MHTCET 15 May 2024 Shift 1 Paper

Concept:The maximum kinetic energy of a photoelectron is given by Einstein's photoelectric equation: $K_{\max} = h\nu - \phi_0$.This kinetic energy is equal to $\frac{1}{2}mv^2$, which gives the velocity of the emitted electron.Explanation:Let the work function of the metal be $\phi_0$.For the first photon, energy is $E_1 = 2\phi_0$.Using $K_{\max} = E - \phi_0$: $K_{\max,1} = 2\phi_0 - \phi_0 = \phi_0$.For the second photon, energy is $E_2 = 3\phi_0$.So $K_{\max,2} = 3\phi_0 - \phi_0 = 2\phi_0$.Since $K_{\max} = \frac{1}{2}mv^2$, we get:$\frac{K_{\max,1}}{K_{\max,2}} = \frac{v_1^2}{v_2^2} = \frac{\phi_0}{2\phi_0} = \frac{1}{2}$.Taking the square root: $\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$.Therefore, the ratio of maximum velocities is $1 : \sqrt{2}$.Answer:Option C: $1 : \sqrt{2}$