The stopping potential as a function of frequ | vedclass

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Question

(B) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by:$e V_s = h 
u - \Phi_0$$V_s = \frac{h}{e} 
u - \frac{\P

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smaller than that of $B$.

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(B) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by:$e V_s = h 
u - \Phi_0$$V_s = \frac{h}{e} 
u - \frac{\Phi_0}{e}$where $\Phi_0$ is the work function and $
u_0$ is the threshold frequency, such that $\Phi_0 = h 
u_0$.From the given graph, the intercept on the frequency axis represents the threshold frequency $
u_0$.Since the threshold frequency $
u_{0,A}$ for surface $A$ is less than the threshold frequency $
u_{0,B}$ for surface $B$ ( $
u_{0,A} it follows that the work function $\Phi_{0,A} = h 
u_{0,A}$ is smaller than the work function $\Phi_{0,B} = h 
u_{0,B}$.Therefore, the work function of $A$ is smaller than that of $B$.

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

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