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

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$8 	imes 10^6 \, m/s$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function.Case $1$: $E_1 = 2hv_0$ and $\Phi = hv_0$.$K_{max1} = 2hv_0 - hv_0 = hv_0 = \frac{1}{2}mv_1^2$,where $v_1 = 4 	imes 10^6 \, m/s$.Case $2$: $E_2 = 5hv_0$ and $\Phi = hv_0$.$K_{max2} = 5hv_0 - hv_0 = 4hv_0 = \frac{1}{2}mv_2^2$.Dividing the two equations:$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4hv_0}{hv_0} = 4$.$\frac{v_2^2}{v_1^2} = 4 \Rightarrow \frac{v_2}{v_1} = 2$.$v_2 = 2v_1 = 2 	imes (4 	imes 10^6 \, m/s) = 8 	imes 10^6 \, m/s$.

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