factoring out \( hc \):
\[ \phi_0 = hc\left(\frac{2}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

6. **Combine the Fractions**:
To combine the fractions, find a common denominator:
\[ \phi_0 = hc\left(\frac{2\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right) \]

7. **Final Expression for Work Function**:
Thus, the work function \( \phi_0 \) can be expressed as:
\[ \phi_0 = \frac{hc(2\lambda_2 - \lambda_1)}{\lambda_1 \lambda_2} \]

### Final Result:
The work function of the metal is:
\[ \phi_0 = \frac{hc(2\lambda_2 - \lambda_1)}{\lambda_1 \lambda_2} \]