Explain the formulas for the energy of an electron in an atom revolving around the nucleus in different orbits.

(A) According to the Bohr model of the atom,an electron revolves around a positively charged nucleus in stable orbits. The electrostatic force of attraction $F_{e}$ between the electron and the nucleus provides the necessary centripetal force $F_{c}$ to keep the electron in its orbit.
For a stable orbit in a hydrogen atom:
$F_{e} = F_{c}$
$\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r^{2}} = \frac{m v^{2}}{r}$
From this,we get the radius relation: $r = \frac{e^{2}}{4 \pi \epsilon_{0} m v^{2}} \quad \dots (1)$
The kinetic energy $K$ of the electron is:
$K = \frac{1}{2} m v^{2} = \frac{e^{2}}{8 \pi \epsilon_{0} r} \quad \dots (2)$
The potential energy $U$ of the electron in the field of the nucleus ($Z=1$ for hydrogen) is:
$U = -\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r} \quad \dots (3)$
The total energy $E$ is the sum of kinetic and potential energy:
$E = K + U = \frac{e^{2}}{8 \pi \epsilon_{0} r} - \frac{e^{2}}{4 \pi \epsilon_{0} r} = -\frac{e^{2}}{8 \pi \epsilon_{0} r}$
The negative sign indicates that the electron is bound to the nucleus.