In a photocell circuit,the stopping potential | vedclass

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Question

(B) For a photoelectron,the Einstein's photoelectric equation is given by:$e V_0 = h 
u - \phi_0 \Rightarrow V_0 = \frac{h}{e} 
u - \frac{\phi_0}{e}

Vedclass · Accepted Answer

$6.0 	imes 10^{-34} \, Js, 2.0 \, eV$

Vedclass · Answer

(B) For a photoelectron,the Einstein's photoelectric equation is given by:$e V_0 = h 
u - \phi_0 \Rightarrow V_0 = \frac{h}{e} 
u - \frac{\phi_0}{e}$Comparing this with the equation of a straight line $y = mx + c$,the slope of the $V_0$ versus $
u$ graph is $\frac{h}{e}$ and the intercept on the $V_0$-axis is $-\frac{\phi_0}{e}$.From the given graph,we take two points $(0.8 	imes 10^{15} \, Hz, 1 \, V)$ and $(1.6 	imes 10^{15} \, Hz, 4 \, V)$.Slope $= \frac{h}{e} = \frac{V_{0_2} - V_{0_1}}{
u_2 - 
u_1} = \frac{4 - 1}{(1.6 - 0.8) 	imes 10^{15}} = \frac{3}{0.8 	imes 10^{15}} = 3.75 	imes 10^{-15} \, V \cdot s$.Now,$h = e 	imes 	ext{Slope} = (1.6 	imes 10^{-19} \, C) 	imes (3.75 	imes 10^{-15} \, V \cdot s) = 6.0 	imes 10^{-34} \, Js$.From the graph,the intercept on the $V_0$-axis is $-2 \, V$. Therefore,$-\frac{\phi_0}{e} = -2 \, V$,which gives the work function $\phi_0 = 2 \, eV$.

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