Photoelectric emission is observed from a met | vedclass

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

Vedclass · Accepted Answer

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

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ is given by:$(KE)_{\max} = h v - h v_0$where $h$ is Planck's constant and $v_0$ is the threshold frequency.For the first case with frequency $v_1$:$(KE_1)_{\max} = h(v_1 - v_0)$For the second case with frequency $v_2$:$(KE_2)_{\max} = h(v_2 - v_0)$Given the ratio $\frac{(KE_1)_{\max}}{(KE_2)_{\max}} = \frac{1}{k}$,we have:$\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{k}$$k(v_1 - v_0) = v_2 - v_0$$k v_1 - k v_0 = v_2 - v_0$$k v_1 - v_2 = k v_0 - v_0$$k v_1 - v_2 = v_0(k - 1)$$v_0 = \frac{k v_1 - v_2}{k - 1}$

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