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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - \Phi$, where $\Phi = hf_0$ is the

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$1 : 2$

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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - \Phi$, where $\Phi = hf_0$ is the work function.For frequency $2f_0$, the maximum kinetic energy is $K_1 = h(2f_0) - hf_0 = hf_0$.Since $K_1 = \frac{1}{2}mv_1^2$, we have $\frac{1}{2}mv_1^2 = hf_0 \implies v_1 = \sqrt{\frac{2hf_0}{m}}$.For frequency $5f_0$, the maximum kinetic energy is $K_2 = h(5f_0) - hf_0 = 4hf_0$.Since $K_2 = \frac{1}{2}mv_2^2$, we have $\frac{1}{2}mv_2^2 = 4hf_0 \implies v_2 = \sqrt{\frac{8hf_0}{m}}$.The ratio $v_1 : v_2 = \sqrt{\frac{2hf_0}{m}} : \sqrt{\frac{8hf_0}{m}} = \sqrt{2} : \sqrt{8} = \sqrt{2} : 2\sqrt{2} = 1 : 2$.

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