Quantum Mechanics and the Atom
In 1927, Clinton Davisson and Lester Germer at Bell Telephone Laboratories in New Jersey accidentally produced electron diffraction when a vacuum accident allowed their nickel target to form large crystals. Scattering 54 eV electrons off the crystal surface, they observed a peak at 50° whose angle matched the de Broglie wavelength λ = h/p = 167 pm with the nickel lattice spacing of 215 pm via the Bragg condition. Clinton Davisson was awarded the Nobel Prize in Physics in 1937. The experiment confirmed de Broglie's 1924 hypothesis that matter has wave-like properties.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Bohr's model worked for hydrogen but failed for helium and all other multi-electron atoms.
Before reading on, answer:
- Why might a model that works for one electron fail for two or more electrons?
- What might happen if we treat the electron as a wave rather than a particle?
- How could a wave-like electron explain why only certain orbits are allowed?
Warm-up: Bohr's model of the hydrogen atom assigned electrons to allowed orbits using which condition?
Know — Wave-Particle Duality
- de Broglie wavelength $\lambda = h/p$
- Matter waves and standing wave orbits
- Davisson-Germer confirmation
Understand — Quantum Model
- Schrödinger equation and orbitals
- Probability distributions ($|\psi|^2$)
- Four quantum numbers
Can Do — Apply Quantum Ideas
- Calculate de Broglie wavelength
- Explain quantisation from standing waves
- Apply Heisenberg uncertainty principle
Core Content
If light can be a particle, can matter be a wave?
In 1927, Clinton Davisson and Lester Germer at Bell Telephone Laboratories aimed 54 eV electrons at a nickel crystal surface and observed a diffraction peak at 50° — the same angle predicted if the electrons had wavelength λ = 167 pm, matching the nickel crystal spacing of 215 pm via the Bragg condition. This was direct experimental proof of what Louis de Broglie had proposed in 1924: that particles have wave-like properties with de Broglie wavelength:
$$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$$where $h$ is Planck's constant, $p$ is momentum, $m$ is mass and $v$ is velocity. This was confirmed experimentally by electron diffraction in the Davisson-Germer experiment (1927), which showed electrons produce interference patterns — definitive evidence of wave behaviour.
De Broglie's insight explained Bohr's quantisation postulate. An allowed orbit is one where the electron's wave forms a standing wave around the nucleus — the circumference must equal an integer number of wavelengths:
$$2\pi r = n\lambda = n\dfrac{h}{mv}$$Rearranging gives $mvr = n\hbar$ — exactly Bohr's angular momentum quantisation condition. Quantisation emerges naturally from wave behaviour rather than being postulated.
Figure 1 — de Broglie standing waves: allowed Bohr orbits are those where the electron wave closes smoothly on itself (integer wavelengths fitting the circumference).
Calculate the de Broglie wavelength of an electron moving at $2.2\times10^6$ m/s.
Given: $h = 6.63\times10^{-34}$ J·s, $m_e = 9.11\times10^{-31}$ kg
de Broglie (1924): λ = h/p = h/(mv) — all matter has a wavelength. Davisson-Germer (1927) confirmed it by electron diffraction from nickel. Standing wave condition: 2πr = nλ naturally gives Bohr's L = nℏ — allowed orbits are those where the electron wave closes on itself with integer wavelengths.
Write the formula and the standing wave derivation of Bohr quantisation — this is a common 3-mark proof question.
An electron moves at $1.0\times10^6$ m/s ($h = 6.63\times10^{-34}$ J·s, $m_e = 9.11\times10^{-31}$ kg). Its de Broglie wavelength is approximately:
Probability replaces certainty
We just saw that de Broglie's matter waves explain Bohr's quantised orbits as standing wave resonances. That raises a question: if electrons are waves, what exactly is waving, and what does the wave tell us? This card answers it → the Schrödinger equation gives wave functions whose squared magnitude |ψ|² is a probability density — electrons occupy 3D orbitals, not exact circular paths.
In 1926, Erwin Schrödinger formulated an equation describing how quantum wave functions evolve. The time-independent Schrödinger equation for hydrogen is:
$$\hat{H}\psi = E\psi$$where $\hat{H}$ is the Hamiltonian operator (total energy), $\psi$ is the wave function, and $E$ is the energy eigenvalue. Solving this for hydrogen yields:
- Quantised energy levels: $E_n = -13.6/n^2$ eV — same as Bohr's result, but now derived from first principles.
- Probability distributions: The electron does not orbit at a fixed radius; instead, $|\psi|^2$ gives the probability per unit volume of finding the electron at any point. These probability clouds are called orbitals.
- Quantum numbers: Four numbers define each electron's state:
Principal quantum number ($n$): $n = 1, 2, 3, \ldots$ — determines energy level and average distance from nucleus.
Angular momentum quantum number ($l$): $l = 0, 1, 2, \ldots, n-1$ — determines orbital shape (s, p, d, f).
Magnetic quantum number ($m_l$): $m_l = -l, -l+1, \ldots, +l$ — determines orientation of the orbital in space.
Spin quantum number ($m_s$): $m_s = \pm\tfrac{1}{2}$ — intrinsic angular momentum of the electron.
The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. This limits each spatial orbital to two electrons (opposite spins) and underpins the structure of the periodic table.
$\lambda = h/p = h/(mv)$ — de Broglie wavelength
$2\pi r = n\lambda$ — standing wave condition (Bohr orbits)
$\Delta x \cdot \Delta p \geq \hbar/2$ — Heisenberg uncertainty principle
$E_n = -13.6/n^2$ eV — hydrogen energy levels (from Schrödinger)
$n, l, m_l, m_s$ — quantum numbers; Pauli exclusion: all four unique per electron
Figure 2 — Bohr orbit (left) places the electron at an exact radius; the quantum orbital (right) describes a probability cloud — $|\psi|^2$ highest near the nucleus for the 1s state.
How many electrons can occupy the $n = 2$ shell? List all allowed combinations of quantum numbers ($n, l, m_l, m_s$) for these electrons.
Schrödinger (1926): Ĥψ = Eψ; |ψ|² = probability density — orbitals, NOT orbits. Four quantum numbers: n (energy/size), l (shape, 0 to n−1), m_l (orientation, −l to +l), m_s (±½). Pauli exclusion: no two electrons same (n,l,m_l,m_s); max 2 per orbital. Shell capacity: 2n². n=2: 8 electrons (2s + 2p).
List the four quantum numbers with their constraints — examiners ask you to count allowed states for a given n.
In the quantum mechanical model, the electron in a hydrogen atom always occupies a definite circular orbit at the Bohr radius.
The $n=3$ shell can hold a maximum of 18 electrons ($2n^2 = 18$).
The Heisenberg uncertainty principle is a limitation of our measuring instruments, not a fundamental property of nature.
Why certainty is physically impossible
We just saw that quantum orbitals replace Bohr's sharp circular orbits with 3D probability clouds. That raises a question: is the fuzziness of an orbital just a practical limitation, or is it fundamental? This card answers it → it is fundamental — Heisenberg's uncertainty principle ΔxΔp ≥ ℏ/2 is built into wave mechanics, not a measurement artefact.
The Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum cannot be smaller than $\hbar/2$:
$$\Delta x \cdot \Delta p \geq \dfrac{\hbar}{2}$$This is a fundamental property of waves: a wave with a well-defined position (narrow wave packet) necessarily has a spread of wavelengths (and hence momenta). It is not caused by clumsy measurements — it is built into the fabric of quantum mechanics.
For electrons in atoms, the uncertainty principle explains why electrons cannot occupy definite Bohr orbits: a precise circular orbit would require knowing both position and momentum exactly, which violates $\Delta x \cdot \Delta p \geq \hbar/2$. Instead, electrons occupy spread-out orbitals.
Orbit vs orbital: Bohr orbits are precise circular paths; quantum orbitals are probability clouds with no definite trajectory. This distinction is frequently tested.
Quantum number constraints: $l$ runs from $0$ to $n-1$; $m_l$ runs from $-l$ to $+l$. The $n=2$ shell has 2s ($l=0$, 2 electrons) and 2p ($l=1$, 6 electrons) — total 8 electrons.
Uncertainty principle: It is a fundamental limit, not a technological limitation. Even a perfect measuring device cannot beat it.
Pauli exclusion: Determines how many electrons each shell holds. $n=1$: 2; $n=2$: 8; $n=3$: 18; general formula $2n^2$.
Heisenberg uncertainty: ΔxΔp ≥ ℏ/2 — fundamental to wave mechanics, not a measurement limitation. A precise circular Bohr orbit requires simultaneous exact x and p, which is impossible. Shell capacity: 2n² electrons (n=1: 2; n=2: 8; n=3: 18). Pauli exclusion + Aufbau principle → periodic table structure.
Record the uncertainty formula and the key phrase "fundamental, not instrumental" — that distinction is tested every year.
An electron has de Broglie wavelength $3.3\times10^{-10}$ m. Using $p = h/\lambda$ with $h = 6.63\times10^{-34}$ J·s, the momentum of the electron in kg·m/s (give the coefficient to 2 sig figs) is _____×10⁻²⁴ kg·m/s.
Three of these statements about quantum mechanics are correct. Pick the odd one out.
Activities
Practice using $\lambda = h/(mv)$ and the standing wave condition
- Calculate the de Broglie wavelength of an electron ($m_e = 9.11\times10^{-31}$ kg) travelling at $2.2\times10^6$ m/s. ($h = 6.63\times10^{-34}$ J·s)
- A proton ($m_p = 1.67\times10^{-27}$ kg) moves at the same speed. Calculate its de Broglie wavelength and compare it to the electron's. Explain the difference.
- Using the standing wave condition $2\pi r = n\lambda$, verify that the ground state ($n=1$) Bohr orbit radius for hydrogen ($r_1 = 5.29\times10^{-11}$ m) is consistent with the de Broglie wavelength of an electron moving at $2.2\times10^6$ m/s.
- A particle has kinetic energy 100 eV ($1$ eV $= 1.6\times10^{-19}$ J). Calculate its de Broglie wavelength if it is an electron.
Apply the four quantum numbers and the Pauli exclusion principle
- List all allowed sets of quantum numbers $(n, l, m_l, m_s)$ for the $n=2$ shell. How many total states are there?
- State the ground-state electron configuration of sodium ($Z=11$) and explain how the Pauli exclusion principle and Aufbau principle determine it.
- Explain why the Heisenberg uncertainty principle means electrons cannot occupy definite Bohr orbits. Use the formula $\Delta x \cdot \Delta p \geq \hbar/2$ in your answer.
- Distinguish between a Bohr orbit and a quantum orbital. Include in your answer: dimensionality, certainty of electron position, and how each handles quantisation.
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ApplyBand 4(3 marks) 1. (a) State de Broglie's hypothesis in the form of an equation, defining all symbols. (b) Calculate the de Broglie wavelength of an electron ($m_e = 9.11\times10^{-31}$ kg) with kinetic energy 150 eV ($1$ eV $= 1.6\times10^{-19}$ J, $h = 6.63\times10^{-34}$ J·s). (c) Explain how de Broglie's hypothesis accounts for Bohr's angular momentum quantisation condition.
1 mark: correct equation with symbols · 1 mark: correct $\lambda$ · 1 mark: standing wave reasoning
AnalyseBand 6(5 marks) 2. (a) Distinguish between a Bohr orbit and a quantum mechanical orbital. (b) State the four quantum numbers and the constraints on their values. (c) How many electrons can occupy the $n=3$ shell? Show your working using quantum numbers. (d) Explain the Heisenberg uncertainty principle and why it is a fundamental limit rather than a measurement limitation. (e) State the Pauli exclusion principle and explain its significance for the structure of the periodic table.
1 mark: orbit/orbital distinction · 1 mark: four quantum numbers + constraints · 1 mark: n=3 capacity with working · 1 mark: uncertainty principle explanation · 1 mark: Pauli + periodic table
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Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks): (a) $\lambda = h/p = h/(mv)$, where $h$ is Planck's constant ($6.63\times10^{-34}$ J·s), $p$ is momentum, $m$ is mass and $v$ is velocity (1 mark). (b) $E_k = \tfrac{1}{2}m_ev^2$, so $v = \sqrt{2E_k/m_e} = \sqrt{2 \times 150 \times 1.6\times10^{-19} / 9.11\times10^{-31}} = \sqrt{5.27\times10^{13}} \approx 7.26\times10^6$ m/s. $\lambda = h/(m_ev) = 6.63\times10^{-34}/(9.11\times10^{-31}\times7.26\times10^6) = 6.63\times10^{-34}/6.61\times10^{-24} \approx 1.0\times10^{-10}$ m (1 mark). (c) For the electron wave to close on itself around an orbit of circumference $2\pi r$, the circumference must equal an integer number of wavelengths: $2\pi r = n\lambda = nh/(mv)$. Rearranging: $mvr = nh/(2\pi) = n\hbar$, which is exactly Bohr's angular momentum quantisation $L = n\hbar$ (1 mark).
Q2 (5 marks): (a) A Bohr orbit is a definite, sharp, two-dimensional circular path — the electron has an exact position and speed at every instant. A quantum orbital is a three-dimensional probability distribution ($|\psi|^2$) with no definite path — it describes the probability per unit volume of finding the electron at a given location (1 mark). (b) Principal $n = 1, 2, 3, \ldots$; Angular momentum $l = 0, 1, \ldots, n-1$; Magnetic $m_l = -l, -l+1, \ldots, +l$; Spin $m_s = +\tfrac{1}{2}$ or $-\tfrac{1}{2}$ (1 mark). (c) For $n=3$: $l$ can be 0 (1 orbital), 1 (3 orbitals), or 2 (5 orbitals) — total 9 orbitals. Each holds 2 electrons (opposite spins). Maximum electrons $= 2\times9 = 18 = 2n^2 = 2\times9 = 18$ (1 mark). (d) The uncertainty principle $\Delta x \cdot \Delta p \geq \hbar/2$ states that the product of positional and momentum uncertainties cannot be zero simultaneously. This is fundamental because a localised wave (small $\Delta x$) must be a superposition of many wavelengths, giving large $\Delta p$. No improvement in instruments can overcome this — it is a property of waves, not of measurement (1 mark). (e) The Pauli exclusion principle states that no two electrons in an atom can share the same set of four quantum numbers. Because $m_s = \pm\tfrac{1}{2}$ only, each spatial orbital holds at most two electrons. This determines how many electrons fill each subshell and shell, producing the repeating pattern of chemical properties that defines the periodic table (1 mark).
At the start you were asked about the Davisson-Germer experiment at Bell Telephone Laboratories in 1927 — 54 eV electrons diffracting from a nickel crystal at the angle predicted by λ = h/p = 167 pm. This proved de Broglie's 1924 matter wave hypothesis. Review your predictions about what wave behaviour means for atomic structure:
- Did you predict that treating electrons as waves produces natural quantisation via the standing wave condition 2πr = nλ? Correct — the Davisson-Germer diffraction pattern is the direct physical evidence that electrons behave as waves, which is why Bohr's postulated orbits are actually standing wave resonances.
- Did you predict multi-electron model failure because electron-electron repulsion is not accounted for? Correct — Bohr's model has no mechanism for electron-electron interactions, which significantly alters energy levels beyond hydrogen.
- Did you predict wave-like electrons explain allowed orbits via constructive interference? Correct — only orbits where the wave closes on itself are stable; non-integer wavelengths produce destructive interference and are forbidden.
Extend: (a) An electron is confined to a box of width $L = 1.0\times10^{-10}$ m (a rough atomic scale). Using the uncertainty principle, estimate the minimum momentum uncertainty and hence the minimum kinetic energy of this electron ($m_e = 9.11\times10^{-31}$ kg, $\hbar = 1.055\times10^{-34}$ J·s). (b) Explain qualitatively why this "zero-point energy" means electrons in atoms can never be completely at rest. (c) How does the Pauli exclusion principle explain why white dwarf stars don't collapse under gravity?
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