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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$,where $E$ is the incident pho

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$12 	imes 10^6 \ m/s$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$,where $E$ is the incident photon energy and $\phi$ is the work function.For the first case: $K_1 = 3\phi - \phi = 2\phi$. Thus,$\frac{1}{2}mv_1^2 = 2\phi$.For the second case: $K_2 = 9\phi - \phi = 8\phi$. Thus,$\frac{1}{2}mv_2^2 = 8\phi$.Taking the ratio of the two equations: $\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{8\phi}{2\phi} = 4$.Therefore,$\frac{v_2^2}{v_1^2} = 4$,which implies $\frac{v_2}{v_1} = 2$.Given $v_1 = 6.6 	imes 10^6 \ m/s$,we have $v_2 = 2 	imes v_1 = 2 	imes 6.6 	imes 10^6 \ m/s = 13.2 	imes 10^6 \ m/s$. Note: Based on the provided options,the intended calculation likely assumes $v_1 = 6 	imes 10^6 \ m/s$ to match option $A$ $(12 	imes 10^6 \ m/s)$. Using $v_1 = 6 	imes 10^6 \ m/s$,$v_2 = 2 	imes 6 	imes 10^6 = 12 	imes 10^6 \ m/s$.

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