The Principal Quantum Number n

What is a quantum number?

When the Schrödinger equation is solved for a hydrogen atom — a single proton bound to a single electron — the solutions are not arbitrary. The requirement that the wavefunction remain normalizable forces certain quantities to take only discrete values. These discrete labels are the quantum numbers. The most fundamental of them is the principal quantum number $n$ .

Allowed values of n

The principal quantum number must satisfy

$n = 1, 2, 3, \ldots$

There is no upper limit in principle, but practical bound states have finite $n$ . The value $n = 1$ is the ground state; $n = 2, 3, \ldots$ are successively higher excited states. The atom is ionised (the electron escapes) when $n \to \infty$ .

The energy formula

Solving the Schrödinger equation for the hydrogen atom yields energies that depend on $n$ alone:

$E_n = -\frac{m_e e^4}{2(4\pi\epsilon_0)^2\hbar^2}\,\frac{1}{n^2}.$

The combination of constants in front evaluates to exactly $13.6\,\text{eV}$ , so in practice we write

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

The negative sign signals a bound state: the electron is trapped by the proton. Energy must be added to the atom to reach higher $n$ , and to ionise it completely one must supply at least $13.6\,\text{eV}$ from the ground state.

Worked examples

Ground state ( $n = 1$ ). $E_1 = -\frac{13.6}{1^2}\,\text{eV} = -13.6\,\text{eV}.$ This is the lowest energy the hydrogen atom can have.

First excited state ( $n = 2$ ). $E_2 = -\frac{13.6}{4}\,\text{eV} = -3.4\,\text{eV}.$ An electron in this state holds $3.4\,\text{eV}$ of binding energy.

Ionisation from the ground state. To free the electron from $n = 1$ , the atom must absorb $E_\infty - E_1 = 0 - (-13.6) = 13.6\,\text{eV}$ . This is the ionisation energy of hydrogen, measured experimentally as $13.598\,\text{eV}$ , confirming the theory.

Spectral lines from n

When an electron transitions from a higher level $n_i$ to a lower level $n_f$ , the atom emits a photon whose energy equals the difference:

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

This is the Rydberg formula. The visible Balmer series (the spectral lines in the visible range) all terminate at $n_f = 2$ ; the first line ( $n_i = 3 \to n_f = 2$ ) appears red at $656\,\text{nm}$ and corresponds to $\Delta E = 13.6(1/4 - 1/9)\,\text{eV} \approx 1.89\,\text{eV}$ .

Why only n matters for the energy (in hydrogen)

In a pure $1/r$ Coulomb potential, the energy turns out to depend exclusively on $n$ . The other quantum numbers ( $\ell$ and $m_\ell$ ) affect the shape of the wavefunction but — in the absence of spin-orbit coupling and external fields — not the energy. This special $n$ -degeneracy is a mathematical consequence of the hidden $SO(4)$ symmetry of the Kepler problem, and it is a distinctive feature of the hydrogen atom that does not hold for multi-electron atoms.

Try it

This is a numerical exercise — return a number. Use the formula $E_n = -13.6\,\text{eV}/n^2$ to find the energy of the $n = 3$ level of hydrogen. Express your answer in eV.

Run your code to see the quantum state.