Why do atoms emit light at discrete wavelengths rather than a continuous spectrum?
AElectrons move in discrete circular orbits, so their orbital speeds take only discrete values
BAtomic energy levels are discrete, so transitions between levels release photons with exactly quantized energies ΔE = hf, corresponding to specific wavelengths
CThe Rydberg formula restricts wavelengths by an empirical rule that happens to give discrete values
DElectrons only emit photons when they collide with other atoms, and collision energies are quantized
The discreteness of spectral lines follows directly from the discreteness of energy levels. An electron transitioning from level n₂ to level n₁ releases exactly ΔE = E_{n₂} − E_{n₁} as a photon. Since ΔE = hf = hc/λ, each specific pair of levels produces a specific wavelength — no other wavelength is possible. Option A describes an older (Bohr) picture that is not the full quantum explanation. Option C reverses the logic: the Rydberg formula is a consequence of quantized energy levels, not an independent empirical constraint.
Question 2 Multiple Choice
A cool hydrogen gas cloud sits between an observer and a hot, bright star emitting a continuous spectrum. What does the observer see?
ABright emission lines at Lyman-series wavelengths superimposed on the continuous spectrum
BA continuous spectrum with no features — cold gas is transparent to all wavelengths
CA continuous spectrum with dark absorption lines at Lyman-series wavelengths, because ground-state electrons absorb photons matching those transitions
DA continuous spectrum with dark lines at Balmer-series wavelengths, because the visible photons are selectively absorbed
At normal temperatures, nearly all hydrogen atoms are in the ground state (n = 1). Photons that match Lyman-series energies (transitions from n = 1 to higher levels) are absorbed, producing dark absorption lines in the continuous stellar spectrum. The Balmer series (n = 1 → n = 2 transitions) would only be absorbed by atoms already in n = 2, which is negligible at room temperature — so Balmer absorption requires hot gas. The key principle: absorption lines appear at exactly the same wavelengths as emission lines for the same transitions.
Question 3 True / False
The Balmer series of hydrogen spectral lines falls in the ultraviolet, because transitions to n = 2 involve large energy differences that produce high-frequency photons.
TTrue
FFalse
Answer: False
The Balmer series (transitions ending on n₁ = 2) falls in the VISIBLE range — this is why hydrogen appears reddish in emission nebulae (the Hα line at 656 nm) and why Balmer lines were the first hydrogen series discovered (visible to the naked eye). The LYMAN series (transitions to n₁ = 1, the ground state) falls in the ultraviolet, because the ground state sits so far below higher levels that those transitions carry more energy. The Paschen series (n₁ = 3) and higher fall in the infrared.
Question 4 True / False
The same set of wavelengths that appear as dark absorption lines in a cool hydrogen gas also appear as bright emission lines in hot hydrogen gas, because the relevant energy differences are the same regardless of transition direction.
TTrue
FFalse
Answer: True
Absorption and emission are mirror processes: the same pair of energy levels (say n = 1 and n = 3) is involved whether a photon is absorbed (electron jumps up) or emitted (electron falls down). The photon energy — and therefore wavelength — is determined by the energy difference ΔE = |E_3 − E_1|, which is the same in both cases. This equivalence is the basis for stellar spectroscopy: astronomers identify elements in stellar atmospheres by matching dark absorption line patterns to known emission spectra of the same elements.
Question 5 Short Answer
Explain why the Lyman, Balmer, and Paschen series each falls in a different part of the electromagnetic spectrum. What property of the transitions determines which series lands in the UV, visible, or IR?
Think about your answer, then reveal below.
Model answer: Each series corresponds to transitions ending on a specific lower level n₁ (Lyman: n₁ = 1, Balmer: n₁ = 2, Paschen: n₁ = 3). The energy of the emitted photon is ΔE = 13.6 eV × (1/n₁² − 1/n₂²). The crucial factor is n₁: transitions ending on n₁ = 1 (the ground state) release the most energy, because the ground state energy (−13.6 eV) is far below the higher levels. These large energy differences correspond to high-frequency, short-wavelength photons in the UV. Transitions ending on n₁ = 2 release less energy (the reference level is higher), landing in the visible. Transitions ending on n₁ = 3 release even less energy, landing in the infrared. Within each series, lines get closer together as n₂ increases, converging at the series limit where n₂ → ∞.
The series structure is the most elegant feature of hydrogen spectroscopy: instead of a random collection of lines, all lines fall into families with a shared lower level. The Rydberg formula makes this explicit — grouping by n₁ groups by final energy level, and the energy of that final level sets the photon energy scale for the whole series.