Two identical photocathodes receive the light of frequencies $$f_1$$ and $$f_2$$ respectively. If the velocities of the photo-electrons coming out are $$v_1$$ and $$v_2$$ respectively, then:

By Einstein's photoelectric equation, the maximum kinetic energy of a photoelectron is $$\frac{1}{2}mv^2 = hf - \phi$$, where $$\phi$$ is the work function of the photocathode.

For the first photocathode: $$\frac{1}{2}mv_1^2 = hf_1 - \phi$$. For the second photocathode (identical, so same $$\phi$$): $$\frac{1}{2}mv_2^2 = hf_2 - \phi$$.

Subtracting the second equation from the first: $$\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$$, which gives $$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$$.