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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi_{0}$,where $E$ is the incident

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$1: 3$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi_{0}$,where $E$ is the incident energy and $\phi_{0}$ is the work function.For the first radiation,$E_{1} = 2\phi_{0}$,so $K_{1} = \frac{1}{2}mv_{1}^{2} = 2\phi_{0} - \phi_{0} = \phi_{0}$.For the second radiation,$E_{2} = 10\phi_{0}$,so $K_{2} = \frac{1}{2}mv_{2}^{2} = 10\phi_{0} - \phi_{0} = 9\phi_{0}$.Taking the ratio of the two equations: $\frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}} = \frac{\phi_{0}}{9\phi_{0}}$.This simplifies to $\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{9}$.Taking the square root on both sides,we get $\frac{v_{1}}{v_{2}} = \frac{1}{3}$.

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

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