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

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$1: \sqrt{2}$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons 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,$E_1 = 2\phi_0$. Therefore,$K.E_1 = 2\phi_0 - \phi_0 = \phi_0$.Since $K.E_1 = \frac{1}{2}mv_1^2$,we have $\frac{1}{2}mv_1^2 = \phi_0$  --- $(1)$For the second case,$E_2 = 3\phi_0$. Therefore,$K.E_2 = 3\phi_0 - \phi_0 = 2\phi_0$.Since $K.E_2 = \frac{1}{2}mv_2^2$,we have $\frac{1}{2}mv_2^2 = 2\phi_0$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{\phi_0}{2\phi_0}$$\frac{v_1^2}{v_2^2} = \frac{1}{2}$Taking the square root on both sides:$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$Thus,the ratio $v_1: v_2$ is $1: \sqrt{2}$.

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other,the maximum velocities of the photoelectrons emitted in the two cases are $v_1$ and $v_2$ respectively. The ratio $v_1: v_2$ is

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