Lights of wavelenths `lambda_(1)=340nm` and `lambda_(2)=540nm` are incident on a metallic surface. If the ratio of the maximum speeds of electrons ejected is 2, the work function of the metal is

Options

A. 2 eV
B. 1.8 eV
C. 1 eV
D. 1.5 eV

Answer

To solve the problem, we need to find the work function of the metal given the wavelengths of incident light and the ratio of the maximum speeds of the ejected electrons.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelengths of light:
- \( \lambda_1 = 340 \, \text{nm} \)
- \( \lambda_2 = 540 \, \text{nm} \)
- Ratio of maximum speeds of ejected electrons:
- \( \frac{v_1}{v_2} = 2 \)

2. **Use the Photoelectric Equation**:
The energy of the photons can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

3. **Write the Energy Equations for Each Wavelength**:
For the first wavelength \( \lambda_1 \):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{hc}{340 \, \text{nm}} \]
For the second wavelength \( \lambda_2 \):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{hc}{540 \, \text{nm}} \]

4. **Apply the Photoelectric Effect Equation**:
The energy of the incident photon is used to overcome the work function \( \phi \) and provide kinetic energy to the ejected electron:
\[ E_1 = \phi + KE_1 \quad \text{and} \quad E_2 = \phi + KE_2 \]
where \( KE = \frac{1}{2}mv^2 \).

5. **Express Kinetic Energy in Terms of Speed**:
From the ratio of speeds:
\[ KE_1 = \frac{1}{2}mv_1^2 \quad \text{and} \quad KE_2 = \frac{1}{2}mv_2^2 \]
Given \( \frac{v_1}{v_2} = 2 \), we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = 2v_2 \implies KE_1 = \frac{1}{2}m(2v_2)^2 = 2mv_2^2 = 4KE_2 \]

6. **Set Up the Equations**:
Substituting \( KE_1 \) and \( KE_2 \) into the energy equations:
\[ \frac{hc}{340} = \phi + 4KE_2 \quad \text{(1)} \]
\[ \frac{hc}{540} = \phi + KE_2 \quad \text{(2)} \]

7. **Eliminate Work Function \( \phi \)**:
From equation (2):
\[ \phi = \frac{hc}{540} - KE_2 \]
Substitute \( \phi \) into equation (1):
\[ \frac{hc}{340} = \left(\frac{hc}{540} - KE_2\right) + 4KE_2 \]
Simplifying gives:
\[ \frac{hc}{340} = \frac{hc}{540} + 3KE_2 \]

8. **Solve for \( KE_2 \)**:
Rearranging:
\[ 3KE_2 = \frac{hc}{340} - \frac{hc}{540} \]
Finding a common denominator (which is \( 340 \times 540 \)):
\[ 3KE_2 = \frac{hc(540 - 340)}{340 \times 540} \]
\[ 3KE_2 = \frac{hc(200)}{340 \times 540} \]
\[ KE_2 = \frac{hc(200)}{3 \times 340 \times 540} \]

9. **Substitute Back to Find Work Function**:
Substitute \( KE_2 \) back into the equation for \( \phi \):
\[ \phi = \frac{hc}{540} - \frac{hc(200)}{3 \times 340 \times 540} \]
Simplifying gives the work function in terms of \( h \) and \( c \).

10. **Calculate Numerical Value**:
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \), convert nm to meters, and calculate \( \phi \).

### Final Answer:
After performing the calculations, we find that the work function \( \phi \) is approximately \( 1.8 \, \text{eV} \).