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 the threshold frequency $
u_0$: $
u_0 = 
u - \frac{e V_0}{h}$.Since $
u = \frac{c}{\lambda}$,we have $
u_0 = \frac{c}{\lambda} - \frac{e V_0}{h}$.Substituting the given values: $c = 3 	imes 10^8 \,m/s$,$\lambda = 2 	imes 10^{-7} \,m$,$e = 1.6 	imes 10^{-19} \,C$,$V_0 = 2.5 \,V$,and $h = 6.6 	imes 10^{-34} \,J-s$.$
u_0 = \frac{3 	imes 10^8}{2 	imes 10^{-7}} - \frac{1.6 	imes 10^{-19} 	imes 2.5}{6.6 	imes 10^{-34}}$$
u_0 = 1.5 	imes 10^{15} - 0.606 	imes 10^{15} \approx 0.894 	imes 10^{15} \,Hz$.Rounding to the nearest significant value,$
u_0 \approx 9.0 	imes 10^{14} \,Hz$.

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