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Atomic Spectra: Photons, Quantized Energy Levels & PES

Understand emission and absorption spectra

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Atomic Spectra: Photons, Quantized Energy Levels & PES

Atomic spectra are the experimental "fingerprints" that revealed the quantum nature of the atom. When an atom absorbs energy, an electron is promoted to a higher quantized level; when it relaxes, a photon is emitted with energy equal to the gap between levels. Because the levels are discrete, only certain photon energies are produced — and those line patterns are unique to each element.

This topic ties together three big ideas tested on the AP Chemistry exam: (1) the quantization of light and energy (E=hνE = h\nu), (2) Bohr-style energy levels for hydrogen, and (3) photoelectron spectroscopy (PES) as direct experimental evidence for shells, subshells, and effective nuclear charge (ZeffZ_\text{eff}).

1. Light, Photons & Energy

Light is both a wave and a stream of particles called photons. Two relationships you will use constantly:

  • Wave relation: c=λνc = \lambda \nu, where c=3.00×108c = 3.00 \times 10^{8} m/s.
  • Photon energy: E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}, where h=6.626×10−34h = 6.626 \times 10^{-34} J·s.

Higher frequency ↔ shorter wavelength ↔ higher energy. The visible range is roughly 400 nm (violet, high E) → 700 nm (red, low E). UV photons (≤ 400 nm) carry enough energy to ionize many atoms; IR photons (≥ 700 nm) are too low-energy for most electronic transitions but excite vibrations.

Worked example. A green photon has λ=532\lambda = 532 nm. Its energy is E=hc/λ=(6.626×10−34)(3.00×108)/(5.32×10−7)=3.74×10−19E = hc/\lambda = (6.626 \times 10^{-34})(3.00 \times 10^{8})/(5.32 \times 10^{-7}) = 3.74 \times 10^{-19} J.

2. Atomic Emission & Absorption Spectra

When isolated atoms in the gas phase interact with light, they produce line spectra, not continuous rainbows.

  • Emission spectrum (bright lines on dark background). Excited electrons relax from a higher level nin_i to a lower level nfn_f and emit photons with ΔE=hν=Ei−Ef\Delta E = h\nu = E_i - E_f.
  • Absorption spectrum (dark lines on a continuous spectrum). Cool atoms absorb only the photons whose energies match an allowed gap, promoting electrons. The "missing" wavelengths are the same ones the hot gas would emit.

The same set of energy gaps controls both processes — emission and absorption are mirror images of each other for a given element. This is why astronomers can identify hydrogen, helium, calcium, and sodium in stars from absorption lines in starlight.

AP tip. A continuous spectrum (a smooth rainbow) requires a hot, dense source — a glowing solid, liquid, or very dense gas. Isolated atoms always give line spectra.

3. The Bohr Model & Quantized Levels (Hydrogen)

Bohr postulated that the electron in hydrogen could occupy only specific orbits with quantized energies:

En=−2.18×10−18 Jn2=−13.6 eVn2,n=1,2,3,…E_n = -\dfrac{2.18 \times 10^{-18}\,\text{J}}{n^{2}} = -\dfrac{13.6\,\text{eV}}{n^{2}}, \quad n = 1, 2, 3, \ldots

The energy is negative (the electron is bound) and gets less negative as nn increases. The n=∞n = \infty limit corresponds to an ionized H atom (E=0E = 0). The first ionization energy of hydrogen is therefore ∣E1∣=2.18×10−18|E_1| = 2.18 \times 10^{-18} J = 13.6 eV per atom = 1310 kJ/mol.

When the electron drops from nin_i to nfn_f, the photon energy is

∣ΔE∣=2.18×10−18(1nf2−1ni2) J.|\Delta E| = 2.18 \times 10^{-18}\left(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right)\,\text{J}.

4. The Hydrogen Spectrum & the Rydberg Formula

The Rydberg formula, an empirical relation from the 1880s that Bohr's model later derived from first principles, predicts every hydrogen line:

1λ=RH(1nf2−1ni2),RH=1.097×107 m−1.\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right), \quad R_H = 1.097 \times 10^{7}\,\text{m}^{-1}.

The lines fall into named series by their ending level:

Seriesnfn_fEM regionFamous lines
Lyman1Ultraviolet121.6 nm (Lyα)
Balmer2Visible656 (Hα), 486 (Hβ), 434 (Hγ), 410 nm (Hδ)
Paschen3Infrared1875 nm
Brackett4Far IR4051 nm

The Balmer lines are the ones you see in a hydrogen discharge tube; they are also why H is the easiest element to identify in stellar spectra.

5. Beyond Hydrogen — Multi-Electron Atoms & PES

Bohr's exact formula only works for one-electron systems (H, He⁺, Li²⁺). For multi-electron atoms, electron–electron repulsion and shielding mean each subshell sits at its own energy. We probe those energies directly with photoelectron spectroscopy (PES).

The PES experiment.

  1. Monochromatic high-energy photons (UV or X-ray) hit the sample.
  2. Each photon ejects one electron.
  3. The instrument measures the kinetic energy of each ejected electron.
  4. By conservation of energy:

  BE=hν−KE  \boxed{\;BE = h\nu - KE\;}

The histogram of BEBE values is the PES spectrum.

Reading a PES spectrum.

  • Peak position (x-axis, binding energy): which subshell. Peaks farther from zero = electrons closer to the nucleus = larger ZeffZ_\text{eff}.
  • Peak height (intensity): how many electrons in that subshell. (1s peak height : 2s peak height : 2p peak height in carbon ≈ 2 : 2 : 2.)

Effective nuclear charge. Inner electrons feel almost the full nuclear charge (Zeff≈ZZ_\text{eff} \approx Z), so their binding energies grow rapidly across a period. Valence electrons are shielded by inner electrons and have much lower binding energies — which is why valence electrons drive chemistry.

AP tip. PES is your most direct evidence for the shell/subshell model of the atom and is a favorite source of "evidence-based reasoning" exam questions. Always relate peak position to ZeffZ_\text{eff} and peak height to electron count.

6. Problem-Solving Workshop

Common templates and the rule of thumb for each:

  • Photon energy from λ\lambda: E=hc/λE = hc/\lambda. Convert nm → m first (1 nm=10−9 m1\,\text{nm} = 10^{-9}\,\text{m}).
  • Hydrogen transition energy: Use either En=−2.18×10−18/n2E_n = -2.18 \times 10^{-18}/n^{2} then take ∣ΔE∣|\Delta E|, or use the Rydberg formula and convert to energy with E=hc/λE = hc/\lambda. Both give the same answer.
  • Identifying a series: Look at nfn_f. nf=1n_f = 1 → Lyman/UV. nf=2n_f = 2 → Balmer/visible. nf=3n_f = 3 → Paschen/IR.
  • PES binding energy: BE=hν−KEBE = h\nu - KE. Convert per-electron J → kJ/mol by multiplying by NA=6.022×1023N_A = 6.022 \times 10^{23} then dividing by 1000.
  • Identifying an element from PES: count electrons (sum of peak heights) → ZZ. Confirm with the pattern of binding energies (1s very high, then a gap, then 2s/2p moderate, then 3s/3p low, etc.).

7. Synthesis & AP Review

Big ideas to leave with:

  • Atoms have quantized electron energies → line spectra.
  • Photons of energy E=hc/λE = hc/\lambda couple electrons between levels (absorbed = up, emitted = down).
  • The Bohr energy formula En=−2.18×10−18/n2E_n = -2.18 \times 10^{-18}/n^{2} J and the Rydberg formula 1/λ=RH(1/nf2−1/ni2)1/\lambda = R_H(1/n_f^{2} - 1/n_i^{2}) describe hydrogen exactly. They do not work for multi-electron atoms.
  • PES gives direct evidence for shells and subshells. Peak height = electron count, peak position = binding energy ≈ ZeffZ_\text{eff}.
  • ZeffZ_\text{eff} increases left-to-right across a period (less shielding) and roughly stays the same down a group, but inner electrons are pulled in much more tightly because they're not shielded.

Common mistakes

  • Using Bohr's formula for anything other than a 1-electron system. It fails badly for He, Li, etc.
  • Forgetting to convert nm to m before plugging into E=hc/λE = hc/\lambda.
  • Confusing emission with absorption diagrams. Emission adds bright lines to a dark background; absorption subtracts dark lines from a continuous background.
  • Assuming peak area in PES is the binding energy. Position is binding energy; height/area is electron count.
  • Identifying the wrong spectral series. Always check nfn_f, not nin_i.

Quick reference card

  • c=λνc = \lambda\nu; Ephoton=hν=hc/λE_\text{photon} = h\nu = hc/\lambda
  • En(H)=−2.18×10−18/n2E_n^{(H)} = -2.18 \times 10^{-18}/n^{2} J =−13.6/n2= -13.6/n^{2} eV
  • 1/λ=RH(1/nf2−1/ni2)1/\lambda = R_H(1/n_f^{2} - 1/n_i^{2}), RH=1.097×107R_H = 1.097 \times 10^{7} m⁻¹
  • BEPES=hν−KEejectedBE_\text{PES} = h\nu - KE_\text{ejected}
  • Lyman / Balmer / Paschen → UV / visible / IR
  • Multiply per-photon J by NAN_A to get J/mol; divide by 1000 for kJ/mol

📚 Practice Problems

1Problem 1easy

❓ Question:

Calculate the energy (in joules) of a single photon of green light with wavelength λ=532\lambda = 532 nm. Use h=6.626×10−34h = 6.626 \times 10^{-34} J·s and c=3.00×108c = 3.00 \times 10^{8} m/s.

💡 Show Solution

Step 1. Convert wavelength to meters: 532 nm=5.32×10−7532\text{ nm} = 5.32 \times 10^{-7} m.

Step 2. Apply E=hc/λE = hc/\lambda:

E=(6.626×10−34)(3.00×108)5.32×10−7=3.74×10−19 JE = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{5.32 \times 10^{-7}} = \boxed{3.74 \times 10^{-19}\text{ J}} per photon.

2Problem 2easy

❓ Question:

Rank the following photons from lowest to highest energy: (a) red, λ=700\lambda = 700 nm; (b) blue, λ=450\lambda = 450 nm; (c) microwave, λ=1\lambda = 1 cm; (d) X-ray, λ=1\lambda = 1 nm.

💡 Show Solution

Energy is inversely proportional to wavelength: E=hc/λE = hc/\lambda.

From longest λ\lambda (lowest E) to shortest λ\lambda (highest E):

microwave (1 cm) < red (700 nm) < blue (450 nm) < X-ray (1 nm)\text{microwave (1 cm) < red (700 nm) < blue (450 nm) < X-ray (1 nm)}

So: c<a<b<d\boxed{\text{c} < \text{a} < \text{b} < \text{d}}.

3Problem 3easy

❓ Question:

Distinguish between an emission spectrum and an absorption spectrum. Why do both reveal the same set of wavelengths for a given element?

💡 Show Solution

Emission spectrum: Excited electrons relax from higher to lower quantized levels and emit photons of energy ΔE=Ei−Ef\Delta E = E_i - E_f. The result is bright lines on a dark background (e.g., the lines from a hydrogen discharge tube).

Absorption spectrum: Ground-state atoms absorb specific photon energies that promote an electron to a higher level. Continuous light passing through such a gas emerges with dark lines on a continuous background at exactly those wavelengths.

Same wavelengths: Both processes are governed by the same set of allowed electronic energy gaps. Emission goes "down," absorption goes "up," but the gaps — and therefore the photon energies — are identical.

4Problem 4medium

❓ Question:

A hydrogen atom transitions from n=5n=5 to n=2n=2. (a) Find the photon energy in J. (b) Find the wavelength in nm. (c) Identify the spectral series and the EM region.

💡 Show Solution

Use En=−2.18×10−18/n2E_n = -2.18 \times 10^{-18}/n^{2} J.

(a) ΔE=E2−E5=−2.18×10−18(1/4−1/25)=−2.18×10−18(0.21)=−4.58×10−19\Delta E = E_2 - E_5 = -2.18 \times 10^{-18}(1/4 - 1/25) = -2.18 \times 10^{-18}(0.21) = -4.58 \times 10^{-19} J. The emitted photon carries ∣ΔE∣=4.58×10−19 J\boxed{|\Delta E| = 4.58 \times 10^{-19}\text{ J}}.

(b) λ=hc/E=(6.626×10−34)(3.00×108)/(4.58×10−19)=4.34×10−7\lambda = hc/E = (6.626 \times 10^{-34})(3.00 \times 10^{8})/(4.58 \times 10^{-19}) = 4.34 \times 10^{-7} m =434 nm= \boxed{434\text{ nm}}.

(c) nf=2n_f = 2 → Balmer series, in the visible region (this is the violet H-γ line).

5Problem 5medium

❓ Question:

Use the Rydberg formula to find the wavelength (in nm) for the n=4→n=1n=4 \to n=1 transition in hydrogen. (RH=1.097×107R_H = 1.097 \times 10^{7} m⁻¹.) Identify the series and EM region.

💡 Show Solution

Step 1. 1λ=RH(112−142)=(1.097×107)(0.9375)=1.028×107\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{1^{2}} - \dfrac{1}{4^{2}}\right) = (1.097 \times 10^{7})(0.9375) = 1.028 \times 10^{7} m⁻¹.

Step 2. λ=1/(1.028×107)=9.72×10−8\lambda = 1/(1.028 \times 10^{7}) = 9.72 \times 10^{-8} m =97.2 nm= \boxed{97.2\text{ nm}}.

Step 3. nf=1n_f = 1 → Lyman series, in the ultraviolet.

6Problem 6medium

❓ Question:

A sodium street lamp emits its characteristic yellow light at λ=589\lambda = 589 nm. (a) What is the photon energy in J? (b) What is the energy in kJ/mol of photons?

💡 Show Solution

(a) E=hc/λ=(6.626×10−34)(3.00×108)/(5.89×10−7)=3.37×10−19 JE = hc/\lambda = (6.626 \times 10^{-34})(3.00 \times 10^{8})/(5.89 \times 10^{-7}) = \boxed{3.37 \times 10^{-19}\text{ J}} per photon.

(b) Multiply by NAN_A then convert to kJ: E=(3.37×10−19)(6.022×1023)=2.03×105E = (3.37 \times 10^{-19})(6.022 \times 10^{23}) = 2.03 \times 10^{5} J/mol =203 kJ/mol= \boxed{203\text{ kJ/mol}}.

This corresponds to the Na 3p → 3s relaxation (the famous "sodium D-line").

7Problem 7medium

❓ Question:

Why does the Bohr model predict the hydrogen spectrum exactly but fail badly for helium and lithium?

💡 Show Solution

Bohr's model assumes a single electron orbiting a point nucleus, with no electron–electron interactions. For one-electron systems (H, He⁺, Li²⁺) this is exact and the formula En=−2.18×10−18 Z2/n2E_n = -2.18 \times 10^{-18}\,Z^{2}/n^{2} J reproduces the observed spectrum.

In He, Li, and beyond, the electrons shield each other from the nucleus and repel each other. The actual energy of each subshell depends on Zeff=Z−SZ_\text{eff} = Z - S (where SS is the shielding) and on the angular momentum quantum number ℓ\ell, splitting subshells (s, p, d, f) that the simple Bohr model treats as degenerate. Photoelectron spectroscopy directly shows these split subshells.

8Problem 8hard

❓ Question:

In a PES experiment, photons of energy 1.50×10−171.50 \times 10^{-17} J eject electrons from a particular orbital with kinetic energy 1.20×10−171.20 \times 10^{-17} J. (a) What is the binding energy per electron in J? (b) Convert that to kJ/mol. (c) Would you expect this to be a core or a valence electron? Justify.

💡 Show Solution

(a) Conservation of energy: BE=hν−KE=1.50×10−17−1.20×10−17=3.0×10−18 JBE = h\nu - KE = 1.50 \times 10^{-17} - 1.20 \times 10^{-17} = \boxed{3.0 \times 10^{-18}\text{ J}} per electron.

(b) (3.0×10−18)(6.022×1023)=1.81×106(3.0 \times 10^{-18})(6.022 \times 10^{23}) = 1.81 \times 10^{6} J/mol =1810 kJ/mol= \boxed{1810\text{ kJ/mol}}.

(c) 1810 kJ/mol is far above typical valence ionization energies (usually 500–2400 kJ/mol for the first IE; valence subshells in second-row elements are generally < 2.4 MJ/mol). A binding energy this large is consistent with a core electron (e.g., 1s of a second- or third-row element) feeling nearly the full nuclear charge.

9Problem 9hard

❓ Question:

A PES spectrum of an unknown neutral atom shows three peaks at binding energies of 11.5 MJ/mol, 1.09 MJ/mol, and 0.578 MJ/mol, with peak heights in the ratio 2 : 2 : 1. Identify the element and write its electron configuration.

💡 Show Solution

Step 1. Peak heights 2 : 2 : 1 → 2 + 2 + 1 = 5 electrons → Z=5Z = 5 → Boron.

Step 2. Match peaks to subshells (highest binding energy = closest to nucleus):

  • 11.5 MJ/mol (height 2) → 1s²
  • 1.09 MJ/mol (height 2) → 2s²
  • 0.578 MJ/mol (height 1) → 2p¹

Configuration: 1s2 2s2 2p1\boxed{1s^{2}\,2s^{2}\,2p^{1}}, which is boron.

10Problem 10hard

❓ Question:

A photon of wavelength λ=95.0\lambda = 95.0 nm strikes a ground-state hydrogen atom. (a) Will the photon be absorbed? Justify with an energy calculation. (b) If absorbed, to what level does the electron go?

💡 Show Solution

Step 1. Convert photon energy to a hydrogen-level ΔE\Delta E. Ephoton=hc/λ=(6.626×10−34)(3.00×108)/(9.50×10−8)=2.09×10−18E_\text{photon} = hc/\lambda = (6.626 \times 10^{-34})(3.00 \times 10^{8})/(9.50 \times 10^{-8}) = 2.09 \times 10^{-18} J.

Step 2. For absorption from n=1n=1 to some nn, ΔE=−2.18×10−18(1/n2−1)=2.18×10−18(1−1/n2)\Delta E = -2.18 \times 10^{-18}(1/n^{2} - 1) = 2.18 \times 10^{-18}(1 - 1/n^{2}).

Set equal: 2.18×10−18(1−1/n2)=2.09×10−182.18 \times 10^{-18}(1 - 1/n^{2}) = 2.09 \times 10^{-18} → 1−1/n2=0.9591 - 1/n^{2} = 0.959 → 1/n2=0.0411/n^{2} = 0.041 → n2≈24.4n^{2} \approx 24.4 → n≈4.94n \approx 4.94.

Step 3. Since nn is not an integer, the photon is not absorbed. Hydrogen only absorbs photons whose energies exactly match an allowed n=1→n=n=1 \to n= integer transition (e.g., n=1→5n=1 \to 5 requires λ=95.0\lambda = 95.0 nm to within rounding; the closest integer is n=5n=5, which actually does match this wavelength to within the precision shown — so a more careful calculation gives n=5n = 5).

Refined answer. Recomputing: for n=5n=5, ΔE=2.18×10−18(1−1/25)=2.09×10−18\Delta E = 2.18 \times 10^{-18}(1 - 1/25) = 2.09 \times 10^{-18} J → λ=9.50×10−8\lambda = 9.50 \times 10^{-8} m = 95.0 nm\boxed{95.0\text{ nm}}. So yes, the photon is absorbed and the electron is promoted to n=5\boxed{n=5} (a Lyman-series absorption).

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💡 Key Test-Day Tips

  • ✓Memorize common polyatomic ions
  • ✓Practice dimensional analysis
  • ✓Know your gas laws

⚠️ Common Mistakes: Atomic Spectra: Photons, Quantized Energy Levels & PES

Avoid these 3 frequent errors

🌍 Real-World Applications: Atomic Spectra: Photons, Quantized Energy Levels & PES

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams

📌 Related Topics in Atomic Structure and Properties

❓ Frequently Asked Questions

What is Atomic Spectra: Photons, Quantized Energy Levels & PES?▾
Understand emission and absorption spectra
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Yes, this page includes 10 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.