1. For photoelectric emission from a photocathode, Einstein’s photoelectric equation is

$hf = \phi + \frac{1}{2}mv^2$

where:

• $h$ = Planck’s constant
• $f$ = frequency of incident light
• $\phi$ = work function of the photocathode
• $m$ = mass of electron
• $v$ = maximum speed of emitted photoelectron

2. Since the two photocathodes are identical, their work function $\phi$ is the same.

For frequency $f_1$:

$hf_1 = \phi + \frac{1}{2}mv_1^2$

For frequency $f_2$:

$hf_2 = \phi + \frac{1}{2}mv_2^2$

3. Subtract the second equation from the first:

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

4. Rearranging,

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

5. Now check the options:

• A: $v_1-v_2 = \sqrt{\frac{2h}{m}(f_1-f_2)}$
  Incorrect, because subtraction of squares does not simplify like this.

• B: $v_1^2+v_2^2 = \frac{2h}{m}(f_1+f_2)$
  Incorrect, because the work function terms would not cancel in addition.

• C: $v_1+v_2 = \sqrt{\frac{2h}{m}(f_1+f_2)}$
  Incorrect.

• D: $v_1^2-v_2^2 = \frac{2h}{m}(f_1-f_2)$
  Correct.

Therefore, the correct option is D.