Radiations of two photons having energies twi | vedclass

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{\max})$ is given by $K.E._{\max} = E - \phi_0$,where $E$ is the

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{\max})$ is given by $K.E._{\max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.For the first case,$E_1 = 2\phi_0$,so $K.E._{\max 1} = 2\phi_0 - \phi_0 = \phi_0$.For the second case,$E_2 = 5\phi_0$,so $K.E._{\max 2} = 5\phi_0 - \phi_0 = 4\phi_0$.The ratio of maximum kinetic energies is $\frac{K.E._{\max 1}}{K.E._{\max 2}} = \frac{\phi_0}{4\phi_0} = \frac{1}{4}$.Since $K.E._{\max} = \frac{1}{2}mv_{\max}^2$,the ratio of velocities is $\frac{v_1}{v_2} = \sqrt{\frac{K.E._{\max 1}}{K.E._{\max 2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

Radiations of two photons having energies twice and five times the work function of a metal are incident successively on a metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be

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