The work functions of metals `A` and `B` are in the ratio `1 : 2`. If light of frequencies `f` and `2 f` are incident on the surfaces of `A` and `B` respectively , the ratio of the maximum kinetic energy of photoelectrons emitted is ( `f` is greater than threshold frequency of `A , 2f` is greater than threshold frequency of `B`)

Options

A. `1 : 1`
B. `1 : 2`
C. `1 : 3`
D. `1 : 4`

Explanation

To solve the problem, we need to find the ratio of the maximum kinetic energy of photoelectrons emitted from metals A and B when light of frequencies \( f \) and \( 2f \) is incident on them, respectively.

### Step-by-Step Solution:

1. **Define Work Functions**:
Let the work function of metal A be \( \phi_A \) and that of metal B be \( \phi_B \). Given that the ratio of their work functions is \( 1:2 \), we can write:
\[ \phi_A = \phi \quad \text{and} \quad \phi_B = 2\phi \]

2. **Use the Photoelectric Equation**:
The maximum kinetic energy (\( K_{max} \)) of the emitted photoelectrons can be calculated using the photoelectric equation:
\[ K_{max} = E - \phi \]
where \( E \) is the energy of the incident photons.

3. **Calculate Kinetic Energy for Metal A**:
For metal A, the frequency of the incident light is \( f \):
\[ E_A = hf \]
Therefore, the maximum kinetic energy of photoelectrons emitted from metal A is:
\[ K_{A} = hf - \phi_A = hf - \phi \]

4. **Calculate Kinetic Energy for Metal B**:
For metal B, the frequency of the incident light is \( 2f \):
\[ E_B = h(2f) = 2hf \]
Thus, the maximum kinetic energy of photoelectrons emitted from metal B is:
\[ K_{B} = 2hf - \phi_B = 2hf - 2\phi \]

5. **Express the Kinetic Energies**:
Now we can express the kinetic energies:
\[ K_A = hf - \phi \]
\[ K_B = 2hf - 2\phi \]

6. **Find the Ratio of Kinetic Energies**:
To find the ratio \( \frac{K_A}{K_B} \):
\[ \frac{K_A}{K_B} = \frac{hf - \phi}{2hf - 2\phi} \]
Simplifying the right-hand side:
\[ \frac{K_A}{K_B} = \frac{hf - \phi}{2(hf - \phi)} = \frac{1}{2} \]

### Final Result:
The ratio of the maximum kinetic energy of photoelectrons emitted from metals A and B is:
\[ \frac{K_A}{K_B} = \frac{1}{2} \]