Photoelectric emission is observed from a met | vedclass

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Question

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - h
u_0$, where $
u$ is the fre

Vedclass · Accepted Answer

$\frac{K
u_1 - 3
u_2}{K - 3}$

Vedclass · Answer

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - h
u_0$, where $
u$ is the frequency of incident light and $
u_0$ is the threshold frequency.For the first case: $K_{max1} = h
u_1 - h
u_0$.For the second case: $K_{max2} = h
u_2 - h
u_0$.The ratio is given as $\frac{K_{max1}}{K_{max2}} = \frac{3}{K}$.Substituting the expressions: $\frac{h
u_1 - h
u_0}{h
u_2 - h
u_0} = \frac{3}{K}$.$K(
u_1 - 
u_0) = 3(
u_2 - 
u_0)$.$K
u_1 - K
u_0 = 3
u_2 - 3
u_0$.$K
u_1 - 3
u_2 = K
u_0 - 3
u_0$.$K
u_1 - 3
u_2 = 
u_0(K - 3)$.Therefore, the threshold frequency $
u_0 = \frac{K
u_1 - 3
u_2}{K - 3}$.

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