In a photoelectric experiment, the slope of t | vedclass

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

Vedclass · Accepted Answer

$4.125 	imes 10^{-15} 	ext{ JsC}^{-1}$

Vedclass · Answer

(C) According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h
u - \phi_0$, where $h$ is Planck's constant, $
u$ is the frequency of incident radiation, and $\phi_0$ is the work function.Since $K_{max} = eV_s$, where $e$ is the charge of an electron and $V_s$ is the stopping potential, we can write:$eV_s = h
u - \phi_0$$V_s = (\frac{h}{e})
u - \frac{\phi_0}{e}$This equation is in the form of a straight line $y = mx + c$, where $y = V_s$, $x = 
u$, and the slope $m = \frac{h}{e}$.Given $h = 6.6 	imes 10^{-34} 	ext{ Js}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$.Slope $m = \frac{6.6 	imes 10^{-34}}{1.6 	imes 10^{-19}} = 4.125 	imes 10^{-15} 	ext{ JsC}^{-1}$.Thus, the correct option is $C$.

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

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