V (stopping potential) is plotted against fra | vedclass

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(C) According to Einstein's photoelectric equation:$K_{max} = \frac{hc}{\lambda} - \phi_0$Since $K_{max} = eV$,where $V$ is the stopping potential:$eV

Vedclass · Accepted Answer

$	heta_1 = 	heta_2$,for any two metals.

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(C) According to Einstein's photoelectric equation:$K_{max} = \frac{hc}{\lambda} - \phi_0$Since $K_{max} = eV$,where $V$ is the stopping potential:$eV = \frac{hc}{\lambda} - \phi_0$$V = \left( \frac{hc}{e} ight) \frac{1}{\lambda} - \frac{\phi_0}{e}$Comparing this with the equation of a straight line $y = mx + c$,the slope $m = \frac{hc}{e}$.Since $h$,$c$,and $e$ are universal constants,the slope of the $V$ vs $\frac{1}{\lambda}$ graph is constant for all metals.Therefore,the angle $	heta$ made by the line with the $\frac{1}{\lambda}$ axis is the same for all metals,i.e.,$	heta_1 = 	heta_2$.

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

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