The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength `lamda` of the incident radiation is changed from 400 nm to 310 nm. The work function of the metal is

Options

A. 0.9 eV
B. 1.7 eV
C. 2.2 eV
D. 3.1 eV

Explanation

To find the work function of the metal based on the given problem, we will use Einstein's photoelectric equation, which relates the kinetic energy of emitted electrons to the energy of the incident radiation 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 electrons (E1) is doubled when the wavelength (λ) of the incident radiation changes from 400 nm to 310 nm.

2. **Use Einstein's Photoelectric Equation**:
The equation states:
\[ K.E. = E - \phi \]
where \( K.E. \) is the kinetic energy of the emitted electrons, \( E \) is the energy of the incident photons, and \( \phi \) is the work function of the metal.

3. **Set Up the Equations**:
- For the first wavelength (λ1 = 400 nm):
\[ E_1 = \frac{hc}{\lambda_1} - \phi \]
- For the second wavelength (λ2 = 310 nm), where the kinetic energy is doubled:
\[ 2E_1 = \frac{hc}{\lambda_2} - \phi \]

4. **Substitute E1 into the Second Equation**:
From the first equation, we can express \( E_1 \):
\[ E_1 = \frac{hc}{\lambda_1} - \phi \]
Substitute this into the second equation:
\[ 2\left(\frac{hc}{\lambda_1} - \phi\right) = \frac{hc}{\lambda_2} - \phi \]

5. **Simplify the Equation**:
Expanding the left side:
\[ \frac{2hc}{\lambda_1} - 2\phi = \frac{hc}{\lambda_2} - \phi \]
Rearranging gives:
\[ \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2} = \phi - 2\phi \]
\[ \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2} = -\phi \]
Thus:
\[ \phi = \frac{hc}{\lambda_2} - \frac{2hc}{\lambda_1} \]

6. **Factor Out hc**:
\[ \phi = hc \left(\frac{1}{\lambda_2} - \frac{2}{\lambda_1}\right) \]

7. **Substitute Values**:
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
- Convert wavelengths from nm to meters:
- \( \lambda_1 = 400 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 310 \times 10^{-9} \, \text{m} \)

Now substitute:
\[ \phi = 6.626 \times 10^{-34} \times 3 \times 10^8 \left(\frac{1}{310 \times 10^{-9}} - \frac{2}{400 \times 10^{-9}}\right) \]

8. **Calculate the Values**:
- Calculate \( \frac{1}{310 \times 10^{-9}} \) and \( \frac{2}{400 \times 10^{-9}} \):
\[ \frac{1}{310 \times 10^{-9}} \approx 3.2258 \times 10^6 \, \text{m}^{-1} \]
\[ \frac{2}{400 \times 10^{-9}} = 5 \times 10^6 \, \text{m}^{-1} \]
- Therefore:
\[ \phi = 6.626 \times 10^{-34} \times 3 \times 10^8 \left(3.2258 \times 10^6 - 5 \times 10^6\right) \]
\[ \phi = 6.626 \times 10^{-34} \times 3 \times 10^8 \times (-1.7742 \times 10^6) \]

9. **Final Calculation**:
- Calculate \( \phi \) and convert to electron volts by dividing by \( 1.6 \times 10^{-19} \):
\[ \phi \approx 3.53 \times 10^{-19} \, \text{J} \]
\[ \phi \text{ (in eV)} \approx \frac{3.53 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.2 \, \text{eV} \]

### Conclusion:
The work function of the metal is approximately **2.2 eV**.