The Bohr Model of Hydrogen

The Core Idea: Quantised Orbits

Imagine an electron orbiting a proton like a planet around the sun, held by electrostatic attraction rather than gravity. Classically, an accelerating charge should radiate energy and spiral into the nucleus. Bohr proposed a radical fix: the electron can only exist in specific, non-radiating circular paths. These paths are defined by a rule: the electron’s angular momentum must be an integer multiple of the reduced Planck constant. This simple constraint forces the electron into a set of discrete "shells," each with a fixed radius and a specific energy. The ground state is the smallest orbit with the lowest energy; excited states are larger orbits with higher (less negative) energies. Transitions between these shells release or absorb photons with energies exactly equal to the difference between the levels.

Deriving the Energy Levels

To find the allowed energies, we balance the centripetal force required for circular motion with the Coulomb attraction. For an electron of mass \(m_e\) and charge \(-e\) orbiting a nucleus of charge \(+Ze\) at radius \(r\), the force balance is \( \frac{m_e v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r^2} \). This gives the kinetic energy \( K = \frac{1}{2}m_e v^2 = \frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r} \). The potential energy is \( U = -\frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} \). The total energy is \( E = K + U = -\frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r} \).

Bohr’s quantisation condition states that angular momentum \( L = m_e v r = n\hbar \), where \( n \) is a positive integer. Solving the force balance and quantisation equations simultaneously for \( r \) yields the radius of the \( n \)-th orbit: \( r_n = \frac{n^2}{Z} a_0 \), where \( a_0 \) is the Bohr radius. Substituting this radius back into the total energy expression gives the famous result for the energy levels of a hydrogen-like atom.

For hydrogen, \( Z=1 \). The negative sign indicates that the electron is bound to the nucleus; energy must be supplied to remove it (ionisation). The energy depends on \( n^2 \), meaning the gap between \( n=1 \) and \( n=2 \) is much larger than between \( n=2 \) and \( n=3 \).

The Bohr Radius and Atomic Scale

The size of the atom is set by the Bohr radius \( a_0 \). This constant arises naturally from the combination of fundamental constants: the permittivity of free space \( \varepsilon_0 \), Planck’s constant \( h \), the electron mass \( m_e \), and the elementary charge \( e \). It is defined as \( a_0 = \frac{\varepsilon_0 h^2}{\pi m_e e^2} \). Numerically, \( a_0 \approx 0.529\,\text{Å} \) (or \( 5.29 \times 10^{-11}\,\mathrm{m} \)).

The radius of the \( n \)-th orbit scales as \( n^2 \). The ground state (\( n=1 \)) has a radius of \( a_0 \). The first excited state (\( n=2 \)) has a radius of \( 4a_0 \). This rapid expansion explains why atoms are mostly empty space. For multi-electron atoms, the effective nuclear charge \( Z_{\mathrm{eff}} \) is less than \( Z \) due to shielding, but the \( n^2 \) dependence remains a key feature of atomic structure.

Connecting to Spectra

When an electron transitions from a higher energy level \( n_i \) to a lower level \( n_f \), it emits a photon. The energy of this photon is the difference between the two levels: \( \Delta E = E_{n_i} - E_{n_f} \). Using the energy formula, \( \Delta E = 13.6\,Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV} \). Since \( E = h\nu \), the frequency of the emitted light is \( \nu = \frac{\Delta E}{h} \).

This expression is mathematically identical to the Rydberg formula derived empirically in the previous lecture. The Rydberg constant \( R_H \) can be expressed in terms of fundamental constants as \( R_H = \frac{m_e e^4}{8\varepsilon_0^2 h^3 c} \). The Bohr model thus provides a theoretical derivation for the Rydberg constant, linking the discrete spectral lines directly to the quantisation of angular momentum.