The Bohr Model

By 1913 the nuclear atom was established but deeply puzzling: classical electrodynamics predicted that an electron orbiting a proton should radiate energy continuously and spiral inward in nanoseconds. Niels Bohr cut through the contradiction with two bold postulates that mixed classical mechanics with an entirely new quantization condition.

Bohr's postulates

1. Stationary states. An electron can occupy only certain allowed orbits without radiating. The allowed orbits are those for which the angular momentum $L$ is an integer multiple of $\hbar = h / (2\pi)$:

$L = mvr = n\hbar, \qquad n = 1, 2, 3, \ldots$

2. Quantum jumps. When an electron transitions between two stationary states, a single photon is emitted or absorbed whose energy equals the difference in orbital energies: $E_\text{photon} = |E_{n_i} - E_{n_f}|$.

Deriving the Bohr radii

Consider a circular orbit of radius $r$. The Coulomb attraction provides the centripetal force:

$\frac{ke^2}{r^2} = \frac{mv^2}{r},$

where $k = 1/(4\pi\varepsilon_0)$. Solving for $v^2$ gives $v^2 = ke^2/(mr)$. Substituting into the quantization condition $mvr = n\hbar$ yields $v = n\hbar/(mr)$, so

$\frac{n^2\hbar^2}{m^2 r^2} = \frac{ke^2}{mr}.$

Solving for $r$:

$r_n = \frac{n^2 \hbar^2}{mke^2} = n^2 a_0,$

where the Bohr radius is defined as

$a_0 = \frac{\hbar^2}{mke^2} \approx 0.0529\,\text{nm}.$

The first orbit ($n = 1$) has radius $a_0$; the second ($n = 2$) has radius $4a_0$; and in general $r_n$ grows as $n^2$.

Deriving the energy levels

The total mechanical energy is the sum of kinetic and potential energy. The potential energy of the electron–proton pair separated by $r$ is $U = -ke^2/r$. Using the circular-orbit relation $ke^2/r = mv^2$, the kinetic energy is $K = mv^2/2 = ke^2/(2r)$. Therefore

$E = K + U = \frac{ke^2}{2r} - \frac{ke^2}{r} = -\frac{ke^2}{2r}.$

Substituting $r = r_n = n^2 a_0$:

$E_n = -\frac{ke^2}{2a_0} \cdot \frac{1}{n^2}.$

The combination $ke^2/(2a_0)$ evaluates numerically to $13.6\,\text{eV}$, giving the celebrated Bohr energy formula:

$\boxed{E_n = -\frac{13.6\,\text{eV}}{n^2}.}$

The ground state ($n = 1$) has energy $-13.6\,\text{eV}$; the first excited state ($n = 2$) has energy $-3.4\,\text{eV}$; and excited states approach zero as $n \to \infty$ (ionization).

Emission and absorption spectra

When an electron drops from level $n_i$ to level $n_f$ (with $n_i > n_f$), the emitted photon carries energy

$E_\text{photon} = E_{n_i} - E_{n_f} = 13.6\,\text{eV}\!\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).$

For the Balmer series ($n_f = 2$, $n_i = 3, 4, 5, \ldots$), the photons fall in the visible range, producing the characteristic red, blue-green, and violet lines of hydrogen that spectroscopists had measured since 1885. The Bohr model explained the entire Balmer series and predicted the ultraviolet Lyman series ($n_f = 1$) and infrared Paschen series ($n_f = 3$) before they were fully mapped, a striking predictive success.

Try it

This is a numerical exercise — return a number. Using Bohr's formula $E_n = -13.6\,\text{eV} / n^2$, compute the energy in eV of the $n = 2$ level of hydrogen.

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