When a metal surface is illuminated by light wavelengths `400 nm` and `250 nm` , the maximum velocities of the photoelectrons ejected are `upsilon` and `2 v` respectively . The work function of the metal is
`( h = "Planck's constant" , c = "velocity of light in air" )`

Options

A. ` 2 hc xx 10^(6) J`
B. ` 1.5 hc xx 10^(6) J`
C. ` hc xx 10^(6) J`
D. ` 0.5 hc xx 10^(6) J`

Explanation

To find the work function of the metal when illuminated by light of wavelengths 400 nm and 250 nm, we can use the photoelectric effect equation derived from Einstein's photoelectric equation:

1. **Understanding the Problem**:
- We have two wavelengths: \( \lambda_1 = 400 \, \text{nm} \) and \( \lambda_2 = 250 \, \text{nm} \).
- The maximum velocities of the photoelectrons are \( v_1 = v \) for \( \lambda_1 \) and \( v_2 = 2v \) for \( \lambda_2 \).

2. **Using the Photoelectric Equation**:
The kinetic energy of the emitted photoelectrons can be expressed as:
\[ K_{\text{max}} = \frac{1}{2} mv^2 = E_{\text{incident}} - \phi \]
where \( E_{\text{incident}} = \frac{hc}{\lambda} \) and \( \phi \) is the work function.

3. **For Wavelength \( \lambda_1 = 400 \, \text{nm} \)**:
\[ K_{\text{max},1} = \frac{1}{2} mv^2 = \frac{hc}{400 \times 10^{-9}} - \phi \]
Let's denote this equation as (1).

4. **For Wavelength \( \lambda_2 = 250 \, \text{nm} \)**:
\[ K_{\text{max},2} = \frac{1}{2} m(2v)^2 = \frac{1}{2} m(4v^2) = 2mv^2 = \frac{hc}{250 \times 10^{-9}} - \phi \]
Let's denote this equation as (2).

5. **Equating the Kinetic Energies**:
From equation (1):
\[ \frac{1}{2} mv^2 = \frac{hc}{400 \times 10^{-9}} - \phi \]
From equation (2):
\[ 2mv^2 = \frac{hc}{250 \times 10^{-9}} - \phi \]

6. **Expressing \( mv^2 \) from Equation (1)**:
Multiply both sides of equation (1) by 2:
\[ mv^2 = \frac{hc}{200 \times 10^{-9}} - 2\phi \]

7. **Substituting into Equation (2)**:
Substitute \( mv^2 \) from above into equation (2):
\[ 2\left(\frac{hc}{200 \times 10^{-9}} - 2\phi\right) = \frac{hc}{250 \times 10^{-9}} - \phi \]

8. **Simplifying the Equation**:
\[ \frac{hc}{100 \times 10^{-9}} - 4\phi = \frac{hc}{250 \times 10^{-9}} - \phi \]
Rearranging gives:
\[ \frac{hc}{100 \times 10^{-9}} - \frac{hc}{250 \times 10^{-9}} = 3\phi \]

9. **Finding a Common Denominator**:
The common denominator for \( 100 \) and \( 250 \) is \( 500 \):
\[ \frac{5hc - 2hc}{500 \times 10^{-9}} = 3\phi \]
This simplifies to:
\[ \frac{3hc}{500 \times 10^{-9}} = 3\phi \]

10. **Solving for Work Function \( \phi \)**:
Dividing both sides by 3:
\[ \phi = \frac{hc}{500 \times 10^{-9}} \]

11. **Final Expression**:
\[ \phi = \frac{hc}{500 \times 10^{-9}} = \frac{hc}{5 \times 10^{-7}} \]