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(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = hv - h v_0$, where $v$ is the frequenc

Vedclass · Accepted Answer

$\frac{kv_2 - v_1}{k - 1}$

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(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = hv - h v_0$, where $v$ is the frequency of incident light and $v_0$ is the threshold frequency.For frequency $v_1$, $K_1 = h v_1 - h v_0$.For frequency $v_2$, $K_2 = h v_2 - h v_0$.Given the ratio of kinetic energies is $K_1 / K_2 = k / 1$, we have $K_1 = k K_2$.Substituting the expressions: $h v_1 - h v_0 = k(h v_2 - h v_0)$.Dividing by $h$: $v_1 - v_0 = k v_2 - k v_0$.Rearranging the terms to solve for $v_0$: $k v_0 - v_0 = k v_2 - v_1$.$v_0(k - 1) = k v_2 - v_1$.Therefore, $v_0 = \frac{k v_2 - v_1}{k - 1}$.

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