Structure of Atom · Lesson 7 of 12

Matter Waves and the Limits of Orbits

“Discover why electron waves and uncertainty rule out a precise orbital path.”

Learning Objectives

• Use the de Broglie relation to connect particle momentum and wavelength. • Compare observable microscopic wave behaviour with negligible macroscopic wavelengths. • Explain electron diffraction and the basis of electron microscopy. • Calculate uncertainty bounds and explain their physical significance. • Connect matter waves and uncertainty with the failure of definite Bohr trajectories.

Extending duality from radiation to matter

Light had already challenged a simple division between particles and waves. De Broglie proposed that matter also has both aspects. An electron can arrive as a localised detection event and yet produce diffraction patterns associated with waves. A model that treats it only as a tiny classical particle therefore misses part of its measured behaviour.

De Broglie connected wavelength with momentum. For non-relativistic motion, momentum is mass multiplied by speed. Large momentum gives a short wavelength; small momentum gives a long wavelength. The equation does not say that an electron has become an electromagnetic wave: a matter wave describes the particle’s quantum behaviour.

De Broglie wavelengthLaTeX

λ = \frac{h}{p} = \frac{h}{m v}

p is momentum. The form p = mv applies in the non-relativistic calculations used here. h is Planck’s constant.

Electron diffraction confirmed the prediction experimentally. An electron microscope uses the wave behaviour of electrons, just as a light microscope relies on the wave behaviour of light. Short electron wavelengths allow much finer spatial detail to be resolved under appropriate conditions. Practical resolution also depends on the instrument; wavelength alone is not the only limitation.

Allowed standing waves around a model orbit— Combining de Broglie wavelength with Bohr’s angular-momentum rule fits a whole number of wavelengths around the circle.

Example — A moving ball

Problem
Find the wavelength of a 0.10 kg ball moving at 10 m/s.

1. $p = m v = 0.10 \times 10 = 1.0 kg m s^{- 1}$ .
2. $λ = h / p = 6.626 \times 1 0^{- 34} m$ .
3.This wavelength is extraordinarily smaller than atomic and nuclear sizes. The ball has quantum behaviour in principle, but an ordinary diffraction experiment cannot reveal a wavelength this small.

Example — An electron from its kinetic energy

Problem
An electron of mass $9.1 \times 1 0^{- 31} kg$ has kinetic energy $3.0 \times 1 0^{- 25} J$ . Find its wavelength.

1.From $K = m v^{2} /2$ , $v = 2 K / m = 6.0 \times 1 0^{- 25} / (9.1 \times 1 0^{- 31}) \approx 812 m s^{- 1}$ .
2.Then $λ = h / (m v) = 6.626 \times 1 0^{- 34} / [(9.1 \times 1 0^{- 31}) (812)] \approx 8.97 \times 1 0^{- 7} m$ .
3.The wavelength is about 897 nm. A slow electron can have a wavelength comparable with optical wavelengths; electrons in practical electron microscopes are usually much faster.

The same calculation can be shortened by substituting the kinetic-energy relation into momentum: $p = 2 m K$ . Therefore $λ = h / 2 m K$ . This is a rearrangement of the same two physical relationships, not a separate principle.

Example — A photon and energy-equivalent mass

Problem
A photon has wavelength $3.6 \overset{˚}{A}$ . Calculate its momentum and its energy-equivalent mass.

1.Use $λ = 3.6 \times 1 0^{- 10} m$ . Photon momentum is $p = h / λ \approx 1.84 \times 1 0^{- 24} kg m s^{- 1}$ .
2.Because $E = h c / λ$ , its energy-equivalent mass is $E / c^{2} = h / (λ c) \approx 6.14 \times 1 0^{- 33} kg$ .
3.A photon’s rest mass remains zero. The calculated energy-equivalent mass is not a rest mass, and the non-relativistic equation p = mv must not be applied to a photon.

Reading a historical calculation carefully

The source uses “mass of a photon” in a worked calculation. Here that phrase is made precise as energy-equivalent mass, E/c². Keeping the distinction avoids teaching a photon as a massive classical particle.

Example — Connecting wave fitting with Bohr quantisation

Problem
Show how the circumference of an allowed Bohr orbit relates to de Broglie wavelength.

1.Bohr’s condition is $m v r = nh / (2 π)$ . Multiply by $2 π / (m v)$ to obtain $2 π r = nh / (m v)$ .
2.De Broglie gives $h / (m v) = λ$ , so $2 π r = nλ$ .
3.A whole number of wavelengths fits around the circumference. This gives a wave interpretation of the quantisation rule, while a full quantum model is still needed to describe electron probability.

The uncertainty principle

A classical trajectory needs both a position and a momentum at each instant. Heisenberg showed that quantum states do not permit these quantities to be simultaneously arbitrarily sharp. Narrowing the position distribution requires a wider momentum distribution, and vice versa.

The source’s microscope illustration gives intuition: short-wavelength light can resolve a small region, but its high-momentum photons interact strongly with an electron. However, uncertainty is deeper than a disturbance caused by a badly designed instrument. It is a property of quantum states, consistent with their wave character; an ideal instrument does not remove it.

Position–momentum uncertaintyLaTeX

Δ x Δ p_{x} \geq \frac{ℏ}{2} = \frac{h}{4 π}

The uncertainties refer to position and momentum in the same direction. ℏ = h/(2π). The inequality gives a lower bound, not an exact value in every state.

Using a speed uncertaintyLaTeX

Δ x Δ v_{x} \geq \frac{h}{4 π m}

Use this form for a particle of fixed mass with non-relativistic momentum. Δv is the uncertainty in one velocity component, not the speed itself.

The mass in the denominator explains why ordinary objects appear to have well-defined positions and speeds. The allowed uncertainty product in position and velocity becomes extremely small for a large mass. For an electron the same product is much larger and can be important on atomic length scales.

Example — Locating an electron closely

Problem
An electron is located within a position uncertainty of $0.1 \overset{˚}{A}$ . Find the minimum corresponding velocity uncertainty.

1.Convert $Δ x = 1.0 \times 1 0^{- 11} m$ . Use $m_{e} = 9.11 \times 1 0^{- 31} kg$ .
2. $Δ v_{x} \geq 6.626 \times 1 0^{- 34} / [4 π (9.11 \times 1 0^{- 31}) (1.0 \times 1 0^{- 11})]$ .
3.The lower bound is $5.79 \times 1 0^{6} m s^{- 1}$ . A very precise position does not coexist with a sharply known atomic-scale trajectory.

Example — Why a golf ball seems classical

Problem
A 40 g golf ball moves at 45 m/s. Taking its velocity uncertainty as 2% of this value, find the minimum position uncertainty.

1. $m = 0.040 kg$ and $Δ v_{x} = 0.02 \times 45 = 0.90 m s^{- 1}$ .
2. $Δ x \geq h / (4 π m Δ v_{x}) = 6.626 \times 1 0^{- 34} / [4 π (0.040) (0.90)]$ .
3.The bound is $1.46 \times 1 0^{- 33} m$ . This is negligible for ordinary measurements of the ball and does not prevent the practical use of a classical trajectory.

Why a definite orbit must be abandoned

A Bohr circle specifies where the electron is and how it moves around a fixed path. Matter-wave evidence shows that the electron cannot be described solely as that kind of particle, and uncertainty prevents assigning simultaneously exact position and momentum to trace such a path.

The replacement is not an electron following an unknown but otherwise ordinary circle. Quantum mechanics predicts probabilities for possible outcomes. We next use a wave function and an orbital to describe this probability structure, while retaining the experimentally successful idea of discrete energy states.

Common mistake

Uncertainty is not a percentage error that can always be eliminated by using a better measuring device. Do not substitute the full electron velocity for Δv unless it is explicitly the stated uncertainty. A photon has zero rest mass even though it carries momentum.

Quick check

If a non-relativistic particle’s momentum doubles, its de Broglie wavelength:

Quick check

Why are wave effects of a moving ball difficult to observe?

Quick check

What does the uncertainty principle restrict?

Quick check

If Δx is reduced by a factor of ten, the lower bound on Δp becomes:

Quick check

Which statement about an electron microscope is correct?

Practice Problems

Find the de Broglie wavelength of an electron moving at $1.6 \times 1 0^{6} m s^{- 1}$ .
A neutron has wavelength 800 pm. Calculate its speed using $m_{n} = 1.675 \times 1 0^{- 27} kg$ .
Compare wavelengths of an electron and a 0.10 kg ball moving at the same speed of $4.37 \times 1 0^{5} m s^{- 1}$ .
Calculate the minimum momentum uncertainty for an electron with $Δ x = 0.002 nm$ .
Derive the relationship between Bohr orbit circumference and de Broglie wavelength.
Explain why a wave-based model and a probability description are needed even if one can draw a neat circular electron path.

Key Takeaways

A particle’s de Broglie wavelength is inversely proportional to momentum. Electron diffraction confirms matter-wave behaviour and supports electron microscopy. Position and corresponding momentum cannot both have arbitrarily small uncertainties. The uncertainty bound is significant on microscopic scales and negligible in ordinary macroscopic tracking. A quantum orbital replaces a definite classical electron path.

Previous · Lesson 6

Bohr’s Model and Hydrogen Energy Levels

Next · Lesson 8

The Quantum Model and Quantum Numbers

Extending duality from radiation to matter

The uncertainty principle

Why a definite orbit must be abandoned

Quiz

Practice Problems

Key Takeaways