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(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h v - \Phi_0$, where $h$ is Planck's c

Vedclass · Accepted Answer

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

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(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h v - \Phi_0$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi_0 = h v_0$ is the work function ($v_0$ is the threshold frequency).For frequency $v_1$, $K_1 = h v_1 - h v_0 = h(v_1 - v_0)$.For frequency $v_2$, $K_2 = h v_2 - h v_0 = h(v_2 - v_0)$.Given the ratio $K_1 : K_2 = 1 : k$, we have $\frac{K_1}{K_2} = \frac{1}{k}$.Substituting the expressions: $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{k}$.This simplifies to $k(v_1 - v_0) = v_2 - v_0$.Expanding the terms: $k v_1 - k v_0 = v_2 - v_0$.Rearranging to solve for $v_0$: $k v_1 - v_2 = k v_0 - v_0 = v_0(k - 1)$.Therefore, $v_0 = \frac{k v_1 - v_2}{k - 1}$.

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is:

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