8G

The nuclear atom

How the nucleus was discovered: Thomson's "plum pudding" model, the Geiger–Marsden alpha-scattering experiment and Rutherford's nuclear model, estimating the size of the nucleus, the limitations of Rutherford's model, and Chadwick's discovery of the neutron using conservation of momentum and energy.

"It was quite the most incredible event that has ever happened to me in my life," Ernest Rutherford said later. "It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." A few alpha particles, fired at gold leaf only a few hundred atoms thick, had bounced almost straight back. The atom was nearly all empty space.

Thomson's model

Once Thomson had shown that atoms contain tiny negative electrons, and atoms are neutral overall, there had to be positive charge somewhere. In 1904 he proposed that the atom is a sphere of evenly spread positive charge, about $10^{-10}$ m across, with electrons embedded in it like plums in a pudding. Most of the atom's mass was assumed to be in the positive sphere. The model was simple and fitted the evidence available at the time.

Three models of the atom: Thomson's plum pudding (1904), Rutherford's nuclear atom (1911), and Bohr's atom with fixed orbits (1913). Not to scale: a real nucleus is about 1/100 000 of the atom's diameter.

The Geiger–Marsden experiment

In 1909, working under Rutherford in Manchester, Hans Geiger and an undergraduate, Ernest Marsden, fired alpha particles from a radium source at a very thin sheet of gold foil, in a vacuum. Alpha particles are helium nuclei: positively charged (+2e), about 7300 times as massive as an electron, and moving at about 5% of the speed of light. They detected where the alphas went by watching tiny flashes of light (scintillations) on a zinc sulfide screen through a microscope, which could be moved around the foil to any angle.

The Geiger–Marsden experiment. Most alpha particles pass straight through the gold foil; a few are deflected through large angles, and about 1 in 8000 bounces back.

The results:

  • Almost all the alpha particles went straight through, or were deflected by less than a degree or so.
  • A small number were deflected through large angles.
  • About 1 in 8000 was deflected by more than 90°, bouncing back towards the source.

In Thomson's model, the positive charge is spread out, so the electric field inside the atom is weak. An alpha particle should be deflected only very slightly by each atom, and the chance of a large deflection, even from many small ones adding up, is vanishingly small. Seeing 1 in 8000 bounce back meant the model was wrong.

Rutherford's nuclear model

In 1911 Rutherford explained the results. The positive charge and almost all the mass of an atom must be concentrated in a tiny central nucleus. Then:

  • Most alphas pass through the empty space far from any nucleus and are hardly deflected.
  • The few that pass close to a nucleus feel a very strong electrostatic repulsion (the field is intense because the charge is concentrated) and are deflected through large angles.
  • Only those heading almost directly at a nucleus bounce back. Because the target area is so small, this is rare.
  • Electrons, being so light, have negligible effect on the massive alpha particle.

Rutherford derived how the number of alphas scattered should depend on the angle, the alpha energy and the foil thickness, assuming only a Coulomb force from a point charge. Geiger and Marsden tested the predictions over the next two years and they matched, strong evidence for the model. In it, electrons orbit the nucleus like planets around the Sun, held by electrostatic attraction.

How small is the nucleus?

An alpha particle aimed straight at a nucleus slows down as it approaches, stops momentarily, and is pushed back. At the turning point all its kinetic energy has become electrical potential energy. The distance of closest approach, $d$, therefore gives an upper limit to the size of the nucleus:

$$K = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{d} \quad\Rightarrow\quad d = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{K}$$

(The expression for electrical potential energy goes beyond the HSC data sheet, but the method uses only energy conservation.)

Worked example: closest approach to a gold nucleus

An alpha particle with 7.7 MeV of kinetic energy heads directly towards a gold nucleus ($Z = 79$). How close does it get? ($\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}$ N m2 C−2)

The question

$K = 7.7 \times 10^{6} \times 1.602 \times 10^{-19} = 1.23 \times 10^{-12}$ J

Kinetic energy in joules

$d = \dfrac{8.99 \times 10^{9} \times (2 \times 1.602 \times 10^{-19}) \times (79 \times 1.602 \times 10^{-19})}{1.23 \times 10^{-12}} = 2.96 \times 10^{-14}$ m $\approx 30$ fm

Closest approach

The gold nucleus must be smaller than about $3 \times 10^{-14}$ m, while a gold atom is about $3 \times 10^{-10}$ m across: the nucleus is less than 1/10 000 of the atom's diameter. Modern measurements give the gold nucleus a radius of about 7 fm.

Interpret
InteractiveFire alphas at a nucleus
Try this
  • With gold selected, set the alpha energy to 7.7 MeV and aim head-on (impact parameter 0). Read the closest approach: about 29.5 fm, compared with a nuclear radius of about 7.0 fm.
  • Increase the impact parameter slowly. How does the deflection angle change? At what aim is it deflected by 90°?
  • Switch to silver, then aluminium, keeping 7.7 MeV. Why does the alpha get closer to the smaller nucleus?
  • With aluminium at 7.7 MeV and a head-on aim, the alpha reaches about 4.9 fm and touches the nucleus. What would happen to Rutherford's Coulomb-only predictions in this case?
  • Find the alpha energy that just reaches the surface of a gold nucleus. Why couldn't Geiger and Marsden's radium alphas do this?

Limitations of Rutherford's model

The nuclear model explained scattering beautifully, but it had serious problems:

  • Atoms should collapse. An orbiting electron is constantly accelerating (centripetal acceleration). Maxwell's theory says an accelerating charge radiates electromagnetic waves, losing energy. The electron should spiral into the nucleus in about $10^{-11}$ s. Atoms are clearly stable.
  • Line spectra are unexplained. A spiralling electron would give out a continuous range of frequencies. Atoms actually emit and absorb only particular wavelengths.
  • The nucleus's mass doesn't add up. A helium nucleus has twice the charge of a hydrogen nucleus but four times the mass. The model said nothing about what made up the extra mass, or what held the protons together against their electrostatic repulsion.
  • No arrangement of electrons. It gave no reason for the chemical properties of elements or the periodic table.

The first two were tackled by Bohr in 1913 (page 8H). The third was solved by the discovery of the neutron.

Chadwick and the neutron

Rutherford proposed in 1920 that the nucleus also contained a neutral particle about the mass of a proton. It was hard to find: with no charge, it doesn't ionise gas or get deflected by fields, so ordinary detectors can't see it.

In 1930, Bothe and Becker found that alpha particles striking beryllium produced a very penetrating, uncharged radiation. They assumed it was high-energy gamma rays. In 1932 Irène Joliot-Curie and Frédéric Joliot showed this radiation knocked protons out of paraffin wax (which is rich in hydrogen) with energies up to about 5.7 MeV. For a gamma ray to give a proton that much energy, it would need about 50 MeV, far more than the reaction could supply.

Chadwick's experiment. Alpha particles from polonium hit beryllium, which emits neutral radiation. This knocks protons out of paraffin wax, and the protons are detected by an ionisation chamber.

James Chadwick, at the Cavendish Laboratory in Cambridge, repeated the experiment and also aimed the radiation at other targets, such as nitrogen gas, measuring the recoil speeds. He argued:

  • The radiation is not deflected by electric or magnetic fields, so it is neutral.
  • If it were photons, the recoil energies of protons and nitrogen nuclei would not fit with conservation of energy and momentum for any single photon energy.
  • If it were a neutral particle of mass $m$, colliding elastically, both recoils fit, and the mass came out at about the mass of a proton.

He concluded that the radiation was neutrons, produced by $^{4}_{2}\text{He} + {}^{9}_{4}\text{Be} \rightarrow {}^{12}_{\ 6}\text{C} + {}^{1}_{0}n$, and won the 1935 Nobel Prize. The nucleus was now understood as protons and neutrons (nucleons), with the mass number $A$ counting both.

Worked example: why a neutron, not a photon?

A neutral particle with mass $m$ and speed $u$ collides head-on and elastically with a stationary proton (mass $m_p$). Show that if $m = m_p$, the proton can take all of the particle's kinetic energy, and explain why this was evidence for the neutron.

The question

Momentum: $mu = mv_1 + m_pv_2$. Kinetic energy: $\tfrac12 mu^2 = \tfrac12 mv_1^2 + \tfrac12 m_pv_2^2$.

Conservation laws

With $m = m_p$ these give $v_1 = 0$ and $v_2 = u$: the incoming particle stops and the proton moves off with all its kinetic energy, like a head-on collision between two snooker balls.

Solve for equal masses

So a neutral particle of about the proton's mass needs only about 5.7 MeV of kinetic energy to produce 5.7 MeV protons, which is reasonable. A massless photon can't transfer energy so efficiently to a heavy proton and would need about 50 MeV. Chadwick's comparison with nitrogen recoils gave the particle's mass as about 1.0 to 1.2 times the proton's.

Interpret
A black and white portrait of a man with a moustache, wearing a suit and tie
Ernest Rutherford, born in New Zealand, who interpreted the Geiger–Marsden results and proposed the nuclear atom.Photo: Bain News Service (restored by Bammesk), Public domain
A black and white photograph of a man in glasses and a three-piece suit, seated in a chair
James Chadwick, who identified the neutron in 1932. He later led the British team in the Manhattan Project.Photo: Los Alamos National Laboratory, Attribution

Test yourself

  1. Which result of the Geiger–Marsden experiment showed that the nucleus is very small?

  2. Why could Thomson's plum pudding model not explain large-angle scattering?

  3. According to classical physics, the main problem with Rutherford's model is that orbiting electrons

  4. Chadwick concluded that the radiation from beryllium was not gamma rays mainly because

  5. The alpha energy is doubled in a head-on approach to a nucleus. The distance of closest approach

Label the Geiger–Marsden experiment.

vacuum chamber (top view)radium source in lead boxalpha particlesgold foil (≈ 400 atoms thick)zinc sulfide screen + microscopeMost went straight through. About 1 in 8000 bounced back by more than 90°.

Put these developments in order.

In the Geiger–Marsden experiment, about 1 in alpha particles was deflected by more than 90°. Rutherford concluded that the positive charge and most of the mass are concentrated in a small . Chadwick showed that radiation from beryllium consisted of uncharged particles with about the same mass as a , which he called .

6 marks

Evaluate the Rutherford model of the atom, with reference to the experimental evidence that led to it and its limitations.