Two photons having energies twice and thrice | vedclass

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

Vedclass · Accepted Answer

$1: \sqrt{2}$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy 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.For the first case,$E_1 = 2\phi_0$,so $K.E_1 = 2\phi_0 - \phi_0 = \phi_0$.For the second case,$E_2 = 3\phi_0$,so $K.E_2 = 3\phi_0 - \phi_0 = 2\phi_0$.Since $K.E. = \frac{1}{2}mv^2$,we have $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$.Substituting the values,$\frac{v_1^2}{v_2^2} = \frac{\phi_0}{2\phi_0} = \frac{1}{2}$.Therefore,the ratio of maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Two photons having energies twice and thrice the work function of a metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively:

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