C. Structure and Properties of Matter · Structure and Properties of MatterThe Quantum Mechanical Model: Matter Waves, Uncertainty and Orbitals
C3.1 C3.2 C2About 70 minutes
Learning goals- I can describe the contributions of de Broglie, Heisenberg and Schrödinger to the quantum mechanical model.
- I can calculate the de Broglie wavelength of a moving particle.
- I can explain the difference between a Bohr orbit and an orbital.
- I can explain why the quantum mechanical model replaced the Bohr model.
Words to know
wave-particle duality
The idea that light and matter both show wave properties and particle properties.
de Broglie wavelength
The wavelength of a moving particle, λ = h/(mv).
uncertainty principle
Heisenberg's principle that position and momentum cannot both be known exactly at the same time.
wave function (ψ)
A mathematical solution of Schrödinger's equation that describes an electron's state; ψ² gives probability.
orbital
A region of space where an electron of a given energy has a high probability of being found.
An electron is not a tiny planet
Bohr's orbits were a brilliant step, but they were wrong in an important way. In the 1920s physicists discovered that electrons behave like waves as well as particles, and that you can never know exactly where an electron is and how it is moving at the same time. The result was the quantum mechanical model, the model of the atom used in chemistry, physics and engineering today.
You do not need the advanced mathematics behind it, but you do need its key ideas, because they explain the periodic table and chemical bonding.
de Broglie: matter waves (1924)
Louis de Broglie reasoned that if light waves can behave like particles, then particles might behave like waves. He proposed that any moving particle has a wavelength
λ = h / (mv)
where m is the mass and v is the speed. For everyday objects the wavelength is far too small to notice. For an electron, it is about the size of an atom. In 1927 experiments showed electrons diffracting off crystals, just like waves, confirming de Broglie's idea. (Electron microscopes depend on this wave behaviour.)
de Broglie's idea also explains Bohr's allowed orbits. A wave trapped in a small space can only exist as a standing wave with a whole number of half-wavelengths, like a guitar string. Only certain wave patterns fit, so only certain energies are allowed.
Heisenberg: the uncertainty principle (1927)
Werner Heisenberg showed that it is impossible to know both the exact position and the exact momentum (mass × velocity) of a particle at the same time. Mathematically, Δx · Δp ≥ h/(4π). To "see" an electron, you must bounce a photon off it, and that photon changes the electron's motion.
This is not a problem with our instruments. It is a basic feature of nature. It means a neat circular orbit, with an exact path at every instant, cannot describe an electron.
Schrödinger: orbitals (1926)
Erwin Schrödinger wrote an equation that treats the electron in an atom as a wave. Its solutions, called wave functions (ψ, psi), describe allowed energy states. The square of the wave function, ψ², gives the probability of finding the electron at each point in space.
An orbital is a region of space around the nucleus where there is a high probability (usually drawn as 90%) of finding an electron of a particular energy. Think of a long-exposure photograph of a moving hummingbird: you cannot say exactly where it is, but you can see where it spends most of its time.
What survives from Bohr?
- Energy levels are quantized. This is still true; the main energy levels are labelled n = 1, 2, 3, ...
- Circular orbits are gone. Electrons occupy 3-D orbitals of different shapes, described by probability.
- Each main level contains sublevels (s, p, d, f) with slightly different energies. This explains the extra lines and splitting in spectra that Bohr could not explain, and it works for every atom, not just hydrogen.
| Bohr model (1913) | Quantum mechanical model (1926 onward) |
|---|
| Electron is | a particle in a circular orbit | a wave-particle described by a wave function |
| Location | exact path at a fixed radius | probability cloud (orbital) |
| Energy | quantized levels n = 1, 2, 3 ... | quantized levels and sublevels (s, p, d, f) |
| Works for | hydrogen (one electron) only | all atoms and molecules |
| Why levels are allowed | assumed | follows from wave behaviour |
Worked exampleExample 1: de Broglie wavelength of an electron
An electron (m = 9.11 × 10⁻³¹ kg) moves at 2.2 × 10⁶ m/s. Find its wavelength and compare it with the size of an atom (about 10⁻¹⁰ m).
- λ = h/(mv) = 6.626 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.2 × 10⁶).
- Denominator: 9.11 × 10⁻³¹ × 2.2 × 10⁶ = 2.004 × 10⁻²⁴ kg·m/s.
- λ = 6.626 × 10⁻³⁴ ÷ 2.004 × 10⁻²⁴ = 3.31 × 10⁻¹⁰ m.
λ ≈ 3.3 × 10⁻¹⁰ m (0.33 nm), about the size of an atom, so wave behaviour matters for electrons in atoms.
Worked exampleExample 2: de Broglie wavelength of a hockey puck
A 0.17 kg hockey puck slides at 40 m/s. Find its wavelength. Why do we never notice it?
- λ = h/(mv) = 6.626 × 10⁻³⁴ ÷ (0.17 × 40).
- mv = 6.8 kg·m/s.
- λ = 6.626 × 10⁻³⁴ ÷ 6.8 = 9.7 × 10⁻³⁵ m.
λ ≈ 9.7 × 10⁻³⁵ m, unimaginably smaller than even a nucleus (10⁻¹⁵ m). No net or goal could ever diffract it, so wave effects are invisible for large objects.
Worked exampleExample 3: Orbit or orbital?
A classmate writes: "In the 1s orbital, the electron travels in a circle 53 pm from the nucleus." Correct the statement.
- 53 pm is Bohr's radius for hydrogen's n = 1 orbit.
- In the quantum model, 53 pm is the most probable distance for a 1s electron, but the electron can be found closer or farther.
- An orbital is a 3-D region of probability; the electron has no definite path.
"A 1s electron is most likely to be found about 53 pm from the nucleus, but it does not follow a path; the orbital is a spherical probability region."
Just for funHeisenberg gets pulled over for speeding. The officer asks, "Do you know how fast you were going?" Heisenberg replies, "No, but I know exactly where I am!"
Hands-on activityMapping a probability cloud
You need: a large sheet of paper (or several taped together), a marker, a small dried bean or a crumpled ball of paper, tape, and a ruler.
- Draw a small cross in the middle of the paper: this is the nucleus. Put the paper on the floor.
- Stand over the cross and, without aiming carefully, drop the bean from about chest height 50 times. Each time, mark a dot where it first lands (ignore bounces if you can).
- Draw circles centred on the cross at 5 cm, 10 cm, 15 cm, 20 cm and so on. Count the dots in each ring (0-5 cm, 5-10 cm, ...).
- Make a bar graph: number of dots against distance.
What to look for: the dots are densest near the centre, but there is no single "track". Your graph of dots per ring probably rises and then falls, because outer rings have more area. That is just like the 1s probability graph, which peaks at 53 pm even though the electron density is highest at the nucleus. Draw a circle that contains about 90% of your dots: that is your "orbital" boundary.
Your turn
1State de Broglie's hypothesis in one sentence.
Every moving particle has a wave nature, with wavelength λ = h/(mv).
2What experimental evidence supported de Broglie's idea?
Electrons were observed to diffract (spread out in a wave pattern) when passed through or reflected from crystals (1927).
3State the Heisenberg uncertainty principle and explain why it rules out Bohr's orbits.
You cannot know both the exact position and exact momentum of an electron at once. A Bohr orbit gives an exact path (position and velocity at every instant), which is impossible.
A 3-D region around the nucleus where there is a high probability (about 90%) of finding an electron with a certain energy.
5Give one idea that the quantum model kept from Bohr and one that it replaced.
Kept: quantized energy levels (n = 1, 2, 3 ...). Replaced: circular orbits (now probability orbitals).
6Find the de Broglie wavelength of a proton (1.67 × 10⁻²⁷ kg) moving at 1.0 × 10⁵ m/s.
λ = 6.626 × 10⁻³⁴ ÷ (1.67 × 10⁻²⁷ × 1.0 × 10⁵) = 6.626 × 10⁻³⁴ ÷ 1.67 × 10⁻²² = 3.97 × 10⁻¹² m.
More practice: warm-up
1Who wrote the wave equation for the electron?
Erwin Schrödinger (1926).
The probability of finding the electron at a point.
3What is the value of Planck's constant?
4Is an electron a wave or a particle?
Both: it shows wave behaviour (diffraction) and particle behaviour (it has mass and charge and hits a screen at one spot).
5What letters name the sublevels?
More practice: on your own
1Why are standing waves a good analogy for the allowed energy levels?
A standing wave can only have a whole number of half-wavelengths between fixed boundaries, so only certain patterns (and energies) are possible, just as only certain electron energies are allowed.
2A tennis ball (0.058 kg) moves at 50 m/s. Find its wavelength.
λ = 6.626 × 10⁻³⁴ ÷ (0.058 × 50) = 6.626 × 10⁻³⁴ ÷ 2.9 = 2.3 × 10⁻³⁴ m.
3What speed would an electron need to have a wavelength of 1.0 × 10⁻⁹ m?
v = h/(mλ) = 6.626 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 1.0 × 10⁻⁹) = 7.3 × 10⁵ m/s.
4Explain the hummingbird analogy for an orbital.
A long-exposure photo of a hummingbird shows a blur where it is most often found, not its exact position at any instant. An orbital similarly shows where the electron is likely to be, not a path.
5Why did scientists accept the quantum mechanical model over the Bohr model?
It explains the spectra and properties of all atoms (not just hydrogen), explains why levels are quantized, and agrees with experiments such as electron diffraction.
Challenge
1If the uncertainty in an electron's position is 1.0 × 10⁻¹⁰ m (the size of an atom), find the minimum uncertainty in its speed. Use Δx · mΔv ≥ h/(4π).
Δv ≥ h ÷ (4π m Δx) = 6.626 × 10⁻³⁴ ÷ (4 × 3.1416 × 9.11 × 10⁻³¹ × 1.0 × 10⁻¹⁰) = 6.626 × 10⁻³⁴ ÷ 1.145 × 10⁻³⁹ = 5.8 × 10⁵ m/s. The uncertainty in speed is enormous, so a definite orbit is meaningless.
2In Bohr's n = 1 orbit, the circumference (2πr, with r = 5.29 × 10⁻¹¹ m) equals exactly one de Broglie wavelength. Find the electron's speed.
λ = 2πr = 2 × 3.1416 × 5.29 × 10⁻¹¹ = 3.324 × 10⁻¹⁰ m. v = h/(mλ) = 6.626 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 3.324 × 10⁻¹⁰) = 2.19 × 10⁶ m/s, which matches Bohr's own result.
3Explain how the quantum model can say the 1s electron density is greatest at the nucleus, while the most probable distance is 53 pm.
Density is probability per unit volume, highest at the centre. But a thin shell at a larger radius has much more volume than a tiny region at the centre. Multiplying density (falling with distance) by shell volume (rising with r²) gives a total probability that peaks at 53 pm.
Quiz
1. What does an orbital represent in the quantum mechanical model?
2. Which scientist proposed that particles such as electrons have wavelengths?
3. The uncertainty principle states that it is impossible to know exactly, at the same time, an electron's:
4. Why don't we notice the wave nature of a moving car?
5. Which idea from the Bohr model is still part of the quantum mechanical model?
6. What is the de Broglie wavelength of an electron moving at 1.0 × 10⁷ m/s?
Big idea (C3.1, C3.2): electrons behave as waves, so they cannot have the neat orbits of the Bohr model. The modern model describes orbitals, regions of probability, while keeping the idea of quantized energy levels.
How to help: this lesson is conceptual and many adults find it strange too, which is fine. Ask your teen to explain the difference between an orbit and an orbital in their own words. The bean-drop activity makes the idea of probability concrete.
Common difficulties:
- Saying the electron "moves around inside the orbital" along a path. It has no definite path.
- Thinking uncertainty is caused by poor equipment; it is a property of nature.
- Calculator errors in λ = h/(mv): put brackets around the denominator.