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

1. Light, Photons & Energy

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

• Wave relation: $c = \lambda \nu$, where $c = 3.00 \times 10^{8}$ m/s.
• Photon energy: $E = h\nu = \dfrac{hc}{\lambda}$, where $h = 6.626 \times 10^{-34}$ 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 $\lambda = 532$ nm. Its energy is $E = hc/\lambda = (6.626 \times 10^{-34})(3.00 \times 10^{8})/(5.32 \times 10^{-7}) = 3.74 \times 10^{-19}$ 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 $n_i$ to a lower level $n_f$ and emit photons with $\Delta E = h\nu = E_i - E_f$.
• 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:

$E_n = -\dfrac{2.18 \times 10^{-18}\,\text{J}}{n^{2}} = -\dfrac{13.6\,\text{eV}}{n^{2}}, \quad n = 1, 2, 3, \ldots$

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

When the electron drops from $n_i$ to $n_f$, the photon energy is

$|\Delta E| = 2.18 \times 10^{-18}\left(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right)\,\text{J}.$

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:

$\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right), \quad R_H = 1.097 \times 10^{7}\,\text{m}^{-1}.$

The lines fall into named series by their ending level:

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:

$\boxed{\;BE = h\nu - KE\;}$

The histogram of $BE$ 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 $Z_\text{eff}$.
• 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 ($Z_\text{eff} \approx Z$), 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 $Z_\text{eff}$ and peak height to electron count.

6. Problem-Solving Workshop

Common templates and the rule of thumb for each:

• Photon energy from $\lambda$: $E = hc/\lambda$. Convert nm → m first ($1\,\text{nm} = 10^{-9}\,\text{m}$).
• Hydrogen transition energy: Use either $E_n = -2.18 \times 10^{-18}/n^{2}$ then take , or use the Rydberg formula and convert to energy with . Both give the same answer.
• Identifying a series: Look at $n_f$. $n_f = 1$ → Lyman/UV. $n_f = 2$ → Balmer/visible. → Paschen/IR.
• PES binding energy: $BE = h\nu - KE$. Convert per-electron J → kJ/mol by multiplying by $N_A = 6.022 \times 10^{23}$ then dividing by 1000.
• Identifying an element from PES: count electrons (sum of peak heights) → $Z$. 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 $E = hc/\lambda$ couple electrons between levels (absorbed = up, emitted = down).
• The Bohr energy formula $E_n = -2.18 \times 10^{-18}/n^{2}$ J and the Rydberg formula $1/\lambda = R_H(1/n_f^{2} - 1/n_i^{2})$ 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 ≈ $Z_\text{eff}$.
• $Z_\text{eff}$ 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 $E = hc/\lambda$.
• 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 $n_f$, not $n_i$.

Quick reference card

• $c = \lambda\nu$; $E_\text{photon} = h\nu = hc/\lambda$
• $E_n^{(H)} = -2.18 \times 10^{-18}/n^{2}$ J eV
• $1/\lambda = R_H(1/n_f^{2} - 1/n_i^{2})$, m⁻¹
• $BE_\text{PES} = h\nu - KE_\text{ejected}$
• Lyman / Balmer / Paschen → UV / visible / IR
• Multiply per-photon J by $N_A$ to get J/mol; divide by 1000 for kJ/mol

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

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.

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

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

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

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

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

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 $Z_\text{eff} = Z - S$ (where $S$ is the shielding) and on the angular momentum quantum number $\ell$, splitting subshells (s, p, d, f) that the simple Bohr model treats as degenerate. Photoelectron spectroscopy directly shows these split subshells.

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

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

Set equal: $2.18 \times 10^{-18}(1 - 1/n^{2}) = 2.09 \times 10^{-18}$ → $1 - 1/n^{2} = 0.959$ → $1/n^{2} = 0.041$ → $n^{2} \approx 24.4$ → $n \approx 4.94$.

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

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