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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$,where $E$ is the energy of th

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$2: 3$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function.For the first case,$E_1 = 5\phi$. Therefore,$K_{max,1} = 5\phi - \phi = 4\phi$.Since $K_{max,1} = \frac{1}{2}mv_1^2$,we have $\frac{1}{2}mv_1^2 = 4\phi$.For the second case,$E_2 = 10\phi$. Therefore,$K_{max,2} = 10\phi - \phi = 9\phi$.Since $K_{max,2} = \frac{1}{2}mv_2^2$,we have $\frac{1}{2}mv_2^2 = 9\phi$.Taking the ratio of the two equations:$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{4\phi}{9\phi}$$\frac{v_1^2}{v_2^2} = \frac{4}{9}$Taking the square root on both sides,we get $\frac{v_1}{v_2} = \frac{2}{3}$.Thus,the ratio of the maximum velocities is $2:3$.

Two streams of photons,possessing energies equal to $5$ and $10$ times the work function of a metal,are incident on the metal surface successively. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases,respectively?

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