Two identical photocathode receive light of frequencies `f_(1)` and `f_(2)`. If the maximum velocities of the photoelectrons (of mass m) coming out are respectively `v_(1)` and `v_(2)` then:

Options

A. `v_(1)-v_(2)=[(2h)/(m)(f_(1)-f_(2))]^(1//2`
B. ` v._(1)^(2)-v._(2)^(2)=(2h)/(m)(f_(1)-f_(2))`
C. `v_(1)+v_(2)=[(2h)/(m)(f_(1)-f_(2))]^(1//2`
D. ` v._(1)^(2)+v._(2)^(2)=(2h)/(m)(f_(1)-f_(2))`

Answer

To solve the problem, we need to analyze the relationship between the maximum velocities of photoelectrons emitted from two identical photocathodes when illuminated with light of different frequencies.

### Step-by-Step Solution:

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

2. **Kinetic Energy in Terms of Velocity**:
The kinetic energy can also be expressed in terms of the maximum velocity \( v \) of the emitted electrons:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its maximum velocity.

3. **Setting Up the Equations**:
For the first photocathode with frequency \( f_1 \) and maximum velocity \( v_1 \):
\[ \frac{1}{2} mv_1^2 = hf_1 - \phi \quad \text{(Equation 1)} \]
For the second photocathode with frequency \( f_2 \) and maximum velocity \( v_2 \):
\[ \frac{1}{2} mv_2^2 = hf_2 - \phi \quad \text{(Equation 2)} \]

4. **Subtracting the Two Equations**:
We can subtract Equation 2 from Equation 1 to eliminate the work function \( \phi \):
\[ \frac{1}{2} mv_1^2 - \frac{1}{2} mv_2^2 = hf_1 - hf_2 \]
This simplifies to:
\[ \frac{1}{2} m (v_1^2 - v_2^2) = h(f_1 - f_2) \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ m(v_1^2 - v_2^2) = 2h(f_1 - f_2) \]

6. **Final Relation**:
Thus, we can express the relationship between the maximum velocities and the frequencies as:
\[ v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2) \]

### Conclusion:
This equation shows how the difference in the squares of the maximum velocities of the photoelectrons relates to the difference in the frequencies of the incident light.