Hydrogen Atom: Orbitals & Energies - Physical Chemistry 2

SECTION 1

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 2s_1/2 and 2p_1/2 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.

SECTION 2

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 ψ_n,ℓ,m(r, θ, φ), whose squared modulus |ψ|² gives the probability density for finding the electron at a given point in space.

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: E_n = −13.6 eV / n². Higher n means higher energy (less negative) and a more diffuse electron cloud.

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.

Magnetic Quantum Number (mₗ)

Specifies the z-component of angular momentum: L_z = 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).

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 (m_s = ±½), each shell accommodates 2n² spin-orbitals.

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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), ℓ 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.

SECTION 3

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.

SECTION 4

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 ψ_n,ℓ,m(r, θ, φ) = R_n,ℓ(r) × Y_ℓ^m(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 = m_em_p/(m_e + m_p) ≈ m_e), ∇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_ℓ^|m| are the associated Legendre polynomials. The spherical harmonics are simultaneous eigenfunctions of L̂² (eigenvalue ℓ(ℓ+1)ℏ²) and L̂_z (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.

ℹ

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 |R_n,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.

SECTION 5

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 R_n,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	n	ℓ	Radial Nodes	Angular Nodes	Shape
1s	1	0	0	0	Sphere
2s	2	0	1	0	Sphere with one spherical node
2p	2	1	0	1	Dumbbell (two lobes)
3s	3	0	2	0	Sphere with two spherical nodes
3p	3	1	1	1	Dumbbell with one radial node
3d	3	2	0	2	Cloverleaf (four lobes)

Orbital classification by quantum numbers and nodal structure

SECTION 6

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)

1
Step 1 — Write the Energy Expression

The energy of the nth level in hydrogen is E_n = −13.6 eV / n². We need the energies of both the initial and final states.
2
Step 2 — Calculate Individual Level Energies

E₄ = −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
3
Step 3 — Find the Energy of the Emitted Photon

The 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
4
Step 4 — Convert to Wavelength

Using 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 Contextualize

The 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.

SECTION 7

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.

Strengths	Limitations
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.

Comparison of the strengths and limitations of the hydrogen atom solution

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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.

SECTION 8

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.

Feature	Hydrogen (Z = 1)	Multi-Electron Atoms
Hamiltonian	Kinetic + single Coulomb term; separable	Kinetic + nuclear Coulomb + e⁻–e⁻ repulsion; not separable
Energy dependence	Eₙ depends only on n	E depends on both n and ℓ (e.g., E₃s < E₃p < E₃d)
Orbital labels	s, p, d, f from spherical harmonics (exact)	Same labels used within the orbital approximation
Electron spin	Adds mₛ = ±½; total 2n² spin-orbitals per shell	Same; Pauli exclusion principle dictates filling order (Aufbau)
Solution method	Exact analytical	Hartree–Fock (iterative SCF), DFT, CI, CCSD(T), etc.

Hydrogen versus multi-electron atoms: what changes and what persists

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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

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?

Show answer

In hydrogen, the potential is purely Coulombic (V ∝ 1/r), which possesses a hidden dynamical symmetry (conservation of the Laplace–Runge–Lenz vector) beyond the obvious rotational symmetry. This additional symmetry renders the energy dependent only on n, not on ℓ, making all subshells within a shell degenerate. In multi-electron atoms, electron–electron repulsion alters the effective potential from a pure 1/r form. Inner electrons shield the nucleus, so the effective nuclear charge Z_eff experienced by an outer electron depends on how deeply its orbital penetrates the core. s orbitals have significant probability density near the nucleus and experience higher Z_eff than p or d orbitals of the same n, lowering their energy. This penetration and shielding effect lifts the ℓ-degeneracy.

Question 2: BASIC CALCULATION

Calculate the ionization energy of a hydrogen atom in the n = 3 state. Express your answer in eV and in kJ/mol.

Show answer

The energy of the n = 3 level is E₃ = −13.6 eV / 3² = −13.6 / 9 = −1.511 eV. Ionization corresponds to bringing the electron from n = 3 to E = 0 (free electron at rest), so the ionization energy is |E₃| = 1.511 eV. Converting to kJ/mol: 1.511 eV × 96.485 kJ/(mol·eV) = 145.8 kJ/mol. This is considerably less than the ground-state ionization energy of 13.6 eV (1312 kJ/mol), reflecting the loosely bound nature of excited-state electrons.

Question 3: INTERMEDIATE

A hydrogen atom emits a photon with wavelength λ = 1026 nm during a transition from an upper state n_i to a lower state n_f. Identify the initial and final quantum numbers and name the spectral series.

Show answer

First, convert wavelength to energy: E = hc/λ = 1240 eV·nm / 1026 nm = 1.209 eV. Using the Rydberg formula ΔE = 13.6(1/n_f² − 1/n_i²), we try n_f = 1 (Lyman series): 13.6(1 − 1/n_i²) = 1.209, so 1/n_i² = 1 − 1.209/13.6 = 1 − 0.0889 = 0.9111, giving n_i ≈ 1.05 — not an integer. Try n_f = 3 (Paschen series): 13.6(1/9 − 1/n_i²) = 1.209, so 1/n_i² = 1/9 − 1.209/13.6 = 0.1111 − 0.0889 = 0.02222, giving n_i² ≈ 45, n_i ≈ 6.7 — not integer. Try n_f = 1 again more carefully with λ = 1025.7 nm: actually this is the Lyman-β line. 13.6(1 − 1/9) = 13.6 × 8/9 = 12.09 eV → λ = 1240/12.09 = 102.6 nm. The wavelength 1026 nm would be in the infrared, so let us try Paschen: n_f = 3, then 13.6(1/9 − 1/n_i²) = 1.209 eV → 1/n_i² = 0.02222 → n_i ≈ 6.7. Trying n_f = 2 (Balmer): 13.6(1/4 − 1/n_i²) = 1.209 → 1/n_i² = 0.25 − 0.0889 = 0.1611 → n_i ≈ 2.49 — not integer. For λ = 1026 nm, try Brackett (n_f = 4): 13.6(1/16 − 1/n_i²) = 1.209 → 1/n_i² = 0.0625 − 0.0889 < 0, impossible. The correct identification is the Paschen-α line: n_i = 4, n_f = 3. Check: 13.6(1/9 − 1/16) = 13.6 × 7/144 = 0.6611 eV, λ = 1240/0.6611 = 1876 nm. For 1026 nm: E = 1.209 eV, and the correct match is actually the Lyman-β line at 102.6 nm (UV). If the problem intends 1026 Å = 102.6 nm, then n_i = 3, n_f = 1, Lyman series (Lyman-β). This is the Lyman-β transition at 102.6 nm in the ultraviolet.

Question 4: APPLIED

The most probable radius for the 1s orbital of hydrogen is a₀ = 0.529 Å. Using the radial distribution function P(r) = r²|R_1,0(r)|², where R_1,0(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.

Show answer

The radial distribution function is P(r) = r²|R_1,0|² = r² × 4/a₀³ ×

e^{−2r/a₀}

= (4/a₀³) r²

e^{−2r/a₀}

. Differentiating: dP/dr = (4/a₀³)[2r

e^{−2r/a₀}

+ r²(−2/a₀)

e^{−2r/a₀}

] = (4/a₀³) r

e^{−2r/a₀}

[2 − 2r/a₀]. Setting dP/dr = 0 and excluding the trivial solutions r = 0 and r = ∞, we require 2 − 2r/a₀ = 0, which gives r = a₀. Thus the most probable distance from the nucleus in the 1s state is exactly one Bohr radius. This elegant result connects the Bohr model's 'orbit radius' to the quantum-mechanical probability distribution: while the electron has no definite orbit, the Bohr radius marks the distance at which finding it is most likely.

Question 5: CRITICAL 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.

Show answer

(a) For a hydrogenic ion with nuclear charge Z, the energy levels are E_n = −Z² × 13.6 eV / n². For He⁺ (Z = 2): E_n = −4 × 13.6 / n² = −54.4 eV / n². (b) Setting −54.4/n_He² = −13.6/n_H² gives n_He² = 4 n_H², so n_He = 2n_H. Every even-numbered level of He⁺ coincides with a hydrogen level: He⁺ n = 2 matches H n = 1 (both at −13.6 eV), He⁺ n = 4 matches H n = 2 (both at −3.4 eV), and so on. (c) This means that certain spectral transitions of He⁺ produce photons at nearly the same wavelengths as hydrogen transitions (the 'Pickering series' of He⁺ interleaves with the Balmer series of hydrogen). Historically, this led to confusion before Bohr's model clarified the Z² dependence. Small differences arise from the different reduced masses of H and He⁺, providing an early experimental test of the reduced-mass correction.

SUMMARY

Summary & Key Concepts

Summary

The hydrogen atom is the only chemically relevant system for which the Schrödinger equation is solved exactly, yielding wave functions ψ_n,ℓ,m = R_n,ℓ(r) × Y_ℓ^m(θ, φ) that define the atomic orbitals. Three quantum numbers govern these solutions: the principal quantum number n determines the energy E_n = −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.