Course 2.1Lesson 2 of 3

Rutherford's nuclear atom and the size of the nucleus

Rutherford fired α particles at gold foil. Most went straight, a few were turned through large angles. So the positive charge and almost all the mass are in a tiny nucleus of radius $R = 1.2\ \text{fm} \times A^{1/3}$

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Try the code and the simulations beside the lesson.

What was the atom like before 1911?

Thomson pictured the atom as a sphere of positive charge, with the electrons placed in it like seeds in a watermelon. In this model the positive charge is spread over the whole atom, and nothing in it is dense or hard.

Rutherford's experiment

In 1909, Hans Geiger and Ernest Marsden worked in Rutherford's laboratory. They placed a source of alpha particles in front of a very thin foil of gold, and they put a screen coated with zinc sulphide around it. Each alpha particle that hit the screen made a tiny flash of light, and they counted the flashes through a microscope.

An alpha particle is a helium nucleus, with charge $+2e$ and a mass of about $4\ \text{u}$ . It is a heavy projectile, travelling at about $2 \times 10^{7}\ \text{m s}^{-1}$ .

What did they see?

Most of the alpha particles went straight through the foil.
A small number were turned through small angles.
A very few, about $1$ in $20\,000$ , were turned back, through angles close to $180^\circ$ .

Rutherford said later that it was as if you fired a large shell at a piece of tissue paper and it bounced back.

What it means

Thomson's atom could not do this. A spread-out positive charge is too weak to turn a fast alpha particle back. The result needs a very strong electric force, and that comes only from a charge that is small in size and large in amount.

Most of the atom is empty space, because most of the particles went straight through.
The positive charge and nearly all of the mass are packed in a very small nucleus. This is why a few particles are turned by large angles.
The electrons are outside the nucleus.
The nucleus is very small compared with the atom, because only a few particles in $20\,000$ came close enough to it.

Paths of alpha particles that pass a gold nucleus at different distances. All come from the left and are pushed away by the positive nucleus. The paths that pass very close to the nucleus bend by a large angle, almost back toward the source. The paths that pass far from the nucleus bend only a little.

The picture shows what happens to the alpha particles. All of them are pushed away by the positive nucleus. The force gets stronger as the particle comes closer, so a particle that passes near the nucleus is turned a lot. A particle that passes far away is hardly turned at all.

How big is the nucleus?

Take a gold nucleus. It has $79$ protons, so the charge is $+79e$ . A head-on alpha particle slows down as it climbs up the electric force, and it stops for a moment at the distance of closest approach $d$ . At that moment all its kinetic energy has become electric potential energy:

\displaystyle KE = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{d} \quad\Rightarrow\quad d = \frac{2Z}{KE} \cdot \frac{e^2}{4\pi\varepsilon_0}

The value of $\dfrac{e^2}{4\pi\varepsilon_0}$ is $1.44\ \text{MeV fm}$ , where $1\ \text{fm} = 10^{-15}\ \text{m}$ is one femtometre (also called one fermi). For alpha particles of $5.5\ \text{MeV}$ :

\displaystyle d = \frac{2 \times 79 \times 1.44}{5.5} = 41\ \text{fm} = 4.1 \times 10^{-14}\ \text{m}

So the alpha particles never came nearer than $4 \times 10^{-14}\ \text{m}$ , and the Coulomb force law still worked. Rutherford concluded that the nucleus is smaller than this, so it is less than about $10^{-14}\ \text{m}$ in radius. The atom itself has a radius of about $10^{-10}\ \text{m}$ .

Later experiments with faster particles found that the radius of a nucleus follows a simple rule:

\displaystyle R = R_0\, A^{1/3}, \qquad R_0 = 1.2 \times 10^{-15}\ \text{m} = 1.2\ \text{fm}

Here $A$ is the mass number. For gold, $A = 197$ , so $R = 1.2 \times 197^{1/3} = 7.0\ \text{fm}$ . The alpha particle with $41\ \text{fm}$ of closest approach was still far outside it.

You should find that for a heavy nucleus such as gold, alpha particles of a few MeV stop far outside it, so the Coulomb law still holds.

The nucleus is very dense

Since the radius goes as $A^{1/3}$ , the volume is proportional to $A$ . The mass is also proportional to $A$ , so every nucleus has the same density:

\displaystyle \rho = \frac{A \times 1.661 \times 10^{-27}\ \text{kg}}{\frac{4}{3}\pi (1.2 \times 10^{-15}\ \text{m})^3 \, A} = 2.3 \times 10^{17}\ \text{kg m}^{-3}

A matchbox full of nuclear matter would have a mass of about $10^{13}\ \text{kg}$ . The width of the atom is about $10^{5}$ times the width of its nucleus. So if the atom were a stadium, the nucleus would be a tiny bead at the centre.

Where the model fails

Rutherford's model was a very large step, but it has two problems.

The atom should collapse. An electron that circles the nucleus is always accelerating. By the laws of electricity, an accelerating charge should give out energy as light. The electron would lose energy and spiral into the nucleus in about $10^{-8}\ \text{s}$ . Real atoms last for ever.
It says nothing about the arrangement of the electrons, and it cannot explain why an atom gives out light of only certain colours.

These two problems lead us to Bohr's model, which we meet in the next course.

Practise with new numbers

Try it in Python

This program finds the radius and the density of five nuclei.

Remember these

Most alpha particles pass through gold foil. A very few are turned back. So the atom is mostly empty, and it has a small, dense, positive nucleus.
The radius of the nucleus is $R = 1.2\ \text{fm} \times A^{1/3}$ . The radius of an atom is about $10^{5}$ times larger.
All nuclei have the same density, about $2.3 \times 10^{17}\ \text{kg m}^{-3}$ .
In Rutherford's model, the orbiting electrons should lose energy and fall into the nucleus. This is why the model had to be improved.