The graph of stopping potential V_s against f | 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$$V_s = \frac{h}{e} 
u - \frac{\phi}

Vedclass · Accepted Answer

$\phi_x < \phi_y$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:$e V_s = h 
u - \phi$$V_s = \frac{h}{e} 
u - \frac{\phi}{e}$Comparing this with the equation of a straight line $y = mx + c$,the intercept on the frequency axis (where $V_s = 0$) is the threshold frequency $
u_0$,given by $
u_0 = \frac{\phi}{h}$,which implies $\phi = h 
u_0$.From the graph,it is clear that the threshold frequency for metal $X$ $(
u_0)$ is less than the threshold frequency for metal $Y$ $(
u_0^{\prime})$,i.e.,$
u_0 Since the work function $\phi$ is directly proportional to the threshold frequency $
u_0$,we have $\phi_x < \phi_y$.

The graph of stopping potential $V_s$ against frequency $\nu$ of incident radiation is plotted for two different metals $X$ and $Y$ as shown in the graph. If $\phi_x$ and $\phi_y$ are the work functions of $X$ and $Y$,respectively,then:

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