MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the incident frequency and the threshold frequency of the metal.Explanation:According to Einstein's photoelectric equation,$$K_{\max} = h(\nu - \nu_0)$$where $h$ is Planck's constant and $\nu_0$ is the threshold frequency of the metallic surface.For the first incident frequency $\nu_1$, the maximum kinetic energy is$$K_1 = h(\nu_1 - \nu_0)$$For the second incident frequency $\nu_2$, the maximum kinetic energy is$$K_2 = h(\nu_2 - \nu_0)$$Given that the ratio of the maximum kinetic energies is$$\frac{K_1}{K_2} = \frac{3}{K}$$Substituting the expressions for $K_1$ and $K_2$, we get$$\frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{3}{K}$$Cross-multiplying,$$K(\nu_1 - \nu_0) = 3(\nu_2 - \nu_0)$$$$K\nu_1 - K\nu_0 = 3\nu_2 - 3\nu_0$$Rearranging the terms,$$K\nu_0 - 3\nu_0 = K\nu_1 - 3\nu_2$$$$\nu_0(K - 3) = K\nu_1 - 3\nu_2$$Therefore, the threshold frequency is$$\nu_0 = \frac{K\nu_1 - 3\nu_2}{K - 3}$$Answer:The threshold frequency of the metallic surface is $\frac{K\nu_1 - 3\nu_2}{K - 3}$, which matches Option D.