PHYSICAL CHEMISTRY 2 • QUANTUM FOUNDATIONS

Hydrogen Atom: Orbitals & Energies

Exact quantum-mechanical solutions for the simplest atom reveal the orbital structure underlying all of chemistry.

Historical Context & Motivation

The hydrogen atom occupies a singular position in the history of physics: it is the only chemically relevant system for which the Schrödinger equation can be solved exactly. Before quantum mechanics, the discrete emission lines of hydrogen — the Balmer series in the visible region, the Lyman series in the ultraviolet — defied explanation by classical electrodynamics. An accelerating electron in a Coulombic orbit should radiate energy continuously and spiral into the nucleus in roughly 10⁻¹¹ seconds, yet atoms are manifestly stable. Resolving this paradox required a fundamentally new theory of matter, and the hydrogen atom served as the proving ground at every stage of its development.

1885
Balmer's Empirical Formula
Johann Balmer discovers that the visible emission wavelengths of hydrogen obey λ = B·n²/(n² − 4), with B = 364.56 nm. Rydberg later generalizes this to all series, introducing the Rydberg constant R.
1913
Bohr's Quantized Orbits
Niels Bohr postulates that electrons occupy stationary orbits with angular momentum quantized in units of ℏ. His model correctly predicts the Rydberg formula but cannot explain fine structure, the Zeeman effect, or multi-electron atoms.
1925
Heisenberg & Schrödinger
Werner Heisenberg formulates matrix mechanics; Erwin Schrödinger independently develops wave mechanics. Schrödinger's equation, applied to the Coulomb potential, yields the exact hydrogen wave functions — the atomic orbitals — together with their energy eigenvalues.
1928
Dirac Equation & Spin
Paul Dirac's relativistic wave equation naturally incorporates electron spin and predicts fine-structure splitting, completing the theoretical picture for hydrogen to spectroscopic precision.
1947
Lamb Shift
Willis Lamb measures a tiny energy difference between the 2s1/2 and 2p1/2 levels, spurring the development of quantum electrodynamics (QED) and revealing that even the 'exact' Schrödinger solution is an approximation to deeper physics.

The central question this lesson addresses is: What are the exact stationary-state wave functions and energies of the hydrogen atom, and how do the quantum numbers n, ℓ, and mₗ govern the spatial shape, angular momentum, and degeneracy of each orbital? Mastering this system is essential because its solutions — the hydrogenic orbitals — form the conceptual and computational basis for understanding multi-electron atoms, molecular orbital theory, and chemical bonding.

Core Principles & Definitions

Solving the hydrogen atom requires casting the time-independent Schrödinger equation in spherical coordinates (r, θ, φ), exploiting the spherical symmetry of the Coulomb potential V(r) = −e²/(4πε₀r). The resulting separation of variables yields three independent differential equations — one radial, two angular — whose solutions are characterized by three quantum numbers. Each allowed combination of these quantum numbers specifies a unique atomic orbital ψn,ℓ,m(r, θ, φ), whose squared modulus |ψ|² gives the probability density for finding the electron at a given point in space.

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Principal Quantum Number (n)

Determines the energy eigenvalue and the overall size of the orbital. Takes positive integer values n = 1, 2, 3, … The energy depends only on n in the non-relativistic treatment: En = −13.6 eV / n². Higher n means higher energy (less negative) and a more diffuse electron cloud.
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Angular Momentum Quantum Number (ℓ)

Governs the magnitude of orbital angular momentum: L = ℏ√[ℓ(ℓ+1)]. Takes integer values from 0 to n − 1. Each value of ℓ corresponds to a subshell labeled s (ℓ=0), p (ℓ=1), d (ℓ=2), f (ℓ=3). The value of ℓ determines the number of angular nodes (ℓ nodal surfaces) and thus the shape of the orbital.
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Magnetic Quantum Number (mₗ)

Specifies the z-component of angular momentum: Lz = mₗℏ. Takes integer values from −ℓ to +ℓ, giving 2ℓ + 1 orientations per subshell. In the absence of an external field all mₗ states within a subshell are degenerate, but an applied magnetic field lifts this degeneracy (the Zeeman effect).
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Degeneracy

For each principal quantum number n, the total number of orbitals is n². This arises because ℓ ranges from 0 to n − 1 and each ℓ level has 2ℓ + 1 values of mₗ. Summing gives Σ(2ℓ+1) = n². Including electron spin (ms = ±½), each shell accommodates 2n² spin-orbitals.
KEY TAKEAWAY
Think of the three quantum numbers as an address system: n tells you which floor of a building you are on (energy shell), identifies the wing of that floor (subshell shape), and mₗ specifies the exact room within the wing (spatial orientation). Together they uniquely label every orbital in the hydrogen atom, while the energy depends only on the floor number — every room on the same floor has the same rent.

Energy Level Diagram

The following diagram illustrates the energy level structure of the hydrogen atom for the first four principal quantum numbers. Notice that within each shell (fixed n), all subshells are degenerate — a unique feature of the pure Coulomb potential. The energy levels converge toward 0 eV as n → ∞, which defines the ionization threshold. Each horizontal line represents a distinct orbital; the number of lines at each n value equals n², confirming the degeneracy pattern discussed above.

Energy level diagram for hydrogen showing shells n = 1 through n = 4. Each horizontal line represents one orbital. The n² degeneracy is visible in the increasing number of lines per shell. Dashed arrows indicate two representative spectral transitions: the Lyman-α line (n = 2 → 1) and the Balmer-Hα line (n = 3 → 2).

Several features of this diagram deserve emphasis. First, the spacings between adjacent levels decrease as 1/n² − 1/(n+1)², which means the levels crowd together as one approaches the ionization continuum. Second, all subshells within a given n are shown at the same height: the 2s and 2p orbitals share the energy E₂ = −3.40 eV. This accidental degeneracy arises from a hidden symmetry of the 1/r potential (related to the conservation of the Laplace–Runge–Lenz vector) and is broken in multi-electron atoms by electron–electron repulsion and penetration effects, which cause s orbitals to drop below p orbitals of the same n.

Mathematical Framework

The hydrogen atom Hamiltonian in the center-of-mass frame consists of the kinetic energy of the electron (with reduced mass μ ≈ mₑ) plus the Coulomb attraction. In spherical coordinates the time-independent Schrödinger equation separates into radial and angular parts, yielding the product form ψn,ℓ,m(r, θ, φ) = Rn,ℓ(r) × Ym(θ, φ). We present the key equations that define this solution.

TIME-INDEPENDENT SCHRÖDINGER EQUATION
Ĥψ = Eψ, where Ĥ = −(ℏ²/2μ)∇² − e²/(4πε₀r)
Here μ is the reduced mass (μ = memp/(me + mp) ≈ me), ∇² is the Laplacian in spherical coordinates, e is the elementary charge, and ε₀ the vacuum permittivity.
ENERGY EIGENVALUES
Eₙ = −(μe⁴)/(2ℏ²(4πε₀)²) × (1/n²) = −13.6 eV / n²
n = 1, 2, 3, … is the principal quantum number. The factor 13.6 eV is one Rydberg of energy (≡ 1 Ry = 2.18 × 10⁻¹⁸ J). The energy depends only on n — not on ℓ or mₗ — giving rise to the n²-fold orbital degeneracy.
RADIAL WAVE FUNCTION
Rₙ,ₗ(r) = −√[(2/(na₀))³ × (n−ℓ−1)! / (2n[(n+ℓ)!]³)] × e^(−r/(na₀)) × (2r/(na₀))^ℓ × L^(2ℓ+1)_(n−ℓ−1)(2r/(na₀))
a₀ = 4πε₀ℏ²/(μe²) ≈ 0.529 Å is the Bohr radius. L denotes the associated Laguerre polynomials. The exponential decay ensures the wave function is normalizable, while the polynomial factor introduces (n − ℓ − 1) radial nodes.
SPHERICAL HARMONICS (ANGULAR PART)
Yₗᵐ(θ, φ) = (−1)ᵐ √[(2ℓ+1)/(4π) × (ℓ−|m|)!/(ℓ+|m|)!] × Pₗ|ᵐ|(cos θ) × eⁱᵐᵠ
P|m| are the associated Legendre polynomials. The spherical harmonics are simultaneous eigenfunctions of L̂² (eigenvalue ℓ(ℓ+1)ℏ²) and L̂z (eigenvalue mₗℏ). They determine the angular shape of orbitals: s orbitals are spherically symmetric (Y₀⁰ = 1/√(4π)), p orbitals have a single nodal plane, and d orbitals have two nodal surfaces.
📐 Radial Probability Distribution
The probability of finding the electron between r and r + dr, regardless of angle, is given by the radial distribution function P(r) = r² |Rn,ℓ(r)|². The extra factor of r² comes from the volume element in spherical coordinates (dV = r² sin θ dr dθ dφ). Consequently, the most probable radius — where P(r) is maximized — is generally not at r = 0, even though the 1s probability density |ψ|² is maximum at the nucleus.

Orbital Shapes & Nodal Structure

The three-dimensional shape of an orbital is dictated by its angular wave function (spherical harmonic) and modulated in extent by the radial wave function. Two types of nodes — surfaces where ψ = 0 — organize this structure. Angular nodes are planes or cones determined by the spherical harmonics; there are ℓ of them. Radial nodes are concentric spheres where the radial function Rn,ℓ passes through zero; there are (n − ℓ − 1) of them. The total number of nodes is always n − 1.

Cross-sectional representations of 1s, 2s, 2p, and 3d orbitals. Radial nodes appear as dashed circles (visible in the 2s orbital), while angular nodes appear as dashed lines through the nucleus (visible in 2p and 3d). Different colors denote regions of opposite sign of the wave function. The lower panel summarizes the node counting rules.
Orbital classification by quantum numbers and nodal structure
OrbitalnRadial NodesAngular NodesShape
1s1000Sphere
2s2010Sphere with one spherical node
2p2101Dumbbell (two lobes)
3s3020Sphere with two spherical nodes
3p3111Dumbbell with one radial node
3d3202Cloverleaf (four lobes)

Worked Example: Photon Emission in Hydrogen

A hydrogen atom initially in the n = 4 state undergoes a transition to the n = 2 state, emitting a single photon. We wish to determine the energy, wavelength, and spectral series of the emitted photon, and to identify the region of the electromagnetic spectrum in which it falls.

Photon Emission: n = 4 → n = 2 (Balmer Series)
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Step 1 — Write the Energy ExpressionThe energy of the nth level in hydrogen is En = −13.6 eV / n². We need the energies of both the initial and final states.
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Step 2 — Calculate Individual Level EnergiesE₄ = −13.6 eV / 4² = −13.6 eV / 16 = −0.850 eV. Similarly, E₂ = −13.6 eV / 2² = −13.6 eV / 4 = −3.400 eV.
E₄ = −0.850 eV, E₂ = −3.400 eV
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Step 3 — Find the Energy of the Emitted PhotonThe photon carries away the energy difference: ΔE = E₄ − E₂ = (−0.850) − (−3.400) = +2.550 eV. The positive sign confirms energy is released (emission). This transition terminates on n = 2, placing it in the Balmer series. Specifically, this is the Hβ line (the second line in the Balmer series, with the upper level n = 4).
ΔE = 2.550 eV
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Step 4 — Convert to WavelengthUsing E = hc/λ, we solve for λ = hc/ΔE. With hc = 1240 eV·nm: λ = 1240 eV·nm / 2.550 eV = 486.3 nm. Alternatively, using the Rydberg equation: 1/λ = R(1/2² − 1/4²) = 1.097 × 10⁷ m⁻¹ × (1/4 − 1/16) = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹, giving λ = 486.1 nm (the small difference arises from rounding).
λ ≈ 486 nm — visible blue-green light (Hβ line)
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Step 5 — Verify and ContextualizeThe wavelength 486 nm falls squarely in the visible spectrum, consistent with the Balmer series (which ranges from 365 nm at the series limit to 656 nm for Hα). This is one of the lines observed by Balmer himself and gives hydrogen discharge tubes their characteristic bluish glow. Note that the quantum numbers ℓ and mₗ of the initial and final orbitals do not affect the photon energy because all subshells within a given n are degenerate in hydrogen; however, electric-dipole selection rules require Δℓ = ±1 for the transition to be allowed.

Strengths & Limitations of the Hydrogenic Model

The exact solution for the hydrogen atom is both a triumph of quantum mechanics and, at the same time, a highly idealized model. Understanding where this model excels and where it breaks down is essential for extending these ideas to real chemical systems.

Comparison of the strengths and limitations of the hydrogen atom solution
StrengthsLimitations
Exact analytical solution — no approximations needed for the non-relativistic single-electron problem.Cannot be solved exactly for multi-electron atoms; electron–electron repulsion terms destroy separability.
Predicts the hydrogen spectrum with extraordinary accuracy (energy levels agree with experiment to parts per million before relativistic/QED corrections).Neglects relativistic effects (spin-orbit coupling, Darwin term, mass-velocity correction) that produce fine structure splitting of ~10⁻⁴ eV.
Provides the orbital basis (s, p, d, f labels) and quantum number framework used throughout chemistry and spectroscopy.The ℓ-degeneracy (2s = 2p in energy) is unique to the 1/r potential and does not hold in multi-electron atoms, where penetration and shielding lift this degeneracy.
Hydrogenic wave functions serve as starting points for variational and perturbation calculations on larger atoms (Slater-type orbitals, STO basis sets).Ignores nuclear structure (finite nuclear size) and quantum electrodynamic effects (Lamb shift ~10⁻⁶ eV), which become relevant at high spectroscopic resolution.
KEY TAKEAWAY
The hydrogen atom is to quantum chemistry what the simple harmonic oscillator is to classical mechanics: a solvable reference system whose solutions — the atomic orbitals — provide the language, intuition, and approximate wave functions for tackling far more complex problems. Just as you might model a complicated potential near its minimum as a parabola, chemists model multi-electron atoms by starting with hydrogenic orbitals and then adding corrections (shielding, exchange, correlation) layer by layer.

Connection to Multi-Electron Atoms & Advanced Theory

Moving from hydrogen to helium and beyond introduces electron–electron repulsion, making the Schrödinger equation analytically unsolvable. The strategies developed to handle this — the orbital approximation, the self-consistent field (SCF) method, and post-Hartree–Fock methods — all build upon the hydrogenic orbital framework. The following table contrasts the hydrogen atom with the general multi-electron case to highlight what changes and what is preserved.

Hydrogen versus multi-electron atoms: what changes and what persists
FeatureHydrogen (Z = 1)Multi-Electron Atoms
HamiltonianKinetic + single Coulomb term; separableKinetic + nuclear Coulomb + e⁻–e⁻ repulsion; not separable
Energy dependenceEₙ depends only on nE depends on both n and ℓ (e.g., E₃s < E₃p < E₃d)
Orbital labelss, p, d, f from spherical harmonics (exact)Same labels used within the orbital approximation
Electron spinAdds mₛ = ±½; total 2n² spin-orbitals per shellSame; Pauli exclusion principle dictates filling order (Aufbau)
Solution methodExact analyticalHartree–Fock (iterative SCF), DFT, CI, CCSD(T), etc.

The most consequential difference is the lifting of ℓ-degeneracy. In multi-electron atoms, s electrons penetrate closer to the nucleus than p electrons of the same n and therefore experience a larger effective nuclear charge. This penetration effect lowers the energy of s relative to p, and p relative to d, producing the familiar ordering (1s < 2s < 2p < 3s < 3p < 4s ≈ 3d …) that governs the periodic table. Relativistic effects — particularly spin-orbit coupling — further split levels in heavy atoms, revealing structure invisible in the non-relativistic hydrogen solution. Courses in advanced quantum chemistry and molecular spectroscopy build directly on the foundation established here.

Practice Problems

PROBLEM 1CONCEPTUAL
Explain why the 3s, 3p, and 3d orbitals of hydrogen are all degenerate (same energy), whereas in multi-electron atoms the 3s orbital lies lower in energy than 3p, which in turn lies lower than 3d. What physical effect is responsible for lifting the degeneracy?
PROBLEM 2BASIC CALCULATION
Calculate the ionization energy of a hydrogen atom in the n = 3 state. Express your answer in eV and in kJ/mol.
PROBLEM 3INTERMEDIATE
A hydrogen atom emits a photon with wavelength λ = 1026 nm during a transition from an upper state ni to a lower state nf. Identify the initial and final quantum numbers and name the spectral series.
PROBLEM 4APPLIED
The most probable radius for the 1s orbital of hydrogen is a₀ = 0.529 Å. Using the radial distribution function P(r) = r²|R1,0(r)|², where R1,0(r) = 2(1/a₀)^(3/2) e^(−r/a₀), verify that the maximum of P(r) occurs at r = a₀ by taking the derivative and setting it to zero.
PROBLEM 5CRITICAL THINKING
The He⁺ ion is a hydrogenic (one-electron) system with nuclear charge Z = 2. (a) Write the expression for the energy levels of He⁺ in terms of the hydrogen energy formula. (b) Determine whether any energy level of He⁺ coincides exactly with an energy level of neutral hydrogen, and if so, identify which levels. (c) Discuss the physical significance of this coincidence for spectroscopic observations.

Summary & Key Concepts

The hydrogen atom is the only chemically relevant system for which the Schrödinger equation is solved exactly, yielding wave functions ψn,ℓ,m = Rn,ℓ(r) × Ym(θ, φ) that define the atomic orbitals. Three quantum numbers govern these solutions: the principal quantum number n determines the energy En = −13.6 eV/n² and overall size; the angular momentum quantum number ℓ controls the orbital shape (s, p, d, f) and contributes ℓ angular nodes; and the magnetic quantum number mₗ specifies the spatial orientation and z-component of angular momentum.

The total number of nodes in any orbital is n − 1, partitioned into (n − ℓ − 1) radial nodes and ℓ angular nodes. The n²-fold degeneracy of each shell is a unique feature of the pure Coulomb potential, broken in multi-electron atoms by penetration and shielding effects. Spectral transitions between levels produce discrete emission and absorption lines — the Lyman, Balmer, and Paschen series — whose wavelengths are given by the Rydberg formula. The hydrogenic orbitals remain the foundation for all atomic and molecular electronic structure calculations, from Hartree–Fock theory to modern density functional approaches.

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