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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by: $(KE)_{\max} = h v - h v_0$,where $v$ is the

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$\frac{k v_1-v_2}{k-1}$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by: $(KE)_{\max} = h v - h v_0$,where $v$ is the frequency of incident radiation and $v_0$ is the threshold frequency.For the first case with frequency $v_1$: $(KE_1)_{\max} = h(v_1 - v_0)$.For the second case with frequency $v_2$: $(KE_2)_{\max} = h(v_2 - v_0)$.Given the ratio of maximum kinetic energies is $1:k$,we have: $\frac{(KE_1)_{\max}}{(KE_2)_{\max}} = \frac{1}{k} = \frac{h(v_1 - v_0)}{h(v_2 - v_0)}$.Cross-multiplying gives: $v_2 - v_0 = k(v_1 - v_0)$.$v_2 - v_0 = k v_1 - k v_0$.Rearranging to solve for $v_0$: $k v_0 - v_0 = k v_1 - v_2$.$v_0(k - 1) = k v_1 - v_2$.$v_0 = \frac{k v_1 - v_2}{k - 1}$.

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