The stopping potential V_0 (in volt) as a fun | vedclass

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Question

(B) According to Einstein's photoelectric equation:$h
u = \phi + eV_0$Rearranging for the stopping potential $V_0$:$V_0 = \frac{h
u}{e} - \frac{\phi

Vedclass · Accepted Answer

$1.66$

Vedclass · Answer

(B) According to Einstein's photoelectric equation:$h
u = \phi + eV_0$Rearranging for the stopping potential $V_0$:$V_0 = \frac{h
u}{e} - \frac{\phi}{e}$From the given graph,the threshold frequency $(
u_0)$ is the frequency at which the stopping potential $V_0$ becomes zero. Looking at the graph,$V_0 = 0$ when $
u = 4 	imes 10^{14} \, Hz$.At this point:$0 = \frac{h
u_0}{e} - \frac{\phi}{e}$$\Rightarrow \phi = h
u_0$Substituting the values:$\phi = (6.63 	imes 10^{-34} \, Js) 	imes (4 	imes 10^{14} \, Hz)$$\phi = 26.52 	imes 10^{-20} \, J$To convert the work function into electron-volts $(eV)$:$\phi (eV) = \frac{26.52 	imes 10^{-20} \, J}{1.6 	imes 10^{-19} \, C}$$\phi = 1.6575 \, eV \approx 1.66 \, eV$

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