In an experiment on photoelectric emission fr | vedclass

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Question

(C) According to Einstein's photoelectric equation: $e V_0 = h 
u - h 
u_0$ Where $V_0$ is the stopping potential,$
u$ is the frequency of incident

Vedclass · Accepted Answer

$9 	imes 10^{14}$

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $e V_0 = h 
u - h 
u_0$ Where $V_0$ is the stopping potential,$
u$ is the frequency of incident light,and $
u_0$ is the threshold frequency. Rearranging for $
u_0$: $
u_0 = 
u - \frac{e V_0}{h} = \frac{c}{\lambda} - \frac{e V_0}{h}$ Given: $\lambda = 2 	imes 10^{-7} \ m$,$V_0 = 2.5 \ V$,$e = 1.6 	imes 10^{-19} \ C$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$. Calculating frequency of incident light: $
u = \frac{3 	imes 10^8}{2 	imes 10^{-7}} = 1.5 	imes 10^{15} \ Hz$. Calculating the work function term: $\frac{e V_0}{h} = \frac{1.6 	imes 10^{-19} 	imes 2.5}{6.6 	imes 10^{-34}} \approx 0.606 	imes 10^{15} \ Hz$. Therefore,$
u_0 = 1.5 	imes 10^{15} - 0.606 	imes 10^{15} = 0.894 	imes 10^{15} \ Hz \approx 9.0 	imes 10^{14} \ Hz$.

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