The work function of a metal is h nu_0. Light | vedclass

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Question

(A) According to Einstein's photoelectric equation,$h 
u - W = K_{	ext{max}}$,where $h$ is Planck's constant,$
u$ is the frequency of incident radi

Vedclass · Accepted Answer

$ 
u > 
u_0 $

Vedclass · Answer

(A) According to Einstein's photoelectric equation,$h 
u - W = K_{	ext{max}}$,where $h$ is Planck's constant,$
u$ is the frequency of incident radiation,$W$ is the work function,and $K_{	ext{max}}$ is the maximum kinetic energy of the emitted photoelectrons.Since kinetic energy cannot be negative,$K_{	ext{max}} \geq 0$.Therefore,$h 
u - W \geq 0$,which implies $h 
u \geq W$.Given that the work function $W = h 
u_0$,the condition becomes $h 
u \geq h 
u_0$.Dividing both sides by $h$,we get $
u \geq 
u_0$.Thus,the photoelectric effect will take place only if the frequency of incident light is greater than or equal to the threshold frequency $
u_0$.

The work function of a metal is $h \nu_0$. Light of frequency $\nu$ falls on this metal. The photoelectric effect will take place only if

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