Atomic Spectra: Models & Estimates (IPhO Modern)
IPhO atomic spectra lesson: hydrogenic model, scaling with Z, wavelength conversion tricks, and fast estimate methods.
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Most olympiad “atomic spectra” questions are really about one idea: the hydrogenic energy scale is set by Z² and 1/n², so line wavelengths come from simple energy differences. The second idea is modeling: you are often asked to justify when a hydrogenic estimate is good (hydrogen, He⁺, high-Z ions) and when screening or structure matters (neutral multi-electron atoms).
- hc ≈ 1240 eV nm
- R ≈ 1.097 × 10⁷ m⁻¹
- Hydrogen ground energy scale: 13.6 eV
1. Definitions (Must Know)
A. Hydrogenic energy levels (one-electron ions)
For a one-electron atom/ion with nuclear charge + Ze (hydrogenic model): Eₙ ≈ -(13.6 eV Z²)/n², n = 1,2,3,…
Ionization energy from the ground state: Eᵢₒₙ ≈ 13.6 eV Z²
B. Photon energy and wavelength
For a transition nᵢ → n_f with nᵢ > n_f: Δ E = |E_(n_f)-E_nᵢ| and the emitted photon satisfies: Δ E = h f = hc/λ
C. Rydberg formula (wavenumber form)
1/λ = RZ²(1/n_f² - 1/nᵢ²)
D. Series names (hydrogen)
- Lyman: n_f = 1 (UV)
- Balmer: n_f = 2 (visible/near-UV)
- Paschen: n_f = 3 (IR)
E. Reduced mass (isotope shift idea)
For higher precision, replace mₑ by the reduced mass μ: μ = mₑm_N/(mₑ + m_N) This slightly changes the effective Rydberg constant (and hence wavelengths).
2. Key Ideas (What Earns Marks)
- Use scaling first. If Z doubles, energy spacings scale by Z².
- Convert to wavelength late. Work in eV, then use hc ≈ 1240 eV nm at the end.
- Series limits are ionization. As nᵢ → ∞, the line approaches a sharp limit related to Eᵢₒₙ.
- Model awareness matters. Hydrogenic results work well for one-electron ions (H, He⁺, Li²⁺) and for high-n outer electrons with an effective charge Z_eff.
- Order-of-magnitude wins marks. Many questions only need the right scale and dependence.
3. Detailed Explanations
A. A fast workflow for line wavelengths
- Identify Z, nᵢ, and n_f.
- Compute Δ E using Eₙ ∝ -Z²/n².
- Convert to wavelength using λ = hc/Δ E.
This is typically faster and less error-prone than plugging into the Rydberg formula directly.
B. Why the series limit is special
For fixed n_f, as nᵢ → ∞: E_nᵢ → 0 So the limit photon energy is: Δ Eₗᵢₘᵢₜ = |E_(n_f)|
For hydrogen (Z = 1), the Balmer limit (n_f = 2) corresponds to |E₂| = 13.6/4 eV = 3.4 eV.
C. When hydrogenic estimates fail (and how to patch them)
For neutral multi-electron atoms:
- inner electrons shield the nucleus, so outer electrons see Z_eff < Z,
- energies are not exactly -13.6Z²/n² because the potential is not purely Coulomb,
- selection rules and splitting become more important.
A standard olympiad-level patch is to treat the outer electron as hydrogenic with an effective charge Z_eff chosen to match one known line or ionization energy.
D. Line splitting scales (very rough)
You can often estimate whether a splitting is negligible by comparing with the main transition energy:
- fine structure is typically smaller by a factor of order α² (with α ≈ 1/137),
- Zeeman splitting scale is Δ E ∼ μ_B B.
4. Common Mistakes
- Forgetting the Z² factor for hydrogenic ions.
- Mixing up nᵢ and n_f (emission requires nᵢ > n_f).
- Converting eV to nm incorrectly (use hc ≈ 1240 eV nm).
- Dropping absolute values and getting a negative photon energy.
- Assuming hydrogenic formulas apply unchanged to neutral multi-electron atoms without an effective charge argument.
5. Exam Tips
- Write the scale first: “Energy levels go like Z²/n².” This earns reasoning marks even before calculation.
- Use λ = hc/Δ E with hc in eV nm to avoid unit mistakes.
- If a question asks for a limit wavelength, take nᵢ → ∞ immediately.
- If you are given one spectral line for a multi-electron atom, use it to calibrate Z_eff and then predict the next one.
6. Worked Examples
A. Balmer alpha line in hydrogen (classic)
Find the wavelength for the hydrogen transition nᵢ = 3 to n_f = 2.
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Use the Rydberg form with Z = 1:
So: λ = 36/5R ≈ 36/(5 × 1.097 × 10⁷) m ≈ 6.56 × 10⁻⁷ m
Therefore λ ≈ 656 nm.
B. Hydrogen-like ion scaling (He⁺ line)
For He⁺ (Z = 2), find the photon energy and wavelength for the transition nᵢ = 2 to n_f = 1.
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Hydrogenic energies: Eₙ = -(13.6 eV Z²)/n² For Z = 2: E₁ = -13.6 × 4 eV = -54.4 eV E₂ = -54.4/4 eV = -13.6 eV
Photon energy: Δ E = |E₁-E₂| = 40.8 eV
Convert to wavelength: λ = hc/(Δ E) ≈ (1240 eV nm)/(40.8 eV) ≈ 30.4 nm
7. Mind Stretchers
A. Isotope shift from reduced mass (order-of-magnitude)
Estimate the fractional change in hydrogen line wavelengths when going from protium (m_N ≈ mₚ) to deuterium (m_N ≈ 2mₚ), using the reduced mass idea.
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For a hydrogenic atom, the Rydberg constant scales like the reduced mass μ: R ∝ μ = mₑm_N/(mₑ + m_N)
For m_N ≫ mₑ: μ ≈ mₑ(1 - mₑ/m_N) So the fractional change in R between m_N = mₚ and m_N = 2mₚ is: (Δ R)/R ≈ mₑ/mₚ - mₑ/2mₚ = mₑ/2mₚ
Numerically, mₑ/mₚ ≈ 1/1836, so: (Δ R)/R ≈ 1/3672 ≈ 2.7 × 10⁻⁴
Since λ ∝ 1/R, the wavelengths shift by the same fraction (with opposite sign).
- The Balmer limit corresponds to ionization from n = 2. Why does the line crowding occur near the limit?
- In multi-electron atoms, why do outer electrons often behave hydrogen-like for high n (Rydberg states)?
8. Practice
- Do 3 fast calculations: one hydrogen Balmer line, one series limit, one hydrogenic Z = 2 or Z = 3 scaling.
- Do 1 modeling question: estimate a multi-electron transition using an effective charge Z_eff.
Syllabus and review details
No official syllabus alignment is listed for this lesson.