Energy of the incident photons on the photose | vedclass

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

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$1 : \sqrt{2}$

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(B) 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, $E_1 = 3W$. Therefore, $K_1 = 3W - W = 2W$.Since $K = \frac{1}{2}mv^2$, we have $\frac{1}{2}mv_1^2 = 2W$, which implies $v_1 = \sqrt{\frac{4W}{m}}$.For the second case, $E_2 = 5W$. Therefore, $K_2 = 5W - W = 4W$.Similarly, $\frac{1}{2}mv_2^2 = 4W$, which implies $v_2 = \sqrt{\frac{8W}{m}}$.The ratio of the velocities is $\frac{v_1}{v_2} = \frac{\sqrt{4W/m}}{\sqrt{8W/m}} = \sqrt{\frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus, the ratio is $1 : \sqrt{2}$.

Energy of the incident photons on the photosensitive metal surface is $3W$ and then $5W$ where '$W$' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

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