Atomic Structure — Bohr Model, Quantum Numbers & Orbitals

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Atomic structure moves from particles to waves. The numericals use a short list of formulas — E = hν, the Bohr energy and radius, the Rydberg equation, λ = h/mv and Δx·Δp ≥ h/4π — and the theory questions come down to quantum numbers and how electrons fill orbitals.

What you will be able to do

• Relate frequency, wavelength, wavenumber and energy, and solve photoelectric-effect problems
• Use Bohr’s formulas for radius, energy and velocity of hydrogen-like species
• Calculate wavelengths of spectral lines and the number of lines from a given level
• Apply de Broglie’s relation and Heisenberg’s uncertainty principle
• Assign quantum numbers, count nodes, and write configurations including the exceptions

The big picture

From particles to waves.

In one line: Electrons are both particles and waves: Bohr’s fixed orbits explain the hydrogen spectrum, but only the quantum-mechanical model of orbitals and quantum numbers explains configurations and the periodic table.

Question type	Formula
Light	c = νλ; ν̄ = 1/λ; E = hν = hc/λ
Photoelectric effect	hν = hν₀ + ½mv²
Bohr (H-like)	rₙ = 52.9 n²/Z pm; Eₙ = −13.6 Z²/n² eV; vₙ = 2.18 × 10⁶ Z/n m s⁻¹
Spectral line	1/λ = Rₕ Z² (1/n₁² − 1/n₂²); Rₕ = 109 677 cm⁻¹
Matter waves	λ = h/mv = h/p
Uncertainty	Δx · Δp ≥ h/4π

Think of it like this: Think of an electron in an atom as a guitar string fixed at both ends. Only certain wavelengths fit — those are the allowed energy levels. Pluck it harder and it jumps to a higher “note”; when it falls back, it gives out light of exactly the energy difference. That is why each element has its own line spectrum, like a fingerprint.

Subatomic particles and early models

What the atom is made of.

Particle	Charge / C	Mass / kg	Discovered by
Electron	−1.602 × 10⁻¹⁹	9.109 × 10⁻³¹	J. J. Thomson (cathode rays, e/m = 1.758820 × 10¹¹ C kg⁻¹)
Proton	+1.602 × 10⁻¹⁹	1.673 × 10⁻²⁷	Goldstein (canal rays)
Neutron	0	1.675 × 10⁻²⁷	Chadwick (1932)

• Millikan’s oil-drop experiment measured the charge of the electron (1.602 × 10⁻¹⁹ C); with e/m this gave its mass.
• Thomson’s model (plum pudding): positive sphere with embedded electrons — could not explain α-particle scattering.
• Rutherford’s α-scattering: most α particles passed straight through gold foil; a few were deflected; about 1 in 20 000 bounced back. So the atom is mostly empty, with a tiny, dense, positive nucleus (radius about 10⁻¹⁵ m, atom about 10⁻¹⁰ m).
• Failure of Rutherford’s model: an orbiting charge should radiate energy and spiral into the nucleus; it also could not explain line spectra.
• Isotopes (same Z, different A: ¹H, ²H, ³H); isobars (same A, different Z: ¹⁴C, ¹⁴N); isotones (same number of neutrons: ¹⁴C, ¹⁶O).

Electromagnetic radiation and quanta

Light as a wave and as a stream of photons.

c = ν λ ν̄ = 1/λ E = hν = hc/λ

c = 3.0 × 10⁸ m s⁻¹; h = 6.626 × 10⁻³⁴ J s

• Order of wavelength: radio > microwave > infrared > visible (400–750 nm) > ultraviolet > X-rays > γ-rays. Energy increases in the opposite order.
• Planck’s quantum theory: energy is emitted or absorbed in discrete packets (quanta) of E = hν. For light these packets are photons.
• Black-body radiation — the shift of peak emission to shorter wavelength at higher temperature — could not be explained by wave theory; Planck’s quanta explained it.

Photoelectric effect

hν = hν₀ + ½ m v²

hν₀ = work function (threshold energy); no electrons are emitted below ν₀, however intense the light

• Electrons are emitted instantly when ν > ν₀.
• The number of electrons depends on intensity; their kinetic energy depends only on frequency.
• Alkali metals (K, Cs) have low work functions and are used in photocells.

Solved example 01 — Photons in a beam

How many photons of wavelength 500 nm carry 1.0 J of energy?

1. E per photon = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (5.00 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J.
2. N = 1.0 / 3.98 × 10⁻¹⁹.

Answer: About 2.5 × 10¹⁸ photons.

Solved example 02 — Photoelectron energy

The threshold frequency of a metal is 7.0 × 10¹⁴ s⁻¹. Find the kinetic energy of electrons emitted by light of frequency 1.0 × 10¹⁵ s⁻¹.

1. KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × (10.0 − 7.0) × 10¹⁴.

Answer: KE ≈ 1.99 × 10⁻¹⁹ J.

The Bohr model

Fixed orbits with quantised angular momentum.

• Electrons move in circular orbits (stationary states) of fixed energy without radiating.
• Angular momentum is quantised: mvr = nh/2π.
• Energy is absorbed or emitted only when an electron jumps between orbits: ΔE = hν.
• Valid only for one-electron species: H, He⁺, Li²⁺, Be³⁺.

rₙ = 52.9 n²/Z pm
Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J = −13.6 Z²/n² eV
vₙ = 2.18 × 10⁶ Z/n m s⁻¹

Radius ∝ n²/Z; energy ∝ −Z²/n²; velocity ∝ Z/n

Key idea: Energy is negative because the electron is bound: zero energy means a free electron at infinity. Ionisation energy of H = +13.6 eV = 2.18 × 10⁻¹⁸ J per atom = 1312 kJ mol⁻¹. Kinetic energy = −E; potential energy = 2E.

Solved example 03 — Bohr radius and energy of He⁺

Find the radius and energy of the first orbit of He⁺.

1. Z = 2, n = 1.
2. r = 52.9 × 1²/2 = 26.45 pm.
3. E = −13.6 × 2²/1² = −54.4 eV.

Answer: r = 26.45 pm; E = −54.4 eV.

Solved example 04 — Ionisation energy of Li²⁺

Find the energy needed to remove the electron from Li²⁺ in its ground state.

1. Z = 3, n = 1: E₁ = −13.6 × 9 = −122.4 eV.
2. Ionisation energy = 0 − E₁.

Answer: 122.4 eV per ion.

Limitations

• Cannot explain spectra of multi-electron atoms.
• Cannot explain splitting of lines in magnetic (Zeeman) or electric (Stark) fields.
• Ignores the dual nature of the electron and contradicts the uncertainty principle (exact orbits).
• Cannot explain chemical bonding or molecular shape.

The hydrogen spectrum

Each line is one jump between two levels.

ν̄ = 1/λ = Rₕ Z² (1/n₁² − 1/n₂²)

Rₕ = 109 677 cm⁻¹ (Rydberg constant); n₂ > n₁

Series	n₁ (lower level)	n₂	Region
Lyman	1	2, 3, 4…	Ultraviolet
Balmer	2	3, 4, 5…	Visible
Paschen	3	4, 5, 6…	Infrared
Brackett	4	5, 6…	Infrared
Pfund	5	6, 7…	Infrared

• Number of lines when an electron falls from level n to the ground state (many atoms): n(n − 1)/2. Between levels n₂ and n₁: (n₂ − n₁)(n₂ − n₁ + 1)/2.
• A single atom with one electron can give at most n − 1 lines in one cascade.
• The series limit (n₂ = ∞) gives the shortest wavelength of a series.

Solved example 05 — The red Balmer line

Find the wavelength of the line for n = 3 → n = 2 in hydrogen.

1. ν̄ = 109 677 × (1/4 − 1/9) = 109 677 × 5/36 = 15 233 cm⁻¹.
2. λ = 1 / 15 233 cm = 6.565 × 10⁻⁵ cm.

Answer: λ ≈ 656.5 nm (red, the Hα line).

Solved example 06 — Counting lines

Electrons in a sample of hydrogen atoms are excited to n = 5. How many lines can appear in the emission spectrum, and how many are in the Balmer series?

1. Total = 5 × 4 / 2 = 10.
2. Balmer lines end on n = 2: from 5, 4 and 3 → 3 lines.

Answer: 10 lines; 3 in the Balmer series.

Solved example 07 — Same line from He⁺

Which transition in He⁺ has the same wavelength as the n = 2 → 1 transition in hydrogen?

1. For H: Z² (1/1 − 1/4) = 3/4.
2. For He⁺ (Z² = 4): 4 (1/n₁² − 1/n₂²) = 3/4 → 1/n₁² − 1/n₂² = 3/16.
3. n₁ = 2, n₂ = 4 gives 1/4 − 1/16 = 3/16.

Answer: n = 4 → n = 2 in He⁺.

Check yourself

• C1. Which series of hydrogen lies in the visible region?
• C2. How many spectral lines appear when electrons fall from n = 4 to n = 1?

Answers at the end.

Dual nature and uncertainty

Electrons behave as waves too.

λ = h / mv = h / p

de Broglie relation; significant only for very small masses

Δx · Δp ≥ h / 4π (Δx · m Δv ≥ h / 4π)

Heisenberg’s uncertainty principle: position and momentum cannot both be known exactly

• Electron diffraction (Davisson–Germer) confirmed the wave nature; the electron microscope uses it.
• For everyday objects λ is far too small to detect, and the uncertainty is negligible.
• The uncertainty principle rules out exact Bohr orbits — only probabilities (orbitals) make sense.
• In Bohr’s orbit, 2πr = nλ: the circumference holds a whole number of de Broglie waves.

Solved example 08 — de Broglie wavelength

Compare the de Broglie wavelengths of (a) an electron moving at 1.0 × 10⁶ m s⁻¹ and (b) a 0.10 kg ball at 10 m s⁻¹.

1. (a) λ = 6.626 × 10⁻³⁴ / (9.109 × 10⁻³¹ × 1.0 × 10⁶) = 7.27 × 10⁻¹⁰ m.
2. (b) λ = 6.626 × 10⁻³⁴ / (0.10 × 10) = 6.6 × 10⁻³⁴ m.

Answer: (a) 727 pm — comparable to atomic sizes, so it matters. (b) 6.6 × 10⁻³⁴ m — undetectable.

Solved example 09 — Uncertainty in velocity

The position of an electron is known to within 0.1 nm. Find the minimum uncertainty in its velocity.

1. Δv ≥ h / (4π m Δx) = 6.626 × 10⁻³⁴ / (4 × 3.142 × 9.109 × 10⁻³¹ × 1.0 × 10⁻¹⁰).
2. = 6.626 × 10⁻³⁴ / 1.145 × 10⁻³⁹.

Answer: Δv ≥ 5.8 × 10⁵ m s⁻¹ — huge, so an exact orbit is meaningless.

Quantum numbers and orbitals

Four numbers describe each electron.

Quantum number	Symbol	Values	Describes
Principal	n	1, 2, 3…	Shell: size and energy
Azimuthal	l	0 to n − 1 (s, p, d, f = 0, 1, 2, 3)	Subshell: shape; orbital angular momentum √[l(l+1)] h/2π
Magnetic	mₗ	−l to +l (2l + 1 values)	Orientation of the orbital
Spin	mₛ	+½, −½	Spin of the electron

• Orbitals in shell n: n²; electrons in shell n: 2n²; orbitals in a subshell: 2l + 1; electrons in a subshell: 2(2l + 1).
• Shapes: s spherical; p dumbbell (px, py, pz along the three axes); d cloverleaf (dxy, dyz, dxz between axes; dx²−y² on the axes; dz² a dumbbell with a ring).
• Nodes: radial nodes = n − l − 1; angular nodes (nodal planes) = l; total nodes = n − 1.
• 1s has no node; 2s has one radial node; 2p has one nodal plane; 3d has two nodal planes and no radial node.

Orbital	n	l	Radial nodes	Angular nodes
1s	1	0	0	0
2s	2	0	1	0
2p	2	1	0	1
3s	3	0	2	0
3p	3	1	1	1
3d	3	2	0	2
4f	4	3	0	3

Solved example 10 — Valid quantum numbers

Which set is not possible: (a) n = 3, l = 2, mₗ = −2; (b) n = 2, l = 2, mₗ = 0; (c) n = 4, l = 0, mₗ = 0?

1. l must be less than n.
2. (b) has l = n = 2.

Answer: (b) is impossible; (a) is a 3d orbital and (c) a 4s orbital.

Check yourself

• C3. How many radial nodes does a 4p orbital have?
• C4. What is the maximum number of electrons with n = 3?

Answers at the end.

Filling orbitals

Aufbau, Pauli and Hund.

• Aufbau principle: orbitals fill in order of increasing energy, given by the (n + l) rule — lower n + l first; if equal, lower n first. Order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s…
• Pauli exclusion principle: no two electrons in an atom have all four quantum numbers the same → at most two electrons per orbital, with opposite spins.
• Hund’s rule: in degenerate orbitals, electrons occupy separate orbitals with parallel spins before pairing.
• Exceptions: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹ — half-filled and filled subshells are extra stable because of symmetry and larger exchange energy (more pairs of parallel-spin electrons can exchange positions).

Solved example 11 — Exchange energy

Why is 3d⁵ 4s¹ (Cr) more stable than 3d⁴ 4s²?

1. In 3d⁵ all five electrons have parallel spins: the number of exchange pairs is 5 × 4 / 2 = 10.
2. In 3d⁴ there are only 4 × 3 / 2 = 6 exchange pairs.
3. More exchanges → more stabilisation; the half-filled set is also symmetrical.

Answer: 3d⁵ 4s¹ gains extra exchange energy and symmetry.

• Magnetic behaviour from configuration: unpaired electrons → paramagnetic; μ = √[n(n + 2)] BM.

Advanced problems

Each needs two ideas from the chapter.

Solved example 12 — Energy to wavelength

How much energy is needed to excite an electron in hydrogen from n = 1 to n = 2, and what wavelength of light does this?

1. ΔE = 2.18 × 10⁻¹⁸ (1 − 1/4) = 1.635 × 10⁻¹⁸ J.
2. λ = hc/ΔE = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / 1.635 × 10⁻¹⁸.

Answer: ΔE = 1.64 × 10⁻¹⁸ J; λ ≈ 1.22 × 10⁻⁷ m = 122 nm (ultraviolet, first Lyman line).

Solved example 13 — Ratio of radii and energies

Find the ratio of (a) the radius of the third orbit of Li²⁺ to the first orbit of H, and (b) the energy of the second orbit of He⁺ to the first orbit of H.

1. (a) r ∝ n²/Z: (9/3) / (1/1) = 3.
2. (b) E ∝ Z²/n²: (4/4) / (1/1) = 1.

Answer: (a) 3 : 1. (b) 1 : 1 — the n = 2 level of He⁺ has the same energy as the n = 1 level of H.

Solved example 14 — Work function in electron-volts

The work function of sodium is 2.3 eV. Will light of wavelength 600 nm eject electrons? (1 eV = 1.602 × 10⁻¹⁹ J)

1. Photon energy = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / (6.00 × 10⁻⁷) = 3.31 × 10⁻¹⁹ J.
2. In eV: 3.31 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 2.07 eV.
3. 2.07 eV < 2.3 eV.

Answer: No — 600 nm photons carry too little energy; no electrons are emitted, however bright the light.

Solved example 15 — Electrons with a given quantum number

How many electrons in an atom can have n = 4 and mₛ = −½? How many can have n = 3, l = 1?

1. n = 4: 16 orbitals → 32 electrons, half with mₛ = −½.
2. n = 3, l = 1: the 3p subshell, 3 orbitals.

Answer: 16 electrons; 6 electrons.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	Brighter light gives faster photoelectrons.	Brighter light gives more electrons; their energy depends only on frequency.	Each photon ejects one electron.
2	Bohr’s model works for He.	Only for one-electron species: H, He⁺, Li²⁺.	Electron–electron repulsion is ignored.
3	Energy levels get farther apart at higher n.	They get closer together (E ∝ −1/n²).	Lines crowd toward the series limit.
4	Balmer series lies in the ultraviolet.	Balmer is visible; Lyman is ultraviolet.	n₁ = 2 for Balmer.
5	The 2p orbital has one radial node.	2p: 0 radial nodes, 1 angular node.	Radial nodes = n − l − 1.
6	l can equal n.	l runs from 0 to n − 1.	Hence no 1p or 2d orbital.
7	4s fills after 3d because 3 < 4.	4s (n + l = 4) fills before 3d (n + l = 5).	Use the (n + l) rule.
8	Cu is [Ar] 3d⁹ 4s².	Cu is [Ar] 3d¹⁰ 4s¹.	A filled d subshell is extra stable.
9	de Broglie waves matter for cricket balls too.	λ is about 10⁻³⁴ m — undetectable.	Only tiny masses show measurable λ.
10	Number of lines from n = 4 to 1 is 3.	Up to 6 lines in a sample (n(n − 1)/2).	Every pair of levels gives one line.
11	Ionisation energy of H is 13.6 J.	13.6 eV per atom = 2.18 × 10⁻¹⁸ J = 1312 kJ mol⁻¹.	Watch the units.
12	Total energy and kinetic energy of an orbiting electron are equal.	KE = −E and PE = 2E.	Virial relation for a Coulomb orbit.

Memory hook: “Radius n²/Z, energy Z²/n², speed Z/n.” · “Lyman UV, Balmer visible, rest infrared.” · “Radial n − l − 1, angular l.” · “n + l small fills first.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The charge on an electron was measured by: (Easy)

• (a) Thomson
• (b) Millikan
• (c) Rutherford
• (d) Chadwick

Q02. The neutron was discovered by: (Easy)

• (a) Rutherford
• (b) Chadwick
• (c) Bohr
• (d) Goldstein

Q03. Radius of the first Bohr orbit of hydrogen: (Easy)

• (a) 52.9 pm
• (b) 26.45 pm
• (c) 105.8 pm
• (d) 0.529 pm

Q04. The Balmer series of hydrogen lies in the: (Easy)

• (a) UV region
• (b) visible region
• (c) IR region
• (d) X-ray region

Q05. Number of orbitals in the n = 3 shell: (Easy)

• (a) 3
• (b) 6
• (c) 9
• (d) 18

Q06. Shape of a p orbital: (Easy)

• (a) spherical
• (b) dumbbell
• (c) cloverleaf
• (d) ring

Q07. Which set of quantum numbers is not allowed? (Easy)

• (a) n=1, l=0, mₗ=0
• (b) n=2, l=1, mₗ=−1
• (c) n=2, l=2, mₗ=0
• (d) n=3, l=2, mₗ=+2

Q08. Energy of the electron in the second orbit of H: (Moderate)

• (a) −13.6 eV
• (b) −3.4 eV
• (c) −1.51 eV
• (d) −6.8 eV

Q09. Number of radial nodes in 3s: (Moderate)

• (a) 0
• (b) 1
• (c) 2
• (d) 3

Q10. Electronic configuration of Cr (Z = 24): (Moderate)

• (a) [Ar] 3d⁴ 4s²
• (b) [Ar] 3d⁵ 4s¹
• (c) [Ar] 3d⁶
• (d) [Ar] 4s² 4p⁴

Q11. Wavelength of the Hα line (n = 3 → 2): (Moderate)

• (a) 486 nm
• (b) 656 nm
• (c) 434 nm
• (d) 122 nm

Q12. In the photoelectric effect, kinetic energy of electrons depends on: (Moderate)

• (a) intensity
• (b) frequency
• (c) exposure time
• (d) area of metal

Q13. Maximum number of lines when electrons fall from n = 6 to n = 1: (Moderate)

• (a) 5
• (b) 10
• (c) 15
• (d) 21

Q14. The orbital with n = 4, l = 2 is: (Moderate)

• (a) 4s
• (b) 4p
• (c) 4d
• (d) 4f

Q15. Which orbital fills first? (Moderate)

• (a) 3d
• (b) 4s
• (c) 4p
• (d) both 3d and 4s together

Q16. de Broglie wavelength is inversely proportional to: (Moderate)

• (a) mass only
• (b) momentum
• (c) energy squared
• (d) frequency

Q17. Energy of the first orbit of He⁺: (Moderate)

• (a) −13.6 eV
• (b) −27.2 eV
• (c) −54.4 eV
• (d) −122.4 eV

Q18. Number of angular nodes in a 4f orbital: (Moderate)

• (a) 1
• (b) 2
• (c) 3
• (d) 4

Q19. Total number of electrons with l = 1 in phosphorus (Z = 15): (Tough)

• (a) 6
• (b) 9
• (c) 12
• (d) 3

Q20. Which transition in He⁺ gives the same wavelength as 2 → 1 in H? (Tough)

• (a) 4 → 2
• (b) 3 → 1
• (c) 2 → 1
• (d) 6 → 3

Q21. Ionisation energy of H is 13.6 eV. That of Li²⁺ is: (Tough)

• (a) 40.8 eV
• (b) 122.4 eV
• (c) 27.2 eV
• (d) 54.4 eV

Q22. Minimum uncertainty in velocity of an electron with Δx = 0.1 nm is about: (Tough)

• (a) 5.8 × 10⁵ m s⁻¹
• (b) 5.8 × 10² m s⁻¹
• (c) 5.8 × 10⁸ m s⁻¹
• (d) 5.8 m s⁻¹

Q23. Orbital angular momentum of a d electron is: (Tough)

• (a) √2 h/2π
• (b) √6 h/2π
• (c) √12 h/2π
• (d) 2 h/2π

Q24. Ratio of radii of the third orbit of Li²⁺ and the first orbit of H: (Tough)

• (a) 1 : 3
• (b) 3 : 1
• (c) 9 : 1
• (d) 1 : 9

Q25. Shortest wavelength in the Lyman series of H: (Tough)

• (a) 91.2 nm
• (b) 121.6 nm
• (c) 656 nm
• (d) 364.6 nm

Answer key with explanations

1. (b) Millikan — Oil-drop experiment.
2. (b) Chadwick — 1932.
3. (a) 52.9 pm — r₁ = 52.9 pm.
4. (b) visible region — n₁ = 2.
5. (c) 9 — n².
6. (b) dumbbell — Two lobes.
7. (c) n=2, l=2, mₗ=0 — l < n.
8. (b) −3.4 eV — −13.6/4.
9. (c) 2 — n − l − 1 = 2.
10. (b) [Ar] 3d⁵ 4s¹ — Half-filled d.
11. (b) 656 nm — Red line.
12. (b) frequency — KE = h(ν − ν₀).
13. (c) 15 — 6 × 5/2.
14. (c) 4d — l = 2 → d.
15. (b) 4s — n + l = 4 vs 5.
16. (b) momentum — λ = h/p.
17. (c) −54.4 eV — Z² = 4.
18. (c) 3 — l = 3.
19. (b) 9 — 2p⁶ + 3p³.
20. (a) 4 → 2 — Z²(1/n₁² − 1/n₂²) equal.
21. (b) 122.4 eV — × Z² = 9.
22. (a) 5.8 × 10⁵ m s⁻¹ — h/(4πmΔx).
23. (b) √6 h/2π — √[l(l+1)] with l = 2.
24. (b) 3 : 1 — n²/Z = 9/3.
25. (a) 91.2 nm — Series limit: 1/λ = Rₕ.

Check-yourself answers

• C1. Balmer.
• C2. Six.
• C3. Two (4 − 1 − 1).
• C4. 18 (2n²).

One-page revision

Read this page the night before the exam.

Particles and models

• e = 1.602 × 10⁻¹⁹ C; mₑ = 9.109 × 10⁻³¹ kg
• Rutherford: tiny dense nucleus
• Isotopes (Z), isobars (A), isotones (n)
• Rutherford fails: radiation, spectra

Light

• c = νλ; E = hν = hc/λ
• Photoelectric: hν = hν₀ + ½mv²
• Intensity → number; frequency → KE
• Visible 400–750 nm

Bohr

• rₙ = 52.9 n²/Z pm
• Eₙ = −13.6 Z²/n² eV = −2.18 × 10⁻¹⁸ Z²/n² J
• mvr = nh/2π
• One-electron species only

Spectrum

• 1/λ = Rₕ Z² (1/n₁² − 1/n₂²)
• Lyman UV, Balmer visible, Paschen IR
• Lines = n(n − 1)/2
• Hα 656 nm; Lyman limit 91.2 nm

Waves and uncertainty

• λ = h/mv
• Δx · mΔv ≥ h/4π
• 2πr = nλ in Bohr orbits
• Large objects: negligible

Quantum numbers

• n, l (0 to n−1), mₗ (2l+1), mₛ ±½
• Radial n − l − 1; angular l
• (n + l) rule; Pauli; Hund
• Cr 3d⁵4s¹, Cu 3d¹⁰4s¹ — exchange energy