Editing Bohr model (section)

=== Derivation ===
In classical mechanics, if an electron is orbiting around an atom with period T, and if its coupling to the electromagnetic field is weak, so that the orbit doesn't decay very much in one cycle, it will emit electromagnetic radiation in a pattern repeating at every period, so that the Fourier transform of the pattern will only have frequencies which are multiples of 1/T.

However, in quantum mechanics, the quantization of angular momentum leads to discrete energy levels of the orbits, and the emitted frequencies are quantized according to the energy differences between these levels. This discrete nature of energy levels introduces a fundamental departure from the classical radiation law, giving rise to distinct spectral lines in the emitted radiation.

Bohr assumes that the electron is circling the nucleus in an elliptical orbit obeying the rules of classical mechanics, but with no loss of radiation due to the [[Larmor formula]].

Denoting the total energy as ''E'', the negative electron charge as ''e'', the positive nucleus charge as ''K=Z|e|'', the electron mass as ''m<sub>e</sub>'', half the major axis of the ellipse as ''a'', he starts with these equations:<ref name="bohr1"/>{{rp|3}}

<div class="center"><math>
\nu=\frac{\sqrt2}{\pi}\frac{\vert E \vert^{\frac{3}{2}}}{\vert e \vert K\sqrt m_e} \ \ \ (1a)
</math></div>

<div class="center"><math>
2a=\frac{\vert e \vert K}{\vert E \vert}\ \ \ (1b)
</math></div>

''E'' is assumed to be negative, because a positive energy is required to unbind the electron from the nucleus and put it at rest at an infinite distance.

Eq. (1a) is obtained from equating the centripetal force to the Coulombian force acting between the nucleus and the electron, considering that <math> E = T + U </math> (where ''T'' is the average kinetic energy and ''U'' the average electrostatic potential), and that for Kepler's second law, the average separation between the electron and the nucleus is ''a''.

Eq. (1b) is obtained from the same premises of eq. (1a) plus the virial theorem, stating that, for an elliptical orbit,

<div class="center"><math>
T = -\frac{1}{2}U\ \ \ (1c).
</math></div>

Then Bohr assumes that <math>\vert E \vert</math> is an integer multiple of the energy of a quantum of light with half the frequency of the electron's revolution frequency,<ref name="bohr1"/>{{rp|4}} i.e.:

<div class="center"><math>
\vert E \vert = nh\frac{\nu}{2}\ \ \ (2).
</math></div>

From eq. (1a,1b,2), it descends:
<div class="center"><math>
\vert E \vert = \frac{2\pi^2m_ee^2K^2}{n^2h^2}\ \ \ (3a)
</math></div>
<div class="center"><math>
\nu = \frac{4\pi^2m_ee^2K^2}{n^3h^3}\ \ \ (3b)
</math></div>
<div class="center"><math>
2a = \frac{n^2h^2}{2\pi^2 m_e \vert e \vert K}\ \ \ (3c).
</math></div>

He further assumes that the orbit is circular, i.e. <math> a = r </math>, and, denoting the angular momentum of the electron as ''L'', introduces the equation:

<div class="center"><math>
\pi L = \frac{T}{\nu}\ \ \ (4).
</math></div>

Eq. (4) stems from the [[virial theorem]], and from the classical mechanics relationships between the angular momentum, the kinetic energy and the frequency of revolution.

From eq. (1c,2,4), it stems:

<div class="center"><math>
L = nL_o,
</math></div>

where:

<div class="center"><math>
L_0 = \frac{h}{2\pi} = \hbar,
</math></div>

that is:

<div class="center"><math>
L = n\hbar.
</math></div>

This results states that the angular momentum of the electron is an integer multiple of the reduced Planck constant.<ref name="bohr1"/>{{rp|15}}

Substituting the expression for the velocity gives an equation for ''r'' in terms of ''n'':
:<math> m_{\text{e}}\sqrt{\dfrac{k_{\text{e}}Ze^2}{m_{\text{e}}r}}r = n\hbar,</math>
so that the allowed orbit radius at any ''n'' is
:<math> r_n = \frac{n^2\hbar^2}{Zk_\mathrm{e} e^2 m_\mathrm{e}}.</math>

The smallest possible value of ''r'' in the hydrogen atom ({{nowrap|''Z'' {{=}} 1}}) is called the Bohr radius and is equal to:
:<math>r_1 = \frac{\hbar^2}{k_\mathrm{e} e^2 m_\mathrm{e}} \approx 5.29 \times 10^{-11}~\mathrm{m}= 52.9~\mathrm{pm} .</math>

The energy of the ''n''-th level for any atom is determined by the radius and quantum number:
:<math> E = -\frac{Zk_\mathrm{e} e^2}{2r_n} = -\frac{Z^2(k_\mathrm{e} e^2)^2 m_\mathrm{e}}{2\hbar^2 n^2} \approx \frac{-13.6Z^2}{n^2}~\mathrm{eV}.</math>

An electron in the lowest energy level of hydrogen ({{nowrap|''n'' {{=}} 1}}) therefore has about 13.6&nbsp;[[electronvolt|eV]] less energy than a motionless electron infinitely far from the nucleus. The next energy level ({{nowrap|''n'' {{=}} 2}}) is −3.4&nbsp;eV. The third ({{nowrap|''n'' {{=}} }}3) is −1.51&nbsp;eV, and so on. For larger values of ''n'', these are also the binding energies of a highly excited atom with one electron in a large circular orbit around the rest of the atom. The hydrogen formula also coincides with the [[Wallis product]].<ref>{{Cite magazine |date=November 17, 2015 |title=Revealing the hidden connection between pi and Bohr's hydrogen model |url=http://physicsworld.com/cws/article/news/2015/nov/17/revealing-the-hidden-connection-between-pi-and-bohrs-hydrogen-model |magazine=[[Physics World]]}}</ref>

The combination of natural constants in the energy formula is called the Rydberg energy (''R''<sub>E</sub>):
:<math> R_\mathrm{E} = \frac{ (k_\mathrm{e} e^2)^2 m_\mathrm{e}}{2 \hbar^2}.</math>

This expression is clarified by interpreting it in combinations that form more [[natural units]]:
: <math>m_\mathrm{e} c^2 </math> is the [[rest mass energy]] of the electron (511&nbsp;keV),
: <math>\frac{k_\mathrm{e} e^2}{\hbar c} = \alpha \approx \frac{1}{137} </math> is the [[fine-structure constant]],
: <math>R_\mathrm{E} = \frac{1}{2} (m_\mathrm{e} c^2) \alpha^2</math>.

Since this derivation is with the assumption that the nucleus is orbited by one electron, we can generalize this result by letting the nucleus have a charge {{nowrap|1=''q'' = ''Ze''}}, where ''Z'' is the atomic number. This will now give us energy levels for hydrogenic (hydrogen-like) atoms, which can serve as a rough order-of-magnitude approximation of the actual energy levels. So for nuclei with ''Z'' protons, the energy levels are (to a rough approximation):
: <math> E_n = -\frac{Z^2 R_\mathrm{E}}{n^2}.</math>

The actual energy levels cannot be solved analytically for more than one electron (see [[n-body problem|''n''-body problem]]) because the electrons are not only affected by the [[atomic nucleus|nucleus]] but also interact with each other via the [[Coulomb force]].

When ''Z'' = 1/''α'' ({{nowrap|''Z'' ≈ 137}}), the motion becomes highly relativistic, and ''Z''<sup>2</sup> cancels the ''α''<sup>2</sup> in ''R''; the orbit energy begins to be comparable to rest energy. Sufficiently large nuclei, if they were stable, would reduce their charge by creating a bound electron from the vacuum, ejecting the positron to infinity. This is the theoretical phenomenon of electromagnetic charge screening which predicts a maximum nuclear charge. Emission of such positrons has been observed in the collisions of heavy ions to create temporary super-heavy nuclei.<ref>{{Cite journal |last1=Müller |first1=U. |last2=de Reus |first2=T. |last3=Reinhardt |first3=J. |last4=Müller |first4=B. |last5=Greiner |first5=W. |date=1988-03-01 |title=Positron production in crossed beams of bare uranium nuclei |url=http://nbn-resolving.de/urn/resolver.pl?urn:nbn:de:hebis:30-29640 |journal=Physical Review A |volume=37 |pages=1449–1455 |bibcode=1988PhRvA..37.1449M |doi=10.1103/PhysRevA.37.1449 |pmid=9899816 |s2cid=35364965 |number=5}}</ref>

The Bohr formula properly uses the [[reduced mass]] of electron and proton in all situations, instead of the mass of the electron,
:<math>m_\text{red} = \frac{m_\mathrm{e} m_\mathrm{p}}{m_\mathrm{e} + m_\mathrm{p}} = m_\mathrm{e} \frac{1}{1 + m_\mathrm{e}/m_\mathrm{p}}.</math>
However, these numbers are very nearly the same, due to the much larger mass of the proton, about 1836.1 times the mass of the electron, so that the reduced mass in the system is the mass of the electron multiplied by the constant 1836.1/(1+1836.1) = 0.99946. This fact was historically important in convincing Rutherford of the importance of Bohr's model, for it explained the fact that the frequencies of lines in the spectra for singly ionized helium do not differ from those of hydrogen by a factor of exactly 4, but rather by 4 times the ratio of the reduced mass for the hydrogen vs. the helium systems, which was much closer to the experimental ratio than exactly 4.

For positronium, the formula uses the reduced mass also, but in this case, it is exactly the electron mass divided by 2. For any value of the radius, the electron and the positron are each moving at half the speed around their common center of mass, and each has only one fourth the kinetic energy. The total kinetic energy is half what it would be for a single electron moving around a heavy nucleus.
:<math> E_n = \frac{R_\mathrm{E}}{2 n^2}</math> (positronium).