Energy Levels of the Hydrogen Atom

Introduction

Solving the radial equation for hydrogen yields one of the most celebrated results in science: the energy levels depend on a single integer and fall as 1/n². The formula reproduces the Bohr model's energies but now rests on a proper wave theory. From it we can calculate the ionisation energy of hydrogen, the energies of hydrogen-like ions and, on the next page, the wavelengths of the whole hydrogen spectrum.

Core explanation

The result. Physically acceptable radial solutions, meaning those that stay finite and go to zero at large r, exist only for particular negative energies:

E n = −μe⁴/(32π²ε₀²ħ²n²) = −hcR H/n², n = 1, 2, 3, …

where the Rydberg constant for hydrogen, expressed as a wavenumber, is R H = 109 677 cm⁻¹. The combination hcR H equals 2.179 × 10⁻¹⁸ J, or 13.6 eV. The first few levels are therefore:

n E n / eV --- --- 1 −13.6 2 −3.40 3 −1.51 4 −0.85 ∞ 0

Where quantisation comes from. For most trial energies, the solution of the radial equation grows exponentially at large r and cannot be normalised. Only when the energy takes one of the values E n does the solution decay properly. This is the same logic as the box, where only certain wavelengths fit between the walls; here the "wall" is the requirement of good behaviour at infinity.

Negative energies. The zero of energy corresponds to an electron and a proton infinitely far apart and at rest. A bound electron has less energy than this, so its energy is negative. The more negative the energy, the more tightly bound the electron.

Convergence. Because of the 1/n² dependence, successive levels crowd together as n increases: the gap from n = 1 to n = 2 is 10.2 eV, but from n = 10 to n = 11 it is less than 0.03 eV. The levels converge on the ionisation limit at E = 0. Above it lies the continuum of positive energies, where the electron is free and any energy is allowed.

Ionisation energy. Removing the electron from the ground state (n = 1) to the limit requires

I = 0 − E₁ = hcR H = 2.179 × 10⁻¹⁸ J = 13.6 eV

Multiplying by the Avogadro constant gives 1312 kJ mol⁻¹, in excellent agreement with experiment.

Hydrogen-like ions. For a single electron bound to a nucleus of charge Ze, the energy contains Z²:

E n = −Z²hcR/n²

He⁺ (Z = 2) has a ground-state energy of −54.4 eV, and Li²⁺ (Z = 3) of −122.4 eV. A larger nuclear charge binds the electron far more tightly.

Only n matters. In hydrogen the energy depends on n but not on l or mₗ. For a given n there are n² orbitals of the same energy, a degeneracy peculiar to the exact 1/r potential.

Formulae

E n = −hcR H/n²; hcR H = 2.179 × 10⁻¹⁸ J = 13.6 eV; R H = 109 677 cm⁻¹; hydrogen-like ions: E n = −Z²hcR/n²; ionisation energy from level n: hcR H/n².

Step-by-step reasoning

To find the energy needed to ionise hydrogen from a given level:

1. Identify n for the starting level. 2. Calculate E n = −13.6 eV/n², including Z² for an ion. 3. The ionisation energy from that level is the magnitude of E n, because the final energy is zero. 4. Convert units as required: 1 eV = 1.602 × 10⁻¹⁹ J, and multiply by 6.022 × 10²³ mol⁻¹ for a molar value.

Visual explanation

Draw a vertical energy axis with the Coulomb funnel. Place a line at −13.6 eV near the bottom, another at −3.40 eV, and then lines crowding closer and closer beneath the zero line at the top. Above zero, shade a continuous band to show the continuum of free-electron states.

Real-world analogy

Think of rungs on a ladder leaning into a deep well, where the rungs get closer together the higher you climb. The top of the well is ground level, the ionisation limit. The deepest rung, n = 1, is by far the hardest to climb out from.

Real-world example

Interstellar clouds near hot stars are ionised by ultraviolet photons with energy above 13.6 eV. When electrons recombine with protons they cascade down the energy levels, emitting light at characteristic wavelengths. The glowing red of nebulae such as the Orion Nebula comes largely from electrons dropping from n = 3 to n = 2.

Why?

Why does E vary as 1/n² rather than n² as in a box? In a box the walls are fixed, so higher states simply squeeze more oscillations into the same space. In the Coulomb well, higher states spread to much larger distances, where the potential is weaker and flatter, so their energies bunch up just below zero instead of spreading apart.

Common misconception

"The electron in the n = 1 level has zero energy because it is in the lowest state." The ground state has the most negative energy, −13.6 eV. Zero energy is the ionisation limit, the top of the ladder, not the bottom.

Worked example

Question: Calculate the energy required to ionise a hydrogen atom that is already in the n = 3 level, in joules and in kJ mol⁻¹.

Reasoning: E₃ = −2.179 × 10⁻¹⁸ J/3² = −2.421 × 10⁻¹⁹ J. Ionisation from n = 3 requires +2.421 × 10⁻¹⁹ J per atom. Per mole: 2.421 × 10⁻¹⁹ × 6.022 × 10²³ = 1.458 × 10⁵ J mol⁻¹.

Answer: 2.42 × 10⁻¹⁹ J per atom, or about 146 kJ mol⁻¹, one-ninth of the ground-state value.

Quick check

1. What is the ground-state energy of the Li²⁺ ion in electronvolts, and why is it so much lower than for hydrogen? Answer: It is −9 × 13.6 = −122.4 eV, because the energy scales with Z² and lithium has Z = 3.

Exam focus

Learn E n = −13.6 eV/n² and its Z² extension. Examiners test unit conversions (eV, J, kJ mol⁻¹) and the reason energies are negative. Always state that the ionisation energy equals minus the energy of the starting level.

Advanced insight

The Rydberg constant for an infinitely heavy nucleus, R∞ = 109 737 cm⁻¹, is one of the most precisely known constants in physics. R H is smaller by the factor μ/m e. Comparing R for hydrogen, deuterium and He⁺ allows the reduced-mass correction to be tested directly, and high-precision hydrogen spectroscopy is used to determine the size of the proton.

Summary

Hydrogen's energy levels are E n = −hcR H/n² = −13.6 eV/n², quantised because only these energies give radial wavefunctions that decay at large distances. Energies are negative relative to the ionisation limit, crowd together as n increases and meet a continuum above zero. The ionisation energy is 13.6 eV, or 1312 kJ mol⁻¹. For hydrogen-like ions energies scale as Z², and in hydrogen they depend only on n.

Practice questions

1. Calculate the energy of the n = 5 level of hydrogen in electronvolts. Answer: E₅ = −13.6/25 = −0.544 eV. 2. What is the energy difference between the n = 1 and n = 2 levels? Answer: −3.40 − (−13.6) = 10.2 eV. 3. Calculate the ionisation energy of He⁺ in its ground state. Answer: Z²hcR = 4 × 13.6 = 54.4 eV. 4. How many orbitals share the energy of the n = 3 level of hydrogen? Answer: n² = 9 orbitals: one 3s, three 3p and five 3d.