Two identical photo-cathodes receive light of frequencies `v_1` and `v_2`. If the velocities of the photoelectrons (of mass m) coming out are `v_1` and `v_2` respectively, then

Options

A. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`
B. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`
C. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`
D. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`

Explanation

To solve the problem, we need to analyze the relationship between the frequency of the light incident on the photo-cathodes and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect states that when light of a certain frequency strikes a photo-cathode, it can eject electrons. The kinetic energy (KE) of the emitted electrons is given by the equation:
\[ KE = h \nu - \phi \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the material.

2. **Setting Up the Equations**:
For the first photo-cathode receiving light of frequency \( \nu_1 \) and emitting electrons with velocity \( v_1 \):
\[ KE_1 = \frac{1}{2} m v_1^2 = h \nu_1 - \phi \quad \text{(Equation 1)} \]
For the second photo-cathode receiving light of frequency \( \nu_2 \) and emitting electrons with velocity \( v_2 \):
\[ KE_2 = \frac{1}{2} m v_2^2 = h \nu_2 - \phi \quad \text{(Equation 2)} \]

3. **Subtracting the Two Equations**:
Now, we will subtract Equation 2 from Equation 1:
\[ \frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2 = (h \nu_1 - \phi) - (h \nu_2 - \phi) \]
This simplifies to:
\[ \frac{1}{2} m (v_1^2 - v_2^2) = h (\nu_1 - \nu_2) \]

4. **Rearranging the Equation**:
We can rearrange the equation to express the difference in velocities:
\[ v_1^2 - v_2^2 = \frac{2h}{m} (\nu_1 - \nu_2) \]

5. **Conclusion**:
The relationship derived shows that the difference in the squares of the velocities of the photoelectrons is directly proportional to the difference in the frequencies of the incident light.

### Final Answer:
The correct relationship is:
\[ v_1^2 - v_2^2 = \frac{2h}{m} (\nu_1 - \nu_2) \]