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Here our problem is to prove that the Bohr hydrogen atom approaches classical conditions when n becomes very large and small quantum jumps are involved.

To prove this let us compute the frequency of a photon that is emitted in the transition between the adjacent state $n_{k}=n$ and $n_{i}=n-1$ when $n\gg 1$.

we define Rydberg’s constant as

$R=\frac{2\pi ^{2}me^{4}}{h^{3}c}$

So,

\({E_k} = \frac{{ch}}{{n_k^2}}R\)

and

\({E_i} = \frac{{ch}}{{n_i^2}}R\)

Therefore the frequency of the emitted photon is

\(\nu = \frac{{n_k^2 – n_i^2}}{{n_k^2n_i^2}}cR = \frac{{\left( {{n_k} + {n_i}} \right)\left( {{n_k} – {n_i}} \right)}}{{n_k^2n_i^2}}cR\)

\({{n_k} – {n_i} = 1}\) , so for $n\gg 1$ we have

\({n_k} + {n_i} \cong 2n\) and \(n_k^2n_i^2 \cong {n^4}\)

Therefore ,

\(\nu = \frac{{2cR}}{{{n^3}}}\)

According to classical theory of electromagnetism , a rotating charge with a frequency $f$ will emit a radiation of frequency $f$. On the other hand , using the Bohr hydrogen model , the orbital frequency of the electron around the nucleus is

\({f_n} = \frac{{{\nu _n}}}{{2\pi {r_n}}} = \frac{{4{\pi ^2}m{e^4}}}{{{n^3}{h^3}}}\)

or

\({f_n} = \frac{{2cR}}{{{n^3}}}\), which is identical to \(\nu \)