The variation of stopping potential (V_0) as | vedclass

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(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = h
u - \phi$where $h$ is Planck's constant,$
u$ i

Vedclass · Accepted Answer

$2.07$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:$eV_0 = h
u - \phi$where $h$ is Planck's constant,$
u$ is the frequency,and $\phi$ is the work function.From the graph,the threshold frequency $
u_0$ (where $V_0 = 0$) is $5 	imes 10^{14} \ Hz$.At the threshold frequency,$h
u_0 = \phi$.Using $h = 6.63 	imes 10^{-34} \ J \cdot s$:$\phi = (6.63 	imes 10^{-34} \ J \cdot s) 	imes (5 	imes 10^{14} \ Hz)$$\phi = 33.15 	imes 10^{-20} \ J$To convert this into electron-volts $(eV)$,divide by the charge of an electron $(1.6 	imes 10^{-19} \ C)$:$\phi = \frac{33.15 	imes 10^{-20}}{1.6 	imes 10^{-19}} \ eV \approx 2.07 \ eV$.

The variation of stopping potential $(V_0)$ as a function of the frequency $\nu \ (\times 10^{14} \ Hz)$ of the incident light for a metal is shown in the figure. The work function of the surface is $........... \ eV$.

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