The stopping potential (V_s) as a function of | vedclass

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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \Phi_0$,where $\Phi_0$ is the work function

Vedclass · Accepted Answer

Smaller than that of $B$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \Phi_0$,where $\Phi_0$ is the work function.This can be rewritten as $V_s = (h/e)
u - (\Phi_0/e)$.Comparing this to the equation of a straight line $y = mx + c$,the intercept on the negative $V_s$ axis is $\Phi_0/e$.The threshold frequency $
u_0$ is the frequency at which the stopping potential is zero,given by $
u_0 = \Phi_0/h$.From the graph,the threshold frequency for surface $A$ is smaller than that for surface $B$ (i.e.,$
u_{0A} Since $\Phi_0 = h
u_0$,a smaller threshold frequency implies a smaller work function.Therefore,the work function of $A$ is smaller than that of $B$.

The stopping potential $(V_s)$ as a function of the frequency $(\nu)$ of the incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graphs show that the work function of $A$ is

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