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Bohr's Energy Levels

Going on, we can calculate the net energy (kinetic plus potential) of an electron in the $n^{\rm th}$ Bohr orbital of the H atom:

$$E_n \; = \; {p_n^2 \over 2 m} \; - \; {1 \over 4 \pi \epsilon_\circ} {e^2 \over r_n}$$

or [again using Eq. (8) to substitute $n \hbar / r_n$ for p_n]

$$E_n \; = \; {n^2 \hbar^2 \over 2 m r_n^2} \; - \; {1 \over 4 \pi \epsilon_\circ} {e^2 \over r_n} .$$

Now we replace r_n with our expression (13) to get

$$E_n \; = \; {n^2 \hbar^2 \over 2 m} \left( m e^2 \over 4 \p . . . . . . r \left( 4 \pi \epsilon_\circ \right)^2 n^2 \hbar^2 \right]$$

which simplifies to

$$E_n \; = \; - \left(1 \over 4 \pi \epsilon_\circ \right)^2 { m e^4 \over n^2 \hbar^2 } \; = \; - {E_0 \over n^2}$$ (24.15)

where $E_0 = 2.18 \times 10^{-18}$ J = 13.6055 eV (where 1 eV = $1.60219 \times 10^{-19}$ J). We have thus reproduced Bohr's explanation for the empirical formulae of Balmer and Rydberg! Note that whereas the energy of confinement of a particle in a box increases as n² (where n-1 is the number of nodes inside the box), the Bohr energy levels of an atom increase as -1/n² (they get less negative and closer together as n increases). Of course, so far all these calculations have been done in the classical (nonrelativistic) limit. If the momenta get big enough (p comparable to or greater than mc) we have to do our calculations differently . . . .