The ratio of kinetic energies of photoelectro | vedclass

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(C) According to Einstein's photoelectric equation: $h v = h v_0 + K.E.$Therefore,$K.E. = h v - h v_0$.Given the ratio of kinetic energies for frequen

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$2v_1 - v_2$

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(C) According to Einstein's photoelectric equation: $h v = h v_0 + K.E.$Therefore,$K.E. = h v - h v_0$.Given the ratio of kinetic energies for frequencies $v_1$ and $v_2$ is $1:2$:$\frac{K.E._1}{K.E._2} = \frac{h v_1 - h v_0}{h v_2 - h v_0} = \frac{1}{2}$Cross-multiplying gives:$2(h v_1 - h v_0) = 1(h v_2 - h v_0)$$2h v_1 - 2h v_0 = h v_2 - h v_0$Rearranging for the threshold frequency $v_0$:$2h v_1 - h v_2 = 2h v_0 - h v_0$$v_0 = 2v_1 - v_2$

The ratio of kinetic energies of photoelectrons emitted from a metal by light of frequencies $v_1$ and $v_2$ is $1 : 2$. The threshold frequency of the metal will be:

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