Rutherford's nuclear atom

Before the nucleus: the plum pudding

J. J. Thomson found the electron in 1897. Atoms are neutral, so he pictured an atom as a ball of positive charge with electrons stuck in it, like plums in a pudding. The ball is about $10^{-10}\ \text{m}$ across.

How would such an atom treat a fast charged particle? The positive charge is spread out, so its pull at any one place is weak. A particle that crosses the ball is bent by a tiny angle only. Even after many atoms in a foil, the small bends mostly cancel and add up to a few degrees.

The experiment of Geiger and Marsden

In 1909 Hans Geiger and Ernest Marsden, working with Ernest Rutherford, fired alpha particles at a thin gold foil. An alpha particle has two protons and two neutrons, so its charge is $+2e$. Their particles had an energy of $5.5\ \text{MeV}$. The foil was about $2.1 \times 10^{-7}\ \text{m}$ thick, a few hundred atoms deep.

A screen coated with zinc sulphide flashed where an alpha particle hit it. A microscope on a turntable counted the flashes at each angle.

What they saw:

• Most alpha particles went straight through, bent by less than a degree.
• A small share were deflected by large angles.
• About one particle in 8000 was deflected by more than $90^\circ$. Some came almost straight back.

Rutherford said it was as strange as firing a heavy shell at tissue paper and seeing it bounce back.

What it means

A big turn needs a strong force on one alpha particle. That means a lot of positive charge packed into a very small space. So Rutherford proposed the nuclear atom:

• All the positive charge, and nearly all the mass, sit in a tiny nucleus at the centre.
• The electrons are outside it, and the atom is mostly empty space.

How close does an alpha particle get?

Aim an alpha particle straight at a nucleus, a head-on hit. The electric force slows it down. At the distance of closest approach, $r_0$, it stops for an instant. All its kinetic energy $K$ has become electric potential energy:

The constant $\dfrac{e^2}{4\pi\varepsilon_0}$ is $1.44\ \text{MeV fm}$. One femtometre, 1 fm, is $10^{-15}\ \text{m}$. With $K$ in MeV, this gives $r_0 = \dfrac{2Z \times 1.44}{K}$ in fm.

Rutherford's own case. Alpha particles of $5.5\ \text{MeV}$ hit gold, with $Z = 79$.

1. Closest approach. $r_0 = \dfrac{2 \times 79 \times 1.44\ \text{MeV fm}}{5.5\ \text{MeV}} = 41\ \text{fm}$, which is $4.1 \times 10^{-14}\ \text{m}$.
2. Compare. A gold nucleus has a radius of only about $7\ \text{fm}$. So the alpha particle stops well outside it. Only the electric force acts.
3. The atom. Its radius is near $0.14\ \text{nm}$. That is about $20{,}000$ times the radius of the nucleus.

Impact parameter and angle

Most alpha particles do not aim at a nucleus. The impact parameter $b$ is the distance from the nucleus to the line along which the alpha particle first moves. A small $b$ is a close pass and a big turn. The angle of deflection $\theta$ follows

• As $b \to 0$, $\theta \to 180^\circ$: a head-on hit bounces straight back.
• At $b = \dfrac{r_0}{2}$, $\theta = 90^\circ$.
• For a large $b$, $\theta$ is tiny. The particle hardly bends.

So only the few alpha particles that pass within about $\dfrac{r_0}{2}$ of a nucleus turn by $90^\circ$ or more. The nucleus is so small that this is rare. That is why only about one particle in 8000 turned back.

Keep the energy and the charge, and slide the impact parameter. Where does the angle pass $90^\circ$? Now raise the energy. Does the curve move up or down?

Why the electrons cannot simply orbit

Rutherford's atom raises a new puzzle. Put the electrons in orbits round the nucleus, like planets round the Sun. An electron on a circle is always accelerating. Classical electromagnetism says an accelerating charge gives out energy as radiation.

So the electron would lose energy and spiral into the nucleus in about $10^{-11}\ \text{s}$. As it fell, it would give out light of every colour in turn. Real atoms last for ever, and they give out only certain colours. We need a new idea. It is Bohr's, in the next lesson.

Practise with new numbers

Try it in Python

The program finds the closest approach of $5.5\ \text{MeV}$ alpha particles to four nuclei, and the size of each nucleus.

Common mistakes

• Forgetting the charge $2e$ of the alpha particle. The top line of $r_0$ has $2Ze^2$, not $Ze^2$.
• Thinking most alpha particles bounce back. Most went straight on. Only about one in 8000 turned back.
• Doubling $\tan\dfrac{\theta}{2}$ by mistake. The formula gives the half angle. Find $\theta$ by doubling the angle.
• Thinking the nucleus is the atom. The atom is about 20,000 times wider than its nucleus.

Remember these

• Geiger and Marsden: most alpha particles go straight, a few turn a lot, about one in 8000 turns back.
• Nuclear atom: all the positive charge and nearly all the mass in a tiny nucleus.
• Closest approach: $r_0 = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}$, with $\dfrac{e^2}{4\pi\varepsilon_0} = 1.44$ MeV fm.
• $\tan\dfrac{\theta}{2} = \dfrac{r_0}{2b}$. A deflection of $90^\circ$ needs $b = \dfrac{r_0}{2}$.
• A classical orbiting electron radiates and spirals in.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which result of the experiment could the plum pudding model not explain?

2. An alpha particle passes a nucleus with an impact parameter $b = \dfrac{r_0}{2}$. By how many degrees is it deflected?

3. The energy of the alpha particles is doubled. What happens to the closest approach in a head-on hit?

4. Why would an electron circling a nucleus, by classical physics, spiral inwards?