In a photoelectric experiment,a graph is draw | vedclass

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Question

(B) The photoelectric equation is given by $eV_s = h
u - \phi$,where $V_s$ is the stopping potential,$h$ is Planck's constant,$
u$ is the frequency,

Vedclass · Accepted Answer

$\frac{e}{h}$

Vedclass · Answer

(B) The photoelectric equation is given by $eV_s = h
u - \phi$,where $V_s$ is the stopping potential,$h$ is Planck's constant,$
u$ is the frequency,and $\phi$ is the work function.Rearranging the equation for $V_s$ as a function of $
u$: $V_s = (\frac{h}{e})
u - \frac{\phi}{e}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = V_s$ and $x = 
u$,the slope of the graph with respect to the $X$-axis is $m = \frac{h}{e}$.However,the question specifies that the angle $	heta$ is made with the $Y$-axis. For a line $y = mx + c$,the slope $m = 	an(\alpha)$ where $\alpha$ is the angle with the $X$-axis. The angle $	heta$ with the $Y$-axis is related to the angle $\alpha$ with the $X$-axis by $	heta = 90^\circ - \alpha$.Therefore,$	an 	heta = 	an(90^\circ - \alpha) = \cot \alpha = \frac{1}{	an \alpha} = \frac{1}{h/e} = \frac{e}{h}$.

In a photoelectric experiment,a graph is drawn with stopping potential along $Y$-axis and the frequency of the incident light along $X$-axis. If the graph is a straight line which makes an angle $\theta$ with $Y$-axis,then $\tan \theta=$ ($h$-Planck's constant,$e$-charge of electron).

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