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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = E - \Phi$,whe

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$1: \sqrt{2}$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.Case $1$: When incident energy $E_1 = 2\Phi$,the maximum kinetic energy is $K_1 = 2\Phi - \Phi = \Phi$.Since $K_1 = \frac{1}{2} m V_1^2$,we have $\frac{1}{2} m V_1^2 = \Phi$,which implies $V_1 = \sqrt{\frac{2\Phi}{m}}$.Case $2$: When incident energy $E_2 = 3\Phi$,the maximum kinetic energy is $K_2 = 3\Phi - \Phi = 2\Phi$.Since $K_2 = \frac{1}{2} m V_2^2$,we have $\frac{1}{2} m V_2^2 = 2\Phi$,which implies $V_2 = \sqrt{\frac{4\Phi}{m}}$.Taking the ratio $V_1: V_2$:$\frac{V_1}{V_2} = \frac{\sqrt{\frac{2\Phi}{m}}}{\sqrt{\frac{4\Phi}{m}}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,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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