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(B) According to Einstein's photoelectric equation,the kinetic energy of emitted photoelectrons is given by $K_{max} = h
u - \phi_0$,where $\phi_0$ i

Vedclass · Accepted Answer

$h/e$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the kinetic energy of emitted photoelectrons is given by $K_{max} = h
u - \phi_0$,where $\phi_0$ is the work function.Since $K_{max} = eV_0$,where $V_0$ is the stopping potential,we can write $eV_0 = h
u - \phi_0$.Rearranging this for the stopping potential,we get $V_0 = \frac{h}{e}
u - \frac{\phi_0}{e}$.This equation is in the form of a straight line $y = mx + c$,where $y = V_0$,$x = 
u$,and the slope $m = \frac{h}{e}$.Therefore,the slope of the graph between the stopping potential $(V_0)$ and the frequency of incident light $(
u)$ is $h/e$.

The slope of the graph between the frequency of incident light and the stopping potential for a given surface is:

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