What an atom had to contain
By 1900 two facts about atoms were known. An atom is electrically neutral, and it contains electrons, which are very light and negatively charged. So an atom must also hold an equal positive charge, and almost all of its mass must be somewhere other than in its electrons. The question was where.
J. J. Thomson, who had discovered the electron, offered an answer in 1904. The positive charge fills the whole atom, a sphere about \(10^{-10}\,\text{m}\) across, like the red flesh of a watermelon. The electrons sit inside it like the seeds. In this picture the charge and the mass are spread thinly through the whole atom.
Probing the atom with alpha particles
To test such a picture, you need a probe small and fast enough to enter an atom. Ernest Rutherford chose the alpha particle, which some radioactive atoms emit. It is the nucleus of a helium atom, with charge \(+2e\) and about 7300 times the mass of an electron. It leaves its source with a kinetic energy of a few MeV, at about \(2 \times 10^7\,\text{m/s}\).
In the experiment a narrow beam of alpha particles struck a gold foil only a few hundred atoms thick. A screen coated with zinc sulphide surrounded the foil. Each alpha particle that hit the screen made a tiny flash. An observer in a dark room counted the flashes through a microscope, angle by angle.
The counts showed three things:
- Almost all the alpha particles passed straight through the foil, or turned by a degree or two.
- A small fraction turned through large angles. About 1 in 8000 turned through more than \(90°\).
- A very few came almost straight back towards the source.
Only a nucleus can turn an alpha particle back
Think about what could turn a fast, heavy alpha particle back. The electrons cannot: an alpha particle meeting an electron is like a cricket ball meeting a mosquito. The positive charge of Thomson's atom cannot either. Spread over a whole atom, its push is weak at every point. It could turn an alpha particle by only a small fraction of a degree.
A strong push needs the alpha particle to come very close to a large charge. So Rutherford concluded in 1911 that all the positive charge, and nearly all the mass, sits in a tiny nucleus at the centre. An alpha particle that happens to pass close to a nucleus feels an enormous repulsion and turns sharply. Everything else in the atom is nearly empty space, which is why most alpha particles pass straight through.
An atom has a tiny central nucleus that holds all its positive charge and nearly all its mass. The electrons move around the nucleus, far outside it, and the atom between them is nearly empty.
The line along which an alpha particle would travel if nothing pushed it passes the nucleus at some distance \(b\), the impact parameter. In the figure above, an alpha particle with \(b = 60\,\text{fm}\) is turned through only \(28°\). One with \(b = 15\,\text{fm}\) is turned through about \(90°\). One aimed straight at the nucleus, \(b = 0\), stops and comes straight back.
Now picture the whole foil. A gold atom is about \(10^{5}\,\text{fm}\) across, so the paths drawn in the figure fill a tiny patch at its centre. An alpha particle aimed at random almost always has a large \(b\) and hardly turns. Only the rare one aimed within a few tens of femtometres of a nucleus turns through a large angle. That is exactly the pattern of the counts.
Rutherford worked out the whole pattern from Coulomb's law. An alpha particle of impact parameter \(b\) turns through an angle \(\theta\) with \(\tan(\theta/2) = d/2b\). Here \(d\) is the distance of closest approach of the next topic. The number reaching an angle \(\theta\) falls as \(1/\sin^4(\theta/2)\), and it grows as \(Z^2\) of the target. Geiger and Marsden's counts followed this law while they changed by a factor of more than a hundred thousand.
In the gold foil experiment, most alpha particles crossed the foil with almost no turn. What does this show?
Which single observation forced Rutherford to put the positive charge in a tiny nucleus?
In 1909 Ernest Marsden was a twenty-year-old student in Rutherford's laboratory. Rutherford and Hans Geiger asked him to check whether any alpha particles bounced back from a gold foil. Nobody expected any. Marsden counted flashes in the dark and found a few, about one in eight thousand. Rutherford later said it was as if a shell fired at a sheet of tissue paper had come back and hit him. Two years later he explained it with the nucleus.
A head-on approach
Now aim an alpha particle straight at a gold nucleus. The two charges repel, so the alpha particle slows as it comes in. Predict first what happens next.
An alpha particle of \(7.7\,\text{MeV}\) heads straight at a gold nucleus. What happens?
Two charges \(q_1\) and \(q_2\) a distance \(r\) apart have electric potential energy \(U = kq_1q_2/r\), with \(k = 9 \times 10^9\,\text{N m}^2/\text{C}^2\) and \(U = 0\) far apart. The chapter on electric potential energy derives this; here we only use it. For an alpha particle, \(q_1 = 2e\). For a nucleus of atomic number \(Z\), \(q_2 = Ze\).
In the figure above, the kinetic energy \(K\) falls as the potential energy \(U\) rises, and their sum stays fixed. The alpha particle stops where all of \(K\) has become \(U\). That point is the distance of closest approach, \(d\). Mechanical energy is conserved, as in the chapter on Potential Energy and Conservation of Energy, so one equation gives \(d\).
\[ K + 0 = 0 + \dfrac{k(2e)(Ze)}{d} \]
Why. far away the potential energy is zero; at the stopping point the kinetic energy is zero. Mechanical energy is the same at both.
\[ d = \dfrac{2Zke^2}{K} \]
Read it. solve for \(d\). A faster alpha particle gets closer, and a larger nuclear charge keeps it further away.
The product \(ke^2\) turns up in every atomic calculation, so it pays to know it in handy units. In SI units it is \(9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 2.3 \times 10^{-28}\,\text{J m}\). In MeV and femtometres this is exactly \(1.44\,\text{MeV fm}\), which is the same as \(1.44\,\text{eV nm}\).
An alpha particle of kinetic energy \(7.7\,\text{MeV}\) is aimed head-on at a gold nucleus, \(Z = 79\). Find the distance of closest approach. What does it say about the size of the gold nucleus?
Head-on, so all the kinetic energy becomes potential energy at the stopping point. \(K = 7.7\,\text{MeV}\) and \(Z = 79\).
\(d = 2Zke^2/K\), with \(ke^2 = 1.44\,\text{MeV fm}\).
\(d = 2 \times 79 \times 1.44 / 7.7 = 227.5/7.7 = 29.5\,\text{fm} = 2.95 \times 10^{-14}\,\text{m}\).
The alpha particle stops \(29.5\,\text{fm}\) from the centre of the gold nucleus. So the nucleus is smaller than about \(3 \times 10^{-14}\,\text{m}\), while the atom is about \(10^{-10}\,\text{m}\) across. Values used: \(ke^2 = 1.44\,\text{MeV fm}\), \(1\,\text{fm} = 10^{-15}\,\text{m}\).
A distance of closest approach is an upper limit on the size of a nucleus: the alpha particle turned back without touching it. Later measurements put the radius of the gold nucleus at about \(7\,\text{fm}\), about fourteen thousand times smaller than the atom. If the nucleus were a marble \(1\,\text{cm}\) across, the atom would be as wide as a cricket ground.
An alpha particle of \(5.0\,\text{MeV}\) is aimed head-on at a gold nucleus (\(Z = 79\)). Take \(ke^2 = 1.44\,\text{MeV fm}\). Find the distance of closest approach, in femtometres.
The kinetic energy of the alpha particles is doubled, and the target stays the same. The distance of closest approach
The radius of a gold nucleus (\(Z = 79\)) is about \(7\,\text{fm}\). What least kinetic energy must an alpha particle have, in MeV, to just reach its surface in a head-on approach? Take \(ke^2 = 1.44\,\text{MeV fm}\).
What keeps the electrons out?
The nucleus attracts the electrons. An electron at rest would simply fall in. So Rutherford pictured the electrons circling the nucleus, as planets circle the Sun, with the electric pull supplying the centripetal force. This is the planetary model of the atom.
Here the planetary picture meets a problem. An electron on a circle has a centripetal acceleration, as in the chapter on Dynamics of Circular Motion. The laws of electricity and magnetism say that an accelerating charge sends out electromagnetic waves. The chapter on electromagnetic waves shows why; here we take it as known. The waves carry away energy.
Follow the chain of Fig 1.1. The electron loses energy, so it falls to a smaller orbit. On a smaller orbit it goes round faster, so it radiates at a higher frequency. The calculation says it reaches the nucleus in about \(10^{-11}\,\text{s}\). So the planetary atom fails twice. It could not last, and on its way in it would give light of every wavelength, a continuous band.
Real atoms do neither. A hydrogen atom lasts for ever if left alone. Hot hydrogen gives light at only a few sharp wavelengths, the red, blue-green and two violet lines of the lower strip. Rutherford's nucleus was right. What was wrong was the classical rule for the electrons around it. The next lesson gives Bohr's repair.
Do not say that the electron falls in because the nucleus attracts it. The attraction is exactly what holds a circular orbit, as the Earth's pull holds the Moon. The classical atom collapses because the circling electron radiates and loses energy.
Rutherford's planetary atom, run by the classical rules of electricity and magnetism, has which of these problems?
Why can the electrons of Rutherford's atom not simply be at rest, and why does circular motion not rescue the model either?
An electron at rest would be pulled straight into the positive nucleus, so it must move, and the natural motion is a circular orbit held by the electric pull. But circular motion has a centripetal acceleration, and an accelerating charge radiates electromagnetic energy. So the circling electron would lose energy and spiral into the nucleus in about \(10^{-11}\,\text{s}\).
- Saying that most alpha particles go straight because the gold nucleus is weak.The nuclear force on a close alpha particle is enormous. Most alpha particles go straight because they never come close to a nucleus.
- Leaving out the factor 2 in \(d = 2Zke^2/K\).The alpha particle carries charge \(2e\); the nucleus carries \(Ze\).
- Taking the distance of closest approach as the radius of the nucleus.It is only an upper limit. The alpha particle turned back without touching the nucleus.