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

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(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.

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