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 need to find the work function (φ) of the metal using the given data about the wavelengths of light and the ratio of the maximum speeds of the photoelectrons.

### Step-by-step Solution:

1. **Understand the relationship between kinetic energy and wavelength**:
The kinetic energy (KE) of photoelectrons can be expressed using the equation:
\[ KE = \frac{hc}{\lambda} - \phi \]
where:
- \( KE \) is the kinetic energy of the photoelectrons,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the metal.

2. **Set up the equations for the two wavelengths**:
Let:
- For \( \lambda_A = 3100 \, \text{Å} \) (first case),
- For \( \lambda_B = 6200 \, \text{Å} \) (second case).

The kinetic energies for the two cases can be written as:
\[ KE_A = \frac{hc}{\lambda_A} - \phi \]
\[ KE_B = \frac{hc}{\lambda_B} - \phi \]

3. **Use the ratio of maximum speeds**:
Given that the maximum speeds of the photoelectrons are in the ratio \( 2:1 \), we know that:
\[ \frac{KE_A}{KE_B} = \left(\frac{v_A}{v_B}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \]
Therefore, we have:
\[ \frac{KE_A}{KE_B} = 4 \]

4. **Substituting the kinetic energy equations**:
From the ratio of kinetic energies, we can write:
\[ \frac{\frac{hc}{\lambda_A} - \phi}{\frac{hc}{\lambda_B} - \phi} = 4 \]

5. **Substituting the values of \( \lambda_A \) and \( \lambda_B \)**:
Plugging in the values:
\[ \frac{\frac{12400}{3100} - \phi}{\frac{12400}{6200} - \phi} = 4 \]

6. **Calculating the fractions**:
Simplifying the fractions:
\[ \frac{4 - \phi}{2 - \phi} = 4 \]

7. **Cross-multiplying**:
Cross-multiplying gives:
\[ 4(2 - \phi) = 4 - \phi \]
Expanding this:
\[ 8 - 4\phi = 4 - \phi \]

8. **Rearranging the equation**:
Rearranging the terms:
\[ 8 - 4 = 4\phi - \phi \]
\[ 4 = 3\phi \]

9. **Solving for φ**:
Dividing both sides by 3:
\[ \phi = \frac{4}{3} \, \text{eV} \approx 1.33 \, \text{eV} \]

10. **Final answer**:
The work function of the metal is approximately \( \phi = 4.3 \, \text{eV} \).