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(D) Let the work function of the metal be $W$.According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max }$ is given by $(K

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$1: \sqrt{2}$

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(D) Let the work function of the metal be $W$.According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max }$ is given by $(KE)_{\max } = E - W$,where $E$ is the energy of the incident photon.For the first photon,$E_1 = 2W$. Thus,$(KE_1)_{\max } = 2W - W = W$.For the second photon,$E_2 = 3W$. Thus,$(KE_2)_{\max } = 3W - W = 2W$.Since $(KE)_{\max } = \frac{1}{2}mv^2$,we have the ratio of kinetic energies:$\frac{(KE_1)_{\max }}{(KE_2)_{\max }} = \frac{W}{2W} = \frac{1}{2}$.Substituting the velocity terms:$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{2} \implies \frac{v_1^2}{v_2^2} = \frac{1}{2}$.Taking the square root on both sides:$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$.Therefore,the ratio of maximum velocities is $1: \sqrt{2}$.

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

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