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. `2hcxx10^(6)` J
B. `1.5hcxx10^(6)J`
C. `hcxx10^(6)` J
D. `0.5hcxx10^(6)`J

Answer

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two different wavelengths of light illuminating a metal surface, which results in the ejection of photoelectrons with different maximum velocities. The first wavelength is \( \lambda_1 = 400 \, \text{nm} \) and the second wavelength is \( \lambda_2 = 250 \, \text{nm} \). The corresponding maximum velocities of the photoelectrons are \( v_1 = v \) and \( v_2 = 2v \).

2. **Using Einstein's Photoelectric Equation**:
The work function \( W_0 \) can be expressed as:
\[ W_0 = E - KE \]
where \( E \) is the energy of the incident light and \( KE \) is the kinetic energy of the emitted electrons.

3. **Calculating Energy for Each Wavelength**:
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
For \( \lambda_1 = 400 \, \text{nm} \):
\[ E_1 = \frac{hc}{400 \times 10^{-9}} \]
For \( \lambda_2 = 250 \, \text{nm} \):
\[ E_2 = \frac{hc}{250 \times 10^{-9}} \]

4. **Calculating Kinetic Energy**:
The kinetic energy of the photoelectrons can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
For the first case:
\[ KE_1 = \frac{1}{2} mv^2 \]
For the second case:
\[ KE_2 = \frac{1}{2} m(2v)^2 = 2mv^2 \]

5. **Setting Up the Equations**:
From the first wavelength:
\[ W_0 = \frac{hc}{400 \times 10^{-9}} - \frac{1}{2} mv^2 \]
From the second wavelength:
\[ W_0 = \frac{hc}{250 \times 10^{-9}} - 2mv^2 \]

6. **Equating the Work Functions**:
Since the work function is the same for both cases, we can set the equations equal to each other:
\[ \frac{hc}{400 \times 10^{-9}} - \frac{1}{2} mv^2 = \frac{hc}{250 \times 10^{-9}} - 2mv^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{hc}{400 \times 10^{-9}} - \frac{hc}{250 \times 10^{-9}} = -\frac{1}{2} mv^2 + 2mv^2 \]
Simplifying the left side:
\[ \frac{hc}{400 \times 10^{-9}} - \frac{hc}{250 \times 10^{-9}} = \frac{hc \cdot (250 - 400)}{100000 \times 10^{-18}} = \frac{-150hc}{100000 \times 10^{-18}} = -\frac{3hc}{2000 \times 10^{-9}} \]
On the right side:
\[ \frac{3}{2} mv^2 \]

8. **Final Equation**:
Thus, we have:
\[ \frac{3}{2} mv^2 = \frac{3hc}{2000 \times 10^{-9}} \]
This simplifies to:
\[ mv^2 = \frac{hc}{2000 \times 10^{-9}} \]

9. **Substituting Back to Find Work Function**:
Now, substituting \( mv^2 \) back into one of the work function equations, we can find \( W_0 \):
\[ W_0 = \frac{hc}{400 \times 10^{-9}} - \frac{1}{2} \left(\frac{hc}{2000 \times 10^{-9}}\right) \]

10. **Calculating the Final Work Function**:
After simplification, we find:
\[ W_0 = \frac{hc}{400 \times 10^{-9}} - \frac{hc}{4000 \times 10^{-9}} = \frac{hc \cdot (10 - 1)}{4000 \times 10^{-9}} = \frac{9hc}{4000 \times 10^{-9}} = \frac{hc}{444.44 \times 10^{-9}} \]

### Final Result:
The work function \( W_0 \) can be expressed in terms of \( hc \) and the numerical values.