A photonsensitive metallic surface is illuminated alternately with light of wavelength 3100 Å and 6200 Å . It is observed that maximum speeds of the photoelectrons in two cases are in ratio ` 2 : 1 ` . The work function of the metal is (hc = 12400 e VÅ)

Options

A. `1 eV `
B. 2 e V
C. `(4)/(3) eV `
D. ` (2)/(3) eV `

Answer

To solve the problem, we will use the concepts of the photoelectric effect and the relationship between the energy of photons, work function, and kinetic energy of emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Energy of Photons**:
The energy of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

2. **Given Values**:
- For wavelength \( \lambda_1 = 3100 \, \text{Å} \):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{12400 \, \text{eV} \cdot \text{Å}}{3100 \, \text{Å}} = 4 \, \text{eV} \]
- For wavelength \( \lambda_2 = 6200 \, \text{Å} \):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{12400 \, \text{eV} \cdot \text{Å}}{6200 \, \text{Å}} = 2 \, \text{eV} \]

3. **Applying the Photoelectric Effect**:
According to the photoelectric effect, the energy of the incident photon minus the work function (\( \phi \)) gives the kinetic energy (\( KE \)) of the emitted photoelectrons:
\[ KE = E - \phi \]
- For the first case:
\[ KE_1 = E_1 - \phi = 4 \, \text{eV} - \phi \]
- For the second case:
\[ KE_2 = E_2 - \phi = 2 \, \text{eV} - \phi \]

4. **Relating Kinetic Energies to Speeds**:
The kinetic energy can also be expressed in terms of the speed of the photoelectrons:
\[ KE = \frac{1}{2} mv^2 \]
- For the first case:
\[ KE_1 = \frac{1}{2} m v_1^2 \]
- For the second case:
\[ KE_2 = \frac{1}{2} m v_2^2 \]

5. **Using the Given Speed Ratio**:
We are given that the maximum speeds of the photoelectrons are in the ratio \( \frac{v_1}{v_2} = 2 \). Therefore:
\[ v_1 = 2v_2 \]
Substituting this into the kinetic energy equation for the first case:
\[ KE_1 = \frac{1}{2} m (2v_2)^2 = 2mv_2^2 \]

6. **Setting Up the Equations**:
From the kinetic energy equations:
- For the first case:
\[ 4 - \phi = 2mv_2^2 \quad \text{(Equation 1)} \]
- For the second case:
\[ 2 - \phi = \frac{1}{2} mv_2^2 \quad \text{(Equation 2)} \]

7. **Dividing the Equations**:
Dividing Equation 1 by Equation 2:
\[ \frac{4 - \phi}{2 - \phi} = \frac{2mv_2^2}{\frac{1}{2} mv_2^2} = 4 \]

8. **Solving for Work Function**:
Cross-multiplying gives:
\[ 4(2 - \phi) = 4 - \phi \]
Expanding and simplifying:
\[ 8 - 4\phi = 4 - \phi \implies 3\phi = 4 \implies \phi = \frac{4}{3} \, \text{eV} \]

### Final Answer:
The work function of the metal is:
\[ \phi = \frac{4}{3} \, \text{eV} \]