If the electron of a hydrogen atom is present | vedclass

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(C) The kinetic energy $(KE)$ of an electron in an orbit is given by $KE = \frac{1}{2} mv^2$. From the electrostatic force balance, $\frac{mv^2}{r} =

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$\frac{-e^2}{2r}$

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(C) The kinetic energy $(KE)$ of an electron in an orbit is given by $KE = \frac{1}{2} mv^2$. From the electrostatic force balance, $\frac{mv^2}{r} = \frac{e^2}{r^2}$, which implies $mv^2 = \frac{e^2}{r}$. Substituting this into the kinetic energy expression, we get $KE = \frac{1}{2} \frac{e^2}{r}$. The potential energy $(PE)$ of the electron is $PE = \frac{-e^2}{r}$. The total energy $(E)$ is the sum of kinetic and potential energy: $E = KE + PE = \frac{1}{2} \frac{e^2}{r} - \frac{e^2}{r} = \frac{-e^2}{2r}$.

If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is

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