Light of frequency `f_1` and `f_2` fall on same metal and the max speed of photo electron is `v_1` and `v_2` resp. and mass is m. Find relation between `v_1` and `v_2`.

Options

A. `v_2^2 - v_1^2 = (h((f_2-f_1)))/2m`
B. `v_2^2 - v_1^2 = (h(f_2-f_1))/m`
C. `v_2^2 - v_1^2 =(2 h(f_2-f_1))/m`
D. `v_2 - v_1 = (2h(f_2-f_1))/m`

Answer

To find the relation between the maximum speeds \( v_1 \) and \( v_2 \) of photoelectrons emitted when light of frequencies \( f_1 \) and \( f_2 \) falls on the same metal, we can use the photoelectric effect principles.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of a photon is given by the equation:
\[ E = h f \]
where \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( f \) is the frequency of the light.

2. **Energy of Photons**:
For the two frequencies \( f_1 \) and \( f_2 \), the energies of the photons can be expressed as:
\[ E_1 = h f_1 \quad \text{and} \quad E_2 = h f_2 \]

3. **Work Function**:
When light falls on the metal, some energy is used to overcome the work function \( \phi \) of the metal. The remaining energy is converted into the kinetic energy of the emitted photoelectrons. The relationship can be expressed as:
\[ E_1 = \phi + K_{max,1} \quad \text{and} \quad E_2 = \phi + K_{max,2} \]
where \( K_{max,1} \) and \( K_{max,2} \) are the maximum kinetic energies of the photoelectrons emitted by frequencies \( f_1 \) and \( f_2 \), respectively.

4. **Kinetic Energy Relation**:
The maximum kinetic energy of a photoelectron can also be expressed in terms of its speed:
\[ K_{max} = \frac{1}{2} m v^2 \]
Therefore, we can write:
\[ K_{max,1} = \frac{1}{2} m v_1^2 \quad \text{and} \quad K_{max,2} = \frac{1}{2} m v_2^2 \]

5. **Substituting Kinetic Energy**:
Substituting the expressions for kinetic energy into the energy equations gives:
\[ h f_1 = \phi + \frac{1}{2} m v_1^2 \quad \text{(1)} \]
\[ h f_2 = \phi + \frac{1}{2} m v_2^2 \quad \text{(2)} \]

6. **Eliminating Work Function**:
By subtracting equation (1) from equation (2), we can eliminate the work function \( \phi \):
\[ h f_2 - h f_1 = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \]
This simplifies to:
\[ h (f_2 - f_1) = \frac{1}{2} m (v_2^2 - v_1^2) \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{h (f_2 - f_1)}{m} = \frac{1}{2} (v_2^2 - v_1^2) \]

8. **Finding the Relation**:
This equation shows the relationship between the frequencies and the speeds of the emitted photoelectrons. If we want to express \( v_2 \) in terms of \( v_1 \) and the frequencies, we can rearrange it further, but it is clear that the relationship involves the differences in frequencies and the squares of the speeds.

### Final Relation:
The derived relationship between the maximum speeds \( v_1 \) and \( v_2 \) is:
\[ \frac{v_2^2 - v_1^2}{f_2 - f_1} = \frac{2h}{m} \]