In the experiment of photoelectric effect, K_ | vedclass

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Question

(B) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi$, where $
u$ is the frequency of incident light and $\Phi$ is the work fun

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi$, where $
u$ is the frequency of incident light and $\Phi$ is the work function of the metal.Case $1$: $K_0 = h
u - \Phi$ --- $(1)$Case $2$: $3K_0 = h(2
u) - \Phi = 2h
u - \Phi$ --- $(2)$Subtracting equation $(1)$ from $(2)$:$(3K_0 - K_0) = (2h
u - \Phi) - (h
u - \Phi)$$2K_0 = h
u$Substituting $h
u = 2K_0$ into equation $(1)$:$K_0 = 2K_0 - \Phi$ implies $\Phi = K_0$Case $3$: If frequency is tripled, $
u' = 3
u$.$K_{max}' = h(3
u) - \Phi = 3(h
u) - \Phi$Substituting $h
u = 2K_0$ and $\Phi = K_0$:$K_{max}' = 3(2K_0) - K_0 = 6K_0 - K_0 = 5K_0$.Therefore, the value is $5$.

In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)

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