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

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$\frac{(n v_1-v_2)}{(n-1)}$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h v - h v_0$,where $h$ is Planck's constant and $v_0$ is the threshold frequency.For frequency $v_1$,$K_1 = h(v_1 - v_0)$.For frequency $v_2$,$K_2 = h(v_2 - v_0)$.Given the ratio of maximum kinetic energies is $K_1 : K_2 = 1 : n$,we have $\frac{K_1}{K_2} = \frac{1}{n}$.Substituting the expressions,we get $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{n}$.$\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{n}$.Cross-multiplying gives $n(v_1 - v_0) = v_2 - v_0$.$n v_1 - n v_0 = v_2 - v_0$.$n v_1 - v_2 = n v_0 - v_0$.$n v_1 - v_2 = v_0(n - 1)$.Therefore,the threshold frequency is $v_0 = \frac{n v_1 - v_2}{n - 1}$.

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

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