Rydberg Formula and Spectral Series
For emission from n_i to n_f with n_i > n_f, the wavenumber is R Z^2(1/n_f^2 - 1/n_i^2), subject to the reduced-mass value of R for the species. Lyman, Balmer, Paschen, Brackett, and Pfund have n_f = 1, 2, 3, 4, and 5.
Why this shows up in the exam
Calculating line wavelengths · Comparing spectra of H and He+ · Identifying a series from its lower level
Learn the idea
Hydrogen-like line wavelengths follow the difference of inverse-square quantum numbers. Every named series fixes the lower landing level. Different starting levels then create a family of lines that crowd toward a limit.
🧠 Memory hook: Series name fixes where the electron lands.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- 1/lambda = R Z² (1/n_f² - 1/n_i²) — hydrogen-like Rydberg formula for n_i greater than n_f
- n_f = 1, 2, 3, 4, 5 — Lyman, Balmer, Paschen, Brackett, and Pfund lower levels
How to approach it
- 1Translate the series name into n_f
- 2Identify n_i from the line order
- 3Apply the positive inverse-square difference
Common slip-ups that cost marks
- •Reversing n_i and n_f
- •Forgetting Z squared for hydrogen-like ions
- •Treating every Balmer line as visible without checking wavelength
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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