Step 1: Recall what light does to the cathode.
When light of frequency $f$ strikes a metal, each photon carries energy $hf$. Part of that energy frees the electron (the work function $\Phi$), and the leftover becomes the electron's kinetic energy. That is Einstein's photoelectric law.
Step 2: Write the equation for the first cathode.
Both cathodes are identical, so they share the same work function $\Phi$. For frequency $f_1$ giving speed $V_1$, \[ hf_1 = \Phi + \tfrac{1}{2} m V_1^2 \]
Step 3: Write the equation for the second cathode.
Similarly, for frequency $f_2$ giving speed $V_2$, \[ hf_2 = \Phi + \tfrac{1}{2} m V_2^2 \]
Step 4: Subtract to cancel the work function.
Since $\Phi$ is the same in both, subtracting the second from the first wipes it out. \[ h(f_1 - f_2) = \tfrac{1}{2} m (V_1^2 - V_2^2) \]
Step 5: Solve for the velocity combination.
Multiply both sides by $2$ and divide by $m$. \[ V_1^2 - V_2^2 = \frac{2h}{m}(f_1 - f_2) \]
Step 6: Match with the options.
Notice the difference of squares matches option 1 exactly, so that is our answer.
\[ \boxed{V_1^2 - V_2^2 = \frac{2h}{m}(f_1 - f_2)} \]