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(D) Let the work function of the metal be $W$.The energy of the first photon is $E_1 = 2W$.The energy of the second photon is $E_2 = 3W$.According to

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$1: \sqrt{2}$

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(D) Let the work function of the metal be $W$.The energy of the first photon is $E_1 = 2W$.The energy of the second photon is $E_2 = 3W$.According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by $(KE)_{\max} = E - W$.For the first photon: $(KE_1)_{\max} = 2W - W = W$.For the second photon: $(KE_2)_{\max} = 3W - W = 2W$.The ratio of maximum kinetic energies is $\frac{(KE_1)_{\max}}{(KE_2)_{\max}} = \frac{W}{2W} = \frac{1}{2}$.Since $(KE)_{\max} = \frac{1}{2}mv^2$,we have $\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{2}$.This simplifies to $\frac{v_1^2}{v_2^2} = \frac{1}{2}$,which gives $\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,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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