A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metals is

Options

A. 1.03 eV
B. 2.11 eV
C. 4.14 eV
D. 2.43 eV

Explanation

To solve the problem, we will use the photoelectric equation and the information provided about the wavelengths and kinetic energies of the emitted photoelectrons.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Wavelength of first light, \( \lambda_1 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Wavelength of second light, \( \lambda_2 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
- Kinetic energy of emitted photoelectrons when \( \lambda_1 \) is incident: \( K_1 = K_{\text{max}} \)
- Kinetic energy of emitted photoelectrons when \( \lambda_2 \) is incident: \( K_2 = 2K_{\text{max}} \)

2. **Write the Photoelectric Equation:**
The photoelectric equation is given by:
\[ K_{\text{max}} = \frac{hc}{\lambda} - \phi_0 \]
where \( \phi_0 \) is the work function of the metal, \( h \) is Planck's constant, and \( c \) is the speed of light.

3. **Apply the Equation for Both Wavelengths:**
- For \( \lambda_1 \):
\[ K_{\text{max}} = \frac{hc}{\lambda_1} - \phi_0 \quad \text{(1)} \]
- For \( \lambda_2 \):
\[ 2K_{\text{max}} = \frac{hc}{\lambda_2} - \phi_0 \quad \text{(2)} \]

4. **Substitute the Kinetic Energies:**
From equation (1):
\[ K_{\text{max}} = \frac{hc}{600 \times 10^{-9}} - \phi_0 \]
From equation (2):
\[ 2K_{\text{max}} = \frac{hc}{400 \times 10^{-9}} - \phi_0 \]

5. **Divide Equation (2) by Equation (1):**
\[ \frac{2K_{\text{max}}}{K_{\text{max}}} = \frac{\frac{hc}{400 \times 10^{-9}} - \phi_0}{\frac{hc}{600 \times 10^{-9}} - \phi_0} \]
Simplifying gives:
\[ 2 = \frac{\frac{hc}{400 \times 10^{-9}} - \phi_0}{\frac{hc}{600 \times 10^{-9}} - \phi_0} \]

6. **Cross Multiply:**
\[ 2\left(\frac{hc}{600 \times 10^{-9}} - \phi_0\right) = \frac{hc}{400 \times 10^{-9}} - \phi_0 \]

7. **Rearranging the Equation:**
\[ 2\frac{hc}{600 \times 10^{-9}} - 2\phi_0 = \frac{hc}{400 \times 10^{-9}} - \phi_0 \]
Bringing terms involving \( \phi_0 \) to one side:
\[ 2\frac{hc}{600 \times 10^{-9}} - \frac{hc}{400 \times 10^{-9}} = \phi_0 \]

8. **Calculate \( \phi_0 \):**
Substitute \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \phi_0 = 2\left(\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\right) - \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \]

9. **Perform the Calculations:**
- Calculate \( \frac{hc}{600 \times 10^{-9}} \) and \( \frac{hc}{400 \times 10^{-9}} \).
- Substitute these values back into the equation to find \( \phi_0 \).

10. **Convert to Electron Volts:**
Finally, convert \( \phi_0 \) from Joules to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)).

### Final Answer:
The work function \( \phi_0 \) is approximately \( 1.03 \, \text{eV} \).

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