MHTCET 20 Apr 2026 Shift 1 Paper

Concept:The photoelectric effect connects the energy of incident photons to the work function and the kinetic energy of emitted photoelectrons.Explanation:For a photocathode, Einstein’s photoelectric equation is:$$h n = \phi + \frac{1}{2} m V^{2}$$Here, $\phi$ is the work function of the cathode material, and $\frac{1}{2} m V^{2}$ is the kinetic energy of the emitted electron.Since both photocathodes are identical, their work functions $\phi$ are equal.For frequency $n_1$, the equation becomes:$$h n_1 = \phi + \frac{1}{2} m V_1^{2}$$For frequency $n_2$, the equation becomes:$$h n_2 = \phi + \frac{1}{2} m V_2^{2}$$Subtract the second equation from the first to eliminate $\phi$:$$h(n_1 - n_2) = \frac{1}{2} m \left(V_1^{2} - V_2^{2}\right)$$Rearranging this gives:$$V_1^{2} - V_2^{2} = \frac{2h}{m} (n_1 - n_2)$$This matches the relation given in option D.Answer:$$V_1^{2} - V_2^{2} = \frac{2h}{m} (n_1 - n_2)$$Therefore, the correct option is D.