Write the slope of the eV_0 - nu graph. | vedclass

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

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h

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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ 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 $K_{max} = eV_0$, where $e$ is the charge of an electron and $V_0$ is the stopping potential, we can write the equation as $eV_0 = h
u - \Phi_0$.Rearranging this to the form $y = mx + c$, we get $V_0 = (h/e)
u - (\Phi_0/e)$.However, for the $eV_0 - 
u$ graph, the equation is $eV_0 = h
u - \Phi_0$.Comparing this with $y = mx + c$, where $y = eV_0$ and $x = 
u$, the slope $m$ is equal to $h$.

Write the slope of the $eV_0 - \nu$ graph.

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