Physics · Atomic and quantum physics · Grade 11

Orbitals of the hydrogen atom

Where is the electron in an atom, and where do the shapes of orbitals come from?

The situation

In chemistry lessons you draw orbitals as “spheres” and “dumbbells”. But an electron does not circle inside the atom along an orbit like a planet — it is a probability “cloud”. The red light of a hydrogen lamp (656 nm) is emitted when an electron passes from one cloud shape to another. Where do the shapes of these clouds come from, and why do they have “node” surfaces where the electron is never found?

Topic

Orbitals of the hydrogen atom. Electron cloud — 3D shapes

Key ideas

Wave function
Quantum numbers
Nodal surfaces

Formulas

ψ_{n l m} (r, θ, φ) = R_{n l} (r) Y_{l m} (θ, φ)

The exact solution of the Schrödinger equation for the hydrogen atom: a radial part and a spherical harmonic.

E_{n} = - \frac{13.6 eV}{n^{2}}

The energy depends only on the principal quantum number $n$ : there are $n^{2}$ orbitals in one level.

P (r) = r^{2} R_{n l}^{2} (r), r_{1 s} = a_{0} = 0.529 Å

The probability of finding the electron in a thin shell at distance $r$ ; for 1s the maximum is at the Bohr radius.

N_{radial} = n - l - 1, N_{angular} = l

The number of nodal surfaces: spheres and planes (cones).

λ = \frac{h c}{E_{n_{2}} - E_{n_{1}}}

A superposition of two states “oscillates” at frequency $Δ E / h$ and emits a photon of the corresponding wavelength.

Theory

The probability cloud

The wave function $ψ$ itself cannot be measured, but $| ψ |^{2} d V$ is the probability of finding the electron in a volume $d V$ . Each dot in the scene is one imagined measurement result: where the dots are dense, the electron is found more often. The dots are refreshed continuously — each measurement gives a new position. The colour shows the sign (phase) of $ψ$ : blue is positive, red is negative. In a chemical bond it is precisely the way these signs add or subtract that matters.

Quantum numbers

n = 1, 2, 3… determines the energy and the size of the cloud ( $⟨ r ⟩ = \frac{a_{0}}{2} [3 n^{2} - l (l + 1)]$ ). l = 0…n−1 gives the shape: s, p, d, f. m = −l…l gives the orientation. The “real” orbitals in the scene ( $p_{x}$ , $d_{x y}$ , …) are combinations of the complex $\pm m$ states — these are exactly the shapes shown in chemistry textbooks. In hydrogen all orbitals with the same $n$ have the same energy.

Nodes

Surfaces where $ψ = 0$ are nodes: the electron is never found there. The number of radial nodes (spheres) is $n - l - 1$ , and the number of angular nodes (planes and cones) is $l$ . For example, in $2 p_{z}$ the plane $z = 0$ is a node. In 1s, $| ψ |^{2}$ is greatest at the nucleus, but $P (r) = 4 π r^{2} | ψ |^{2}$ — because the volume of a shell grows as $r^{2}$ — is greatest at $r = a_{0}$ . Use the “Half section” to look inside the cloud.

The central field and the radial equation

The Coulomb field of the nucleus is centrally symmetric — the potential depends only on $r$ . So the Schrödinger equation separates into variables: $ψ = R_{n l} (r) Y_{l m} (θ, φ)$ . The radial part obeys a one-dimensional equation in which an effective potential appears: $V_{e f} = - \frac{e^{2}}{4 π ε_{0} r} + \frac{ℏ^{2} l (l + 1)}{2 m r^{2}}$ . The second term is a “centrifugal barrier”: for $l > 0$ the electron is almost never near the nucleus ( $R \propto r^{l}$ ), whereas s orbitals can reach the nucleus itself. Finite solutions exist only for $E_{n} = - 13.6 / n^{2}$ eV; in one level $n$ , $n^{2}$ orbitals have the same energy — this is degeneracy ( $2 n^{2}$ with spin). Removing the electron completely from the ground state — ionisation — takes 13.6 eV, and from level $n$ it takes $13.6 / n^{2}$ eV. For hydrogen-like ions (He⁺, Li²⁺) $E_{n} = - 13.6 Z^{2} / n^{2}$ eV and the cloud is $Z$ times smaller (the Pickering series — in the “Atomic structure and spectra” experiment). The nucleus moves too: instead of the electron mass the reduced mass $μ = m_{e} M / (m_{e} + M)$ is used — which is why the deuterium lines are shorter in wavelength than those of hydrogen by ≈ 0.03 %; H. Urey discovered deuterium in late 1931 from exactly this shift.

Angular momentum and magnetic moment

$l$ and $m$ are eigenvalues of the angular momentum operators: ${\hat{L}}^{2} Y_{l m} = l (l + 1) ℏ^{2} Y_{l m}$ , ${\hat{L}}_{z} Y_{l m} = m ℏ Y_{l m}$ ; the spherical functions $Y_{l m}$ are the eigenfunctions of these operators. Therefore $| L | = \sqrt{l (l + 1)} ℏ$ — for $l > 0$ this is never equal to $l ℏ$ : the direction of the vector cannot be completely definite. In a complex state with $m \neq 0$ the probability current circulates about the $z$ axis — this is a current inside the atom — and it produces a magnetic moment: $μ_{z} = - m μ_{B}$ , $μ_{B} = e ℏ / 2 m_{e} = 9.27 \cdot 10^{- 24}$ J/T (the Bohr magneton); the orbital gyromagnetic ratio is $μ / L = e / 2 m_{e}$ . The real orbitals in the scene, such as $p_{x}$ and $d_{x y}$ , are linear combinations (sum or difference) of the $+ m$ and $- m$ states: they have no circulating current, a measurement of $L_{z}$ gives $+ | m | ℏ$ or $- | m | ℏ$ with equal probability, while $L^{2}$ is exactly defined. Electron spin, the total angular momentum and fine structure (the Dirac formula, the Lamb shift) are in the “Spin, the Zeeman effect and MRI” experiment.

Superposition and the model

If two states are added, $| ψ |^{2}$ oscillates at frequency $ν = Δ E / h$ : the cloud “sloshes” and, like an antenna, emits a photon with $λ = h c / Δ E$ (1s + 2p: 121.6 nm, Lyman-α; 2p + 3d: 656 nm, red H-α). In 2s + 2p the energies are equal — the cloud does not oscillate (an sp hybrid). Model: the proton mass is taken as infinite ( $E_{1} = - 13.6057$ eV); spin and fine structure are ignored. The dots are sampled exactly from $| ψ |^{2}$ (the radius by inverse-transform sampling, the direction by rejection sampling). Time is slowed down by a factor of $5 \cdot 10^{15}$ .

Did you know?

Erwin Schrödinger published his wave equation in 1926 and in his very first paper applied it to the hydrogen atom, reproducing the energy levels of the Bohr model.
The Bohr radius $a_{0} \approx 0.529$ Å $= 0.0529$ nm is the most probable distance of the electron from the nucleus in the ground state of the hydrogen atom.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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