The Rydberg formula
1/λ = R × Z² × (1/n₁² − 1/n₂²), with R = 1.097 × 10⁷ m⁻¹. The energy of the photon is ΔE = 13.6 × Z² × (1/n₁² − 1/n₂²) eV, and λ (nm) ≈ 1240 ÷ ΔE (eV).
Hydrogen's spectral series
| Series | n₁ | Region | Longest λ (nm) | Shortest λ (nm) |
|---|---|---|---|---|
| Lyman | 1 | Ultraviolet | 122 | 91 |
| Balmer | 2 | Visible | 656 | 365 |
| Paschen | 3 | Infrared | 1,875 | 820 |
| Brackett | 4 | Infrared | 4,051 | 1,459 |
| Pfund | 5 | Infrared | 7,458 | 2,279 |
The longest wavelength in a series is the first line (n₂ = n₁ + 1); the shortest is the series limit (n₂ = ∞). Only the Balmer series is visible: Hα is red (656 nm), Hβ blue-green (486 nm), Hγ and Hδ violet (434 and 410 nm).
How many lines?
An electron falling from level n to the ground state can give n(n − 1) ÷ 2 different lines, because it can drop in any combination of steps. From n = 4 that's 6 lines. A single atom gives at most n − 1 lines at once.
To find the radius and energy of each level, use the Bohr model calculator.
Questions students ask
What is the wavelength of the first line of the Balmer series?
About 656 nm (red), from n = 3 to n = 2.
What is the shortest wavelength in the Lyman series?
About 91.2 nm, the series limit, when n₂ = ∞. It's 1/R.
Which series of hydrogen lies in the visible region?
The Balmer series, where electrons fall to n = 2.
How many spectral lines are possible from n = 5 to the ground state?
5 × 4 ÷ 2 = 10 lines.