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(B) Let the work function of the metal be $\Phi = h
u_0$. Given energies of incident photons are $E_1 = 3\Phi$ and $E_2 = 9\Phi$. According to Einste

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$1:2$

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(B) Let the work function of the metal be $\Phi = h
u_0$. Given energies of incident photons are $E_1 = 3\Phi$ and $E_2 = 9\Phi$. According to Einstein's photoelectric equation,the maximum kinetic energy is $K_{\max} = E - \Phi$. For the first case: $K_{\max, 1} = \frac{1}{2}mv_1^2 = 3\Phi - \Phi = 2\Phi$. For the second case: $K_{\max, 2} = \frac{1}{2}mv_2^2 = 9\Phi - \Phi = 8\Phi$. Taking the ratio of the two equations: $\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{2\Phi}{8\Phi}$. $\frac{v_1^2}{v_2^2} = \frac{1}{4}$. Taking the square root on both sides,we get $\frac{v_1}{v_2} = \frac{1}{2}$.

Two photons with energies three times and nine times the work function of a metal surface are incident on it. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases?

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