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Q.
In a hydrogen atom the total energy of electron is
Atoms
Solution:
The kinetic energy of the electron in hydrogen atom are
$K = \frac{1}{2}mv^{2} $
$= \frac{e^{2}}{8\pi \varepsilon_{0}r} \left[\because v^{2} = \frac{e^{2}}{4\pi\varepsilon_{0} mr}\right]$
Electrostatic potential energy,
$U =\frac{-e^{2}}{4 \pi \varepsilon_{0} r}$
The total energy $E$ of the electron in a hydrogen atom is
$E = K + U, E $
$ = \frac{e^{2}}{8\pi\varepsilon_{0} r} + \left(\frac{-e^{2}}{4\pi\varepsilon_{0} r}\right) $
$ = - \frac{e^{2}}{8\pi \varepsilon_{0} r}$
here negative sign shows that electron is bound to the nucleus.