From the figure describing the photoelectric | vedclass

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Question

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi$,where $h$ is Planck's constant,$
u$

Vedclass · Accepted Answer

Maximum kinetic energy for both the metals depends linearly on the frequency.

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\phi$ is the work function.This can be rewritten as $V_s = (h/e)
u - (\phi/e)$.This is the equation of a straight line $y = mx + c$,where the slope $m = h/e$ is constant for all metals.Since the slopes of the lines for $Na$ and $Al$ are the same (both make an angle $	heta$ with the frequency axis),the rate of change of stopping potential with respect to frequency is the same for both metals.Therefore,for the same change in frequency $\Delta
u$,the change in stopping potential $\Delta V_s = (h/e)\Delta
u$ will be the same for both $Na$ and $Al$.

From the figure describing the photoelectric effect,we may infer correctly that:

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