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(B) The stopping potential $(V_s)$ depends on the frequency of the incident light,not on the intensity of the light.Changing the distance of the light

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(B) The stopping potential $(V_s)$ depends on the frequency of the incident light,not on the intensity of the light.Changing the distance of the light source changes the intensity of the light incident on the photocell,but it does not change the frequency of the photons.According to Einstein's photoelectric equation: $eV_s = h
u - \phi$,where $
u$ is the frequency of incident light and $\phi$ is the work function.Since the frequency $
u$ remains constant when the distance is changed,the stopping potential $V_s$ remains unchanged.Therefore,the stopping potential remains $0.6 \ V$.

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?

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