Course 2.3Lesson 1 of 4

Waves of matter and the uncertainty principle

Every moving particle has a wavelength, $\lambda = \frac{h}{mv}$

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Try the code and the simulations beside the lesson.

If light can be a particle, can a particle be a wave?

In the last course, we saw that light shows two faces. It is a wave in interference, and it is a stream of photons in the photoelectric effect. In $1924$ , a young French student, Louis de Broglie, asked a bold question. If light waves can act as particles, can particles of matter act as waves?

The de Broglie wavelength

For a photon, $E = h\nu = \dfrac{hc}{\lambda}$ , and the momentum is $p = \dfrac{E}{c}$ . So $p = \dfrac{h}{\lambda}$ , or $\lambda = \dfrac{h}{p}$ . De Broglie said that the same relation is true for any particle of mass $m$ and speed $v$ :

\displaystyle \lambda = \frac{h}{mv}

Here $h$ is Planck's constant. Because $h$ is so small, the wavelength is noticeable only when the mass is tiny.

Object	Mass	Speed	Wavelength
Electron in an atom	$9.11 \times 10^{-31}\ \text{kg}$	$2.19 \times 10^{6}\ \text{m s}^{-1}$	$3.3 \times 10^{-10}\ \text{m}$
Proton at the same speed	$1.67 \times 10^{-27}\ \text{kg}$	$2.19 \times 10^{6}\ \text{m s}^{-1}$	$1.8 \times 10^{-13}\ \text{m}$
Cricket ball	$0.16\ \text{kg}$	$40\ \text{m s}^{-1}$	$1.0 \times 10^{-34}\ \text{m}$

The wavelength of the electron, $330\ \text{pm}$ , is about the size of an atom, so it matters a great deal inside the atom. The wavelength of the cricket ball is far smaller than a nucleus, and no instrument could ever find it. This is why we never see a ball diffract.

Evidence

In $1927$ , Clinton Davisson and Lester Germer fired electrons at a crystal of nickel. The electrons were scattered in a diffraction pattern, with bright and dark places, exactly as X-rays would be. Only waves do this. Beams of neutrons and of atoms behave in the same way.

The electron microscope uses this fact. Electrons that have been accelerated through $100\,000\ \text{V}$ have a wavelength of only about $4\ \text{pm}$ , much shorter than visible light. A shorter wavelength can show much finer detail, so we can see single molecules.

The wave explains Bohr's rule

Bohr had to assume that the angular momentum is $mvr = n\hbar$ . De Broglie's idea shows why. A wave that goes around a circle must join up with itself. If it does not, the crests and troughs cancel after a few turns and the wave dies away. So the circle must hold a whole number of wavelengths:

\displaystyle 2\pi r = n\lambda = n\,\frac{h}{mv} \quad\Rightarrow\quad mvr = n\,\frac{h}{2\pi}

This is exactly Bohr's condition.

Two circles with a nucleus at the centre. In the left picture, an electron wave with exactly 3 wavelengths goes round the circle and joins up with itself. In the right picture, a wave with 3.5 wavelengths goes round the circle, and its two ends do not meet.

You should find that four times the voltage halves the wavelength.

The uncertainty principle

Suppose we want to find where an electron is. To see it, we must hit it with light of a short wavelength. But a photon of a short wavelength has a large momentum, and it kicks the electron, so its speed changes. The more precisely we find the position, the more we disturb the momentum.

In $1927$ , Werner Heisenberg stated this exactly. It is not possible to know both the position and the momentum of a small particle precisely at the same time.

\displaystyle \Delta x \times \Delta p \geq \frac{h}{4\pi}

Here $\Delta x$ is the uncertainty in the position, and $\Delta p = m\,\Delta v$ is the uncertainty in the momentum. This is not about poor instruments. It follows from the wave nature of matter, and no better instrument can remove it.

An electron. If we know the position of an electron to $\Delta x = 100\ \text{pm} = 10^{-10}\ \text{m}$ , about the width of an atom, then

\displaystyle \Delta v \geq \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times 9.11 \times 10^{-31} \times 10^{-10}} = 5.8 \times 10^{5}\ \text{m s}^{-1}

This is almost as large as the speed of the electron in the atom, $2.19 \times 10^{6}\ \text{m s}^{-1}$ . So we cannot say that the electron follows a path.

A cricket ball. If $m = 0.16\ \text{kg}$ and we know its position to $1\ \text{mm}$ , then $\Delta v \geq 3.5 \times 10^{-31}\ \text{m s}^{-1}$ . This is too small to matter. The principle is important only for very small particles.

What this means for the atom

Bohr's orbit says that the electron has a known position and a known speed at all times. The uncertainty principle says that this cannot be true. We must give up the idea of a path. What we can find is the probability of finding the electron in each region of space. This idea gives us the orbital, in the next lesson.

Practise with new numbers

Try it in Python

This program finds the de Broglie wavelength of four objects, and the uncertainty in the speed of an electron.

Remember these

$\lambda = \dfrac{h}{mv}$ . The wavelength is large enough to notice only for electrons, atoms and other very small particles.
Davisson and Germer saw electrons diffract. The electron microscope uses this fact.
A whole number of wavelengths fits around Bohr's orbit, so $mvr = n\hbar$ .
$\Delta x \times \Delta p \geq \dfrac{h}{4\pi}$ . We cannot know both the position and the momentum exactly, and an electron has no fixed path.