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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of emitted electrons is given by $K_{\max} = \frac{1}{2} m v_

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$2$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of emitted electrons is given by $K_{\max} = \frac{1}{2} m v_{\max}^2 = E - \phi_0$,where $E$ is the incident photon energy and $\phi_0$ is the work function.For the first case,$E_1 = 4 	ext{ eV}$ and $\phi_0 = 2 	ext{ eV}$:$\frac{1}{2} m v_1^2 = 4 - 2 = 2 	ext{ eV} \quad \dots(i)$For the second case,$E_2 = 2.5 	ext{ eV}$ and $\phi_0 = 2 	ext{ eV}$:$\frac{1}{2} m v_2^2 = 2.5 - 2 = 0.5 	ext{ eV} \quad \dots(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{2}{0.5}$$\frac{v_1^2}{v_2^2} = 4$Taking the square root on both sides:$\frac{v_1}{v_2} = \sqrt{4} = 2$

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

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