The energy of the incident photon on a metal | vedclass

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an emitted photoelectron is given by $K_{max} = E - W$,wher

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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 an emitted photoelectron is given by $K_{max} = E - W$,where $E$ is the energy of the incident photon and $W$ is the work function.For the first case,$E_1 = 3W$. Thus,$K_1 = \frac{1}{2}mv_1^2 = 3W - W = 2W$.For the second case,$E_2 = 5W$. Thus,$K_2 = \frac{1}{2}mv_2^2 = 5W - W = 4W$.Taking the ratio of the two equations:$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{2W}{4W} = \frac{1}{2}$.Therefore,$\frac{v_1^2}{v_2^2} = \frac{1}{2}$,which implies $\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$.The ratio of velocities is $1: \sqrt{2}$.

The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

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