The graph shows the stopping potential as a f | vedclass

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Question

(B) According to Einstein's photoelectric equation,$eV_0 = h
u - \Phi$,where $V_0$ is the stopping potential,$
u$ is the frequency,$h$ is Planck's c

Vedclass · Accepted Answer

less than $B$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,$eV_0 = h
u - \Phi$,where $V_0$ is the stopping potential,$
u$ is the frequency,$h$ is Planck's constant,and $\Phi$ is the work function.Rearranging gives $V_0 = (h/e)
u - (\Phi/e)$.The threshold frequency $
u_0$ is the frequency at which the stopping potential $V_0$ is zero,so $
u_0 = \Phi/h$,or $\Phi = h
u_0$.From the graph,the intercept on the frequency axis represents the threshold frequency $
u_0$.Since the intercept for surface $A$ is to the left of the intercept for surface $B$,we have $
u_{0A} Therefore,the work function $\Phi_A = h
u_{0A}$ is less than the work function $\Phi_B = h
u_{0B}$.

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

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