Hydrogen Atom

The hydrogen atom is the only atom solved exactly in quantum mechanics. Its energy levels depend on a single quantum number n, while the full wave function requires three quantum numbers (n, ℓ, mℓ) plus spin.

Key Concepts

Energy levels: Eₙ = -13.6/n² eV
Quantum numbers: n (principal), ℓ (angular momentum), mℓ (magnetic), ms (spin)
Bohr radius: a₀ = 0.0529 nm
Spectral series: Lyman (n→1), Balmer (n→2), Paschen (n→3)
Degeneracy: each level n has n² spatial states (2n² with spin)

Key Equations

Energy levels

E_n = -\frac{13.6\text{ eV}}{n^2}

Bohr radius

a_0 = \frac{\hbar^2}{me^2/(4\pi\epsilon_0)} \approx 0.0529\text{ nm}

Photon energy (transition)

h\nu = E_{n_i} - E_{n_f} = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\text{ eV}

Rydberg formula

\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

Worked Example

Example Problem

Problem

Find the wavelength of light emitted in the n=3 → n=2 transition (Hα line).

Solution

ΔE = 13.6(1/4 - 1/9) = 13.6×5/36 = 1.889 eV. λ = hc/ΔE = 1240 nm·eV/1.889 eV = 656 nm (red, Balmer α).

Practice

Exercises

7 problems

1 of 7

Click on the n = 2 orbital in the Bohr model diagram below to read off its energy. What is E₂ for hydrogen in eV?

Click on the n = 2 orbit (or its electron) to read off E₂. The energy scale on the right panel shows all levels. Then enter E₂ in eV.

E₂ = eV

2 of 7

Use the energy level diagram to find the ionization energy of hydrogen — the energy needed to remove the electron from the n=1 ground state.

The diagram shows hydrogen energy levels. The red arrow marks the ionization transition n=1 → free. Press Launch to animate the electron escaping, then enter the ionization energy in eV.

ΔE = eV

3 of 7

Find the wavelength of the n=4→n=2 (Hβ) Balmer transition in nm.