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### Bohr hydrogen atom approaches classical condition

Question
prove that the Bohr hydrogen atom approaches classical conditions when n becomes very large and small quantum jumps are involved.
Solution
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 $\nu$