1. Use Einstein’s photoelectric equation

For a photo-cathode, $hf = \phi + \frac{1}{2}mv^2$ where $\phi$ is the work function.

Since the two photo-cathodes are identical, their work function is the same.

So for frequency $f_1$ and electron speed $v_1$: $hf_1 = \phi + \frac{1}{2}mv_1^2 \quad ...(1)$

For frequency $f_2$ and electron speed $v_2$: $hf_2 = \phi + \frac{1}{2}mv_2^2 \quad ...(2)$

2. Subtract the two equations

Subtracting $(2)$ from $(1)$: $h(f_1-f_2)=\frac{1}{2}m(v_1^2-v_2^2)$

Multiply both sides by $\dfrac{2}{m}$: $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$

3. Match with the options

This is exactly Option A.

4. Check other options

• B: Incorrect, because photoelectric equation gives relation in terms of $v^2$, not directly $v_1+v_2$.
• C: Incorrect, because adding the two equations gives $h(f_1+f_2)=2\phi+\frac{1}{2}m(v_1^2+v_2^2),$ so the work function does not cancel.
• D: Incorrect, because subtraction leads to $v_1^2-v_2^2$, not $v_1-v_2$.

Hence, the correct answer is A.