Lesson 3 · 25 min

The quantum model: orbitals and quantum numbers

Electrons are waves too. Rotate hydrogen orbitals computed from the Schrödinger equation, count their nodes, and learn what n, l, mₗ and mₛ each control.

In this lesson

Use the de Broglie relation to explain why electrons are waves and why standing waves quantise energy
State what the quantum numbers n, l, mₗ and mₛ mean and which values they take, and judge whether a set is allowed
Describe the shapes of s, p, d and f orbitals and count their radial and angular nodes
Read a radial distribution plot: where the most probable radius is, and why 2s has an inner bump

Bohr pictured the electron as a tiny planet circling the nucleus and simply declared that only certain orbits were “allowed”. The model got hydrogen’s lines right but failed as soon as it met helium — and it could not say why only those orbits were allowed. The answer came from an idea that sounds absurd: the electron is also a wave.

Electrons are waves too

In 1924 de Broglie argued that if light is both wave and particle, particles such as electrons should also have a wavelength — the larger the momentum, the shorter the wavelength. Three years later Davisson and Germer fired electrons at a nickel crystal and saw the diffraction pattern only a wave can make.

A wave confined to a finite space can only form standing waves: a guitar string sounds its fundamental and whole-number overtones, nothing in between. An electron wave around a nucleus is no different — only states that close on themselves with a whole number of wavelengths survive. Quantisation is not a rule imposed from outside; it is what confinement does to waves. Strikingly, the de Broglie wavelength of the electron in Bohr’s $n = 1$ orbit is exactly that orbit’s circumference, $2\pi a_0 = 0.332$ nm.

\lambda = \frac{h}{mv}

For everyday objects

mv

is so large that the wavelength is unmeasurably small; only for light particles such as electrons does the wave nature show.

$\lambda$: de Broglie wavelength
$h$: Planck’s constant
$m$: particle mass (electron 9.109 × 10⁻³¹ kg)
$v$: speed

In 1926 Schrödinger wrote down the equation the electron wave obeys. Its solution $\psi$ (the wavefunction) cannot be observed directly, but $|\psi|^2$ gives the probability density of finding the electron at each point. By the uncertainty principle the electron has no definite path, so “orbital” takes on a new meaning: an atomic orbital is a wavefunction describing where the electron may be found, not a route it travels.

Four quantum numbers

Four quantum numbers describe each electron; the first three fix an orbital
Symbol	Name	Allowed values	What it sets
n	principal	1, 2, 3, …	the shell; energy and size
l	angular momentum	0, 1, …, n − 1	subshell and shape: s, p, d, f
mₗ	magnetic	−l, …, 0, …, +l	orientation in space
mₛ	spin	+½, −½	direction of the electron’s spin

Source: P. Atkins, J. de Paula & J. Keeler, Atkins’ Physical Chemistry, 11th ed., Topic 8A.

N_\mathrm{orb}(l) = 2l + 1 \qquad N_\mathrm{orb}(n) = \sum_{l=0}^{n-1}(2l + 1) = n^2 \qquad N_\mathrm{e}(n) = 2n^2

For example

n = 3

l = 0, 1, 2

, giving

1 + 3 + 5 = 9

orbitals and at most 18 electrons.

$N_\mathrm{orb}$: number of orbitals (in a subshell or a whole shell)
$N_\mathrm{e}$: most electrons a shell can hold
$n,\ l$: principal and angular quantum numbers

Predict first

How many nodes (surfaces where the wavefunction is zero, radial plus angular) does a 3p orbital have?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

0 1 2 3

Orbital shapes and nodes

s orbitals ( $l = 0$ ) are spherical; 1s, 2s and 3s nest inside one another, separated by spherical radial nodes.
p orbitals ( $l = 1$ ) are three “dumbbells” along x, y and z, with a nodal plane between the lobes.
d orbitals ( $l = 2$ ) are mostly four-leaf clovers; $d_{z^2}$ is two lobes along z plus a “waist ring”, and its nodes are a pair of cones.
f orbitals ( $l = 3$ ) have three angular nodes and even more intricate shapes.

A node is where $\psi$ changes sign. The sign of the wavefunction (often drawn in two colours) cannot itself be observed, but when orbitals on two atoms overlap to form a bond, like signs add and opposite signs cancel — and the character of the chemical bond depends on exactly that.

N_r = n - l - 1 \qquad N_a = l \qquad N_r + N_a = n - 1

$N_r$: radial nodes: spheres at fixed distances from the nucleus
$N_a$: angular nodes: planes or cones through the nucleus

Hydrogen wavefunctions · quantum numbers · nodes

Orbital explorer

Principal n

Angular l

Orbital

View

Loading 3D view…

Drag to rotate · scroll or pinch to zoom · arrow keys rotate too

+where ψ > 0 (orange, “+”)
−where ψ < 0 (blue, “−”)

Orbital: $2\mathrm{p}_{z}$
Quantum numbers: n = 2 · l = 1 · mₗ = 0
Radial nodes n − l − 1: 0
Angular nodes l: 1the xy plane
Energy (hydrogen): E₂ = −3.400 eV set by n alone: every orbital in the shell has it
Most probable radius: 4.00 a₀ = 212 pm
Mean radius ⟨r⟩: 5.0 a₀ = 265 pm

Shell n = 2: l = 0, 1, so 4 orbitals holding up to 8 electrons

Radial distribution P(r) = r²R² · Shading shows the sign of R(r)

Drag to rotate

2p_z

and see why its lobes have opposite colours; switch to the cross-section and compare the hollow rings (radial nodes) of 1s, 2s and 3s; then find the cone-shaped nodes of

3d_{z^2}

. How does the peak of the radial distribution move as n grows?

The radial distribution $P(r) = r^2R^2$ answers “how far from the nucleus?”. For 1s the peak sits exactly at the Bohr radius $a_0 = 52.9$ pm; for orbitals with $l = n - 1$ (1s, 2p, 3d, 4f) it falls at $n^2a_0$ — Bohr’s orbit radii return as probability peaks. Besides its main outer peak, 2s has a small bump close to the nucleus: part of the time a 2s electron “penetrates” inside the 1s electrons and feels more nuclear charge — which is why 2s lies below 2p in many-electron atoms.

E_n(\mathrm{H}) = -\frac{13.6\ \mathrm{eV}}{n^2}

In hydrogen the energy depends on n alone: 2s and 2p are degenerate, as are 3s, 3p and 3d. In many-electron atoms it also depends on l.

$E_n$: energy of shell n (zero at the ionisation limit)

Worked example

Describe the 4d orbitals

For the 4d subshell, give n, l and the values of mₗ, then count the orbitals, the maximum number of electrons and the nodes of each orbital.

Concept check

Which set of quantum numbers $(n, l, m_l, m_s)$ is impossible?

(2, 1, 0, +½) (3, 2, −2, −½) (2, 2, 1, +½) (4, 0, 0, −½)

Try the calculation

How many electrons can the n = 4 shell hold at most?

Try the calculation

An electron moves at $1.0 \times 10^{6}$ m/s. What is its de Broglie wavelength in nm? ( $h = 6.626 \times 10^{-34}$ J·s, $m_e = 9.109 \times 10^{-31}$ kg)

Model limits Where this model stops

The wavefunctions here are exact solutions of the Schrödinger equation for hydrogen (one electron). Many-electron atoms have no exact orbitals: each electron moves in the average field of the others, so the shapes carry over but energies also depend on l — which is why 4s fills before 3d (next lesson).
Spin comes from relativistic quantum mechanics (the Dirac equation); here it simply enters as a fourth quantum number.
Real orbitals such as $p_x$ and $d_{xy}$ combine the complex wavefunctions with $m_l = \pm1$ and $\pm2$ ; orbitals with a definite $m_l$ are actually ring-shaped.
The point cloud is a deterministic sample of $|\psi|^2$ with a finite number of dots, so its edge looks sharper than it is: an electron cloud has no true boundary.

Key takeaways

Electrons behave as waves ( $\lambda = h/mv$ ); a bound electron wave can only form standing waves, so its energy is quantised.
An orbital is a wavefunction ψ and $|\psi|^2$ is the probability density of finding the electron; there is no definite path.
Quantum numbers: n = 1, 2, …; l = 0…n − 1 (s p d f); mₗ = −l…+l; mₛ = ±½. Shell n has n² orbitals and holds up to 2n² electrons.
Nodes: n − l − 1 radial and l angular, n − 1 in all; the most probable 1s radius is a₀ = 52.9 pm.

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Data and references

L. de Broglie, Recherches sur la théorie des quanta, thesis, Paris (1924); Ann. Phys. (Paris) 3, 22 (1925).
C. Davisson & L. H. Germer, “Diffraction of Electrons by a Crystal of Nickel”, Phys. Rev. 30, 705 (1927).
E. Schrödinger, “Quantisierung als Eigenwertproblem”, Ann. Phys. 79, 361 (1926).
D. J. Griffiths & D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge (2018), §4.2 (hydrogen wavefunctions).
P. Atkins, J. de Paula & J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford (2018), Topics 8A–8B.
E. Tiesinga et al., CODATA Recommended Values of the Fundamental Physical Constants: 2018, Rev. Mod. Phys. 93, 025010 (2021).