The Bohr Model and Hydrogen Spectra

Key Ideas

• Quantized energy levels: an electron's energy in hydrogen is E_n = −R_H/n², where R_H = 2.18 × 10⁻¹⁸ J and n = 1, 2, 3, … (n = 1 is the ground state, lowest energy).
• Emission: an electron drops from a higher level (n_high) to a lower level (n_low) and releases a photon of energy ΔE = hν.
• Absorption: an electron absorbs a photon and jumps up to a higher level.
• Line spectra: each element has a unique set of lines (its "fingerprint"), used to identify elements.
• Spectral series: transitions ending at n = 1 (Lyman, UV), n = 2 (Balmer, visible), n = 3 (Paschen, infrared).
• Rydberg equation: relates wavelength to the two levels involved.
• Limitations: Bohr's model works only for one-electron species (H, He⁺, Li²⁺); it cannot explain multi-electron atoms, electron spin, or why certain transitions are more intense.

Equations and Variables

• E_n = −R_H/n² = −2.18 × 10⁻¹⁸ J / n²
• ΔE = hν = E_final − E_initial
• 1/λ = R (1/n_low² − 1/n_high²), with R = 1.097 × 10⁷ m⁻¹

How It Works

In Bohr's picture, an electron in hydrogen is confined to circular orbits with only certain allowed energies. The lowest energy state (n = 1) is the ground state; higher n values are excited states. Because energies are quantized, the electron cannot spiral smoothly — it jumps instantaneously between levels.

When an electron falls from a higher to a lower level, it emits a photon whose energy equals the gap: ΔE = hν = hc/λ. Solving for λ gives the Rydberg formula. Every pair of levels produces one specific wavelength, so a gas of hydrogen heated until its electrons are excited, then allowed to relax, emits a series of discrete lines — not a smear of color.

Absorption is the reverse: a continuous beam passed through cool hydrogen gas has specific wavelengths removed, leaving dark lines against the rainbow. The dark lines sit at exactly the same wavelengths as the bright emission lines, because the same energy gaps are involved.

Worked Example

Calculate the wavelength (in nm) of the photon emitted when a hydrogen electron falls from n = 3 to n = 2 (the red line of the Balmer series).

Step 1 — Use the Rydberg equation with n_low = 2, n_high = 3:

1/λ = (1.097 × 10⁷ m⁻¹)(1/2² − 1/3²) = (1.097 × 10⁷ m⁻¹)(1/4 − 1/9)

1/4 − 1/9 = 0.2500 − 0.1111 = 0.1389

1/λ = (1.097 × 10⁷ m⁻¹)(0.1389) = 1.524 × 10⁶ m⁻¹

Step 2 — Invert: λ = 1 / (1.524 × 10⁶ m⁻¹) = 6.56 × 10⁻⁷ m = 656 nm

This is the famous red Hα line of hydrogen at 656 nm — exactly where the red line appears in hydrogen's visible spectrum.

Check via energy. ΔE = 2.18 × 10⁻¹⁸ J (1/4 − 1/9) = 2.18 × 10⁻¹⁸ × 0.1389 = 3.03 × 10⁻¹⁹ J. Then λ = hc/ΔE = (6.626 × 10⁻³⁴)(3.00 × 10⁸)/(3.03 × 10⁻¹⁹) = 6.56 × 10⁻⁷ m. Same answer.

Common Confusions

• Sign of ΔE. Emission releases energy (ΔE is negative, photon energy is the positive magnitude); absorption is positive. Many students misplace the minus sign.
• Emission vs. absorption spectra. Emission = bright colored lines on dark; absorption = dark lines on a continuous rainbow. Same wavelengths, opposite appearance.
• Which n is "initial"? In the Rydberg form 1/λ = R(1/n_low² − 1/n_high²), n_high is always the larger number, so the bracket is positive.
• Bohr orbits ≠ orbitals. Bohr's circular orbits were replaced by probability-based orbitals in quantum mechanics — don't carry "fixed circular path" language forward.
• R_H vs. R. The energy constant (2.18 × 10⁻¹⁸ J) and the wavelength constant (1.097 × 10⁷ m⁻¹) are different numbers with different units.