Questions: Rydberg Constant and Spectroscopic Line Formula

Question 1

A hydrogen-like ion He⁺ (Z = 2) undergoes an electron transition from n = 3 to n = 2. Compared to the same transition in neutral hydrogen, what happens to the wavelength of the emitted photon?

• A. The wavelength is identical — the transition n = 3 → 2 is fixed by quantum numbers, independent of the element
• B. The wavelength is longer — helium is heavier and its electrons move more slowly
• C. The wavelength is shorter — He⁺ has Z = 2, replacing R with R·Z² = 4R, giving a larger 1/λ and thus a smaller wavelength
• D. The wavelength cannot be predicted by the Rydberg formula because He⁺ is not hydrogen

Answer: (C)

For hydrogen-like ions, the Rydberg formula becomes 1/λ = R·Z²(1/n₁² − 1/n₂²). With Z = 2, the effective Rydberg constant is four times larger. For the n = 3 → 2 transition: 1/λ = 4R(1/4 − 1/9) = 4R · 5/36 = 5R/9, compared to 1/λ = R · 5/36 for hydrogen. The photon from He⁺ has 4× the inverse wavelength and thus 1/4 the wavelength — much shorter than the corresponding hydrogen line. The intuition: Z = 2 binds the electron tighter, compressing all energy levels by Z², which increases all transition energies and shifts spectral lines to shorter wavelengths.

Question 2

The Balmer series was discovered before the Lyman or Paschen series. Which explanation best accounts for this historical fact?

• A. The Balmer series has more spectral lines than the other series, making it easier to detect statistically
• B. The Balmer series corresponds to the highest-energy transitions and was thus most easily detected by early equipment
• C. The Balmer series (n₁ = 2) falls in or near visible light, making it directly accessible to early spectroscopists using optical instruments and to astronomers observing starlight without infrared or ultraviolet detectors
• D. Balmer personally measured all lines in his series; the others were theoretically predicted before being observed

Answer: (C)

The named spectral series correspond to different choices of the final level n₁. Lyman (n₁ = 1) emits in the ultraviolet, invisible to the naked eye and to early optical detectors. Paschen (n₁ = 3) and higher series emit in the infrared. The Balmer series (n₁ = 2) uniquely falls in visible and near-visible wavelengths, which is why it was the first discovered — it was literally the most visible. Astronomers could see Balmer lines in stellar spectra with prism spectroscopes, long before UV or IR detectors existed. The formula was found empirically for Balmer lines, and the Rydberg generalization came after.

Question 3

The Rydberg constant R∞ has exactly the same value for most hydrogen isotopes (protium, deuterium, tritium), since it is a universal constant derived from fundamental physics.

• T. True
• F. False

Answer: False

False. R∞ (with the ∞ subscript) is derived assuming infinite nuclear mass. Real isotopes have finite nuclear mass, requiring a reduced-mass correction: the actual Rydberg constant for a specific isotope is R = R∞ × μ/mₑ, where μ is the reduced mass of the electron-nucleus system. Since deuterium has a heavier nucleus than protium, its reduced mass is slightly larger, and its spectral lines are shifted to slightly shorter wavelengths. This isotope shift is small but measurable and was historically important: in 1932, Urey identified deuterium by detecting this predicted shift in hydrogen's spectral lines.

Question 4

Within a given spectral series (fixed n₁), the spectral lines crowd closer together as n₂ increases, converging toward a series limit at a finite wavelength corresponding to ionization from that shell.

• T. True
• F. False

Answer: True

True. The Rydberg formula gives 1/λ = R(1/n₁² − 1/n₂²). As n₂ increases, 1/n₂² → 0, so 1/λ approaches R/n₁² from below — a fixed value that corresponds to the ionization wavelength from shell n₁. The spacing between successive lines decreases: the gap between n₂ = 3 and n₂ = 4 is larger than the gap between n₂ = 100 and n₂ = 101. This convergence is directly visible in spectra: lines pile up toward a sharp ionization limit, beyond which lies the continuous spectrum corresponding to ionized electrons.

Question 5

Explain how the Rydberg formula is derived from the Bohr model, and what physical process each spectral line corresponds to.

Model answer: In the Bohr model, electrons occupy discrete energy levels Eₙ = −13.6/n² eV. When an electron transitions from a higher level n₂ to a lower level n₁, it releases a photon whose energy equals the difference: ΔE = E_{n₁} − E_{n₂} = 13.6(1/n₁² − 1/n₂²) eV. Using E = hc/λ to convert energy to wavelength and substituting the Bohr energy formula, the result is 1/λ = R(1/n₁² − 1/n₂²), where the Rydberg constant R bundles together the electron mass, charge, Planck's constant, and permittivity of free space. Each spectral line corresponds to a specific downward transition between two energy levels. The named series (Lyman, Balmer, Paschen) group lines by their common final level n₁.

The formula is not empirical curve-fitting: it follows mathematically from the Bohr energy quantization condition. The fact that it agrees with observation to high precision was a major confirmation of the Bohr model, and later derivation from quantum mechanics confirmed it more rigorously.