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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by:$(K.E.)_{max} = h v - h v_0$where $h$ is Plan

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

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)_{max}$ is given by:$(K.E.)_{max} = h v - h v_0$where $h$ is Planck's constant,$v$ is the frequency of incident light,and $v_0$ is the threshold frequency.For frequencies $v_1$ and $v_2$,we have:$(K.E.)_1 = h v_1 - h v_0$  ....$(1)$$(K.E.)_2 = h v_2 - h v_0$  ....$(2)$Given the ratio of maximum kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{1}{K}$.Dividing equation $(1)$ by $(2)$:$\frac{1}{K} = \frac{h v_1 - h v_0}{h v_2 - h v_0} = \frac{v_1 - v_0}{v_2 - v_0}$Cross-multiplying:$v_2 - v_0 = K(v_1 - v_0)$$v_2 - v_0 = K v_1 - K 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}$

Photoemission from a metal surface takes place for frequencies $v_1$ and $v_2$ of incident rays. If the ratio of the maximum kinetic energy of photoelectrons is $1:K$,then the threshold frequency of the metallic surface is

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