Uncertainty relation results in rejection of semi classic Bohr model for hydrogen atom

Show that the uncertainty relation forces us to reject the semi classical Bohr model for the hydrogen atom
In Bohr atom model we deal with the electron as a classical particle. The allowed orbits are defined by the quantization rules:
The radius r of a circular orbit and the momentum $p = mv$ of the rotating electron must satisfy $pr = n\hbar (n = 1,2,3,....)$. To consider an electron’s model in classical terms , the uncertainties in its position and momentum must be negligible when compared to  $r$ and $p$. In other words $\Delta x \ll r$ and $\Delta p \ll p$
This implies
$\frac{{\Delta x}}{r}\frac{{\Delta p}}{p} \ll 1$                     (1)
On the other hand , the uncertainty relation imposes
$\frac{{\Delta x}}{r}\frac{{\Delta p}}{p} \ge \frac{\hbar }{{rp}} \Rightarrow \frac{{\Delta x\Delta p}}{{rp}} \ge \frac{1}{n}$                           (2)
So, equation 1 is incompatible with 2 , unless $n \gg 1$

Bohr hydrogen atom approaches classical condition

 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
$R=\frac{2\pi ^{2}me^{4}}{h^{3}c}$
\({E_k} = \frac{{ch}}{{n_k^2}}R\)
\({E_i} = \frac{{ch}}{{n_i^2}}R\)
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
\({f_n} = \frac{{{\nu _n}}}{{2\pi {r_n}}} = \frac{{4{\pi ^2}m{e^4}}}{{{n^3}{h^3}}}\)
\({f_n} = \frac{{2cR}}{{{n^3}}}\), which is identical to \(\nu \)

Popular Posts