Bohr atom model

Max Planck and Albert Einstein had already shown, while explaining black-body radiation and the photoelectric effect, that electromagnetic radiation is not emitted or absorbed continuously but in discrete packets ("quanta") of energy $h\nu$, where $\nu$ is the frequency of the radiation and $h=6.626\times10^{-34}$ J s is Planck's constant. Niels Bohr's key move was to extend this same idea of quantisation to the energy of an electron bound inside an atom, producing a model built on four postulates (stated here for the hydrogen atom, but equally applicable to any single-electron species):

1. Quantised energy. The energies of electrons in an atom are quantised - an electron cannot have just any energy, only certain allowed values.

2. Stationary orbits. An electron revolves around the nucleus only in certain fixed circular paths of definite energy, called stationary orbits.

3. Quantised angular momentum. An electron can occupy only those orbits for which its angular momentum $mvr$ is an integer multiple of $h/2\pi$:

Here $n$, later called the principal quantum number, labels which orbit the electron is in.

4. Transitions and photon emission. So long as the electron stays in one of these stationary orbits, it does not lose energy - directly contradicting classical electromagnetism, but resolving the stability problem left open by Rutherford's model. When the electron jumps from a higher-energy orbit ($E_2$) down to a lower-energy orbit ($E_1$), the energy difference is released as a single photon of radiation whose frequency is fixed by

Conversely, an electron absorbs a photon of exactly the right energy to jump from a lower orbit up to a higher one.

Results for a hydrogen-like species. Applying these four postulates to any one-electron ("hydrogen-like") system - hydrogen itself, or an ion such as $\text{He}^+$ or $\text{Li}^{2+}$ that has been stripped down to a single electron - lets you solve for the radius and the energy of the $n$th orbit in closed form:

$E_n=-\left(\frac{1312.8\,Z^2}{n^2}\right)\text{kJ mol}^{-1}\qquad\text{— eq. (2.5)}$ …