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Question

(C) According to Einstein's photoelectric equation: $KE = h
u - h
u_0$,where $h
u_0$ is the work function.Initially: $KE_1 = h
u - h
u_0 \ldots(1

Vedclass · Accepted Answer

Increased but more than doubled of the previous $KE$

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $KE = h
u - h
u_0$,where $h
u_0$ is the work function.Initially: $KE_1 = h
u - h
u_0 \ldots(1)$When frequency is doubled: $KE_2 = h(2
u) - h
u_0 = 2h
u - h
u_0 \ldots(2)$From equation $(1)$,$h
u = KE_1 + h
u_0$. Substituting this into equation $(2)$:$KE_2 = 2(KE_1 + h
u_0) - h
u_0$$KE_2 = 2KE_1 + 2h
u_0 - h
u_0$$KE_2 = 2KE_1 + h
u_0$Since $h
u_0 > 0$,the new kinetic energy $KE_2$ is greater than $2KE_1$.

When the frequency of light incident on a metallic plate is doubled,the $KE$ of the emitted photoelectrons will be:

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