Playground

Hydrogen energy level diagram with transition arrows and photon wavelength.

Variables

SymbolNameSIDimensionRange
EnE_nEnergy of level noutput
Bound-state energy of electron in level n
JM·L²·T⁻²-2.18e-18 – 0
nnPrincipal quantum number
Energy level index
dimensionless11 – 20

Deep dive

Derivation
Bohr postulated quantized angular momentum L = nℏ for circular orbits. Combining centripetal force = Coulomb attraction (m_e v²/r = ke²/r²) with L = m_e v r = nℏ yields r_n = n²a_0 (a_0 ≈ 0.529 Å) and E_n = −ke²/(2r_n) = −13.6 eV/n². The full Schrödinger solution in a Coulomb potential reproduces the same energies.
Experimental verification
Hydrogen Balmer series (n→2) measured by Ångström and Balmer in the 1880s; fits 1/λ = R(1/n_f² − 1/n_i²) with R = 1.097×10⁷ m⁻¹ to 8 decimal places. Lyman, Paschen, Brackett, and Pfund series confirmed the formula. Spectral lines of stars are how we know the universe is made mostly of hydrogen.
Common misconceptions
  • Electrons do not literally orbit the nucleus — quantum mechanics gives probability clouds, not classical paths.
  • The Bohr model fails for atoms with more than one electron (ignores electron-electron repulsion).
  • Negative energy means *bound*, not negative kinetic energy. Ionization sets E = 0.
Real-world applications
  • Stellar spectroscopy — chemical composition and redshift of stars and galaxies.
  • Hydrogen masers and atomic clocks.
  • Lasers (e.g., HeNe, hydrogen plasma lines).
  • Astrophysical 21 cm hydrogen line — used to map galactic structure.

Worked examples

Energy of n = 2 level

Given:
n:
2
Find: E_2
Solution

E_2 = −13.6/4 = −3.4 eV

n=2 → n=1 transition wavelength (Lyman α)

Given:
n_i:
2
n_f:
1
Find: λ
Solution

ΔE = 10.2 eV → λ = 1240/10.2 ≈ 121.6 nm (UV)

Scenarios

What if…
  • scenario:
    What if the nucleus had Z = 2 (helium ion He⁺)?
    answer:
    E_n = −13.6 × Z²/n² = −54.4/n² eV. Ground state is 4× deeper, ionization energy 54.4 eV — observable in stellar coronas.
  • scenario:
    What if the electron were replaced by a muon (200× heavier)?
    answer:
    Muonic hydrogen has E_n scaled by m_μ/m_e ≈ 207 → ground state ≈ −2.8 keV, and Bohr radius ~200× smaller. Used to probe proton charge radius.
  • scenario:
    What if n were not restricted to integers?
    answer:
    The classical orbits would form a continuum, atoms would be unstable, and the sharp spectral lines we observe would be impossible — quantization is essential.
Limiting cases
  • condition:
    n = 1
    result:
    E_1 = −13.6 eV
    explanation:
    Ground state — most tightly bound; ionization energy of hydrogen.
  • condition:
    n → ∞
    result:
    E_n → 0
    explanation:
    Continuum threshold — electron is just barely free (ionization).
  • condition:
    Z > 1 (hydrogenic ions)
    result:
    E_n = −13.6 Z²/n² eV
    explanation:
    Higher nuclear charge binds the electron Z² more tightly (e.g., He⁺: −54.4 eV ground state).

Context

Niels Bohr · 1913

Bohr postulated quantized orbits to explain hydrogen's discrete spectrum, ushering in old quantum theory.

Hook

Why are atomic spectra made of sharp lines instead of a rainbow?

Find the energy of the n=2 level of hydrogen and the wavelength of the n=2 → n=1 transition.

Dimensions: [E_n] = [m_e e⁴]/[ε_0² h²] → energy ✓ (Rydberg constant has units of inverse length when expressed via 1/λ)
Validity: Exact for one-electron hydrogenic atoms (H, He⁺, Li²⁺, ...) ignoring fine structure. Multi-electron atoms require quantum mechanics with electron-electron interactions; relativistic corrections are needed for heavy nuclei.

Related formulas

Photoelectric EffectSchrödinger Equation (Time-Independent)