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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = \frac{hc}{\lambda} - \phi_0 = \frac{h

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$1: 2$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:$K_{\max} = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have:$\frac{1}{2}mv_{\max}^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$$v_{\max} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}$For the first beam with $\lambda_1 = \frac{\lambda_0}{3}$:$v_1 = \sqrt{\frac{2hc}{m} \left( \frac{3}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{2}{\lambda_0}}$For the second beam with $\lambda_2 = \frac{\lambda_0}{9}$:$v_2 = \sqrt{\frac{2hc}{m} \left( \frac{9}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{8}{\lambda_0}}$The ratio of the maximum velocities is:$\frac{v_1}{v_2} = \sqrt{\frac{2/\lambda_0}{8/\lambda_0}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

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