The relation between the stopping potential ( | vedclass

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(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E._{\max})$ of an emitted photoelectron is given by: $K.E._{\max} =

Vedclass · Accepted Answer

$V_0 = \frac{hv}{e} - \frac{\phi}{e}$

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(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E._{\max})$ of an emitted photoelectron is given by: $K.E._{\max} = hv - \phi$ where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\phi$ is the work function of the metal. Since the stopping potential $(V_0)$ is related to the maximum kinetic energy by the equation $K.E._{\max} = eV_0$, we can substitute this into the photoelectric equation: $eV_0 = hv - \phi$ Dividing both sides by the charge of an electron $(e)$, we get: $V_0 = \frac{hv}{e} - \frac{\phi}{e}$ This equation represents a linear relationship between the stopping potential $(V_0)$ and the frequency $(v)$.

The relation between the stopping potential $(V_0)$ and frequency $(v)$ is correctly represented in $[\phi = \text{Work function}]$

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