130–131, 136Nuclear and Particle PhysicsCore

The Nuclear Atom

In 1909 Geiger and Marsden fired alpha particles at gold foil. Most went straight through, but about one in several thousand bounced back. That result overturned the plum-pudding model and revealed that almost all of an atom’s mass sits in a tiny, positively charged nucleus\textbf{nucleus}.

35 min Topic 8 · Nuclear and Particle Physics

What you'll be able to do

  • Describe the alpha-particle scattering experiment and interpret its results
  • Explain how the results support the nuclear model over the plum-pudding model
  • Estimate nuclear size from the distance of closest approach
  • Use nuclear notation ZAX^{A}_{Z}\mathrm{X} and describe isotopes
1

The scattering experiment

A narrow beam of alpha particles from a radioactive source was directed at a very thin gold foil in a vacuum. A movable detector — originally a zinc sulfide screen viewed through a microscope — counted scintillations at different angles.

The vacuum matters: alpha particles are strongly ionising and would be stopped within a few centimetres of air. The foil must be thin so that each alpha particle is scattered by at most one nucleus.

The observations were: most alpha particles passed straight through with little or no deflection; a small fraction were deflected through large angles; and a very small number, roughly 1 in 8000, were scattered back through more than 90∘90^{\circ}.

Tip — Learn the three observations and pair each with its conclusion — that is how six-mark questions on this experiment are marked.

2

What the results mean

Most passed straight through\textbf{Most passed straight through}: the atom is mostly empty space.

Some deflected through large angles\textbf{Some deflected through large angles}: there is a concentration of positive charge that repels the positive alpha particles strongly.

A very few bounced back\textbf{A very few bounced back}: that charge, and most of the atom’s mass, is concentrated in a very small volume. Only a massive, compact centre could reverse an alpha particle’s motion, and the rarity of back-scattering shows how small a target it presents.

In the plum-pudding model, positive charge was spread thinly through the atom. Its field would be far too weak anywhere to deflect a fast alpha particle through a large angle, so the model could not explain the results.

3

Estimating the size of the nucleus

For an alpha particle heading straight at a nucleus, it slows as it approaches, stops momentarily, and is pushed back. At that turning point all its initial kinetic energy has become electric potential energy.

Setting Ek=kQαQNrE_{k} = \dfrac{kQ_{\alpha}Q_{N}}{r} gives the distance of closest approach. The nucleus must be smaller than this, so it gives an upper limit on nuclear radius.

The charge of a nucleus with proton number ZZ is ZeZe, and an alpha particle carries 2e2e.

Ek=kQαQNr=k(2e)(Ze)rE_{k} = \dfrac{kQ_{\alpha}Q_{N}}{r} = \dfrac{k(2e)(Ze)}{r}
Closest approach, head-on collision.
Example 1A 5.0 MeV alpha particle approaches a silver nucleus (Z=47Z = 47) head-on. Find its distance of closest approach.
1Ek=5.0×1.60×10−13=8.0×10−13E_{k} = 5.0 \times 1.60\times10^{-13} = 8.0\times10^{-13} J.
2r=8.99×109×2×47×(1.60×10−19)28.0×10−13r = \dfrac{8.99\times10^{9} \times 2 \times 47 \times (1.60\times10^{-19})^{2}}{8.0\times10^{-13}}.
Answerr=2.7×10−14 mr = 2.7\times10^{-14}\ \mathrm{m}

Tip — Convert MeV to joules first: 1 MeV=1.60×10−13 J1\ \mathrm{MeV} = 1.60\times10^{-13}\ \mathrm{J}.

4

Nuclear notation and better probes

A nucleus is written ZAX^{A}_{Z}\mathrm{X}, where ZZ is the proton number and AA is the nucleon number (protons plus neutrons). The number of neutrons is A−ZA - Z.

Isotopes\textbf{Isotopes} are nuclei of the same element — same ZZ — with different numbers of neutrons. Carbon-12 and carbon-14 are both carbon, with 6 and 8 neutrons.

Alpha particles are limited as probes: they are affected by the strong nuclear force if they get close enough, and they cannot reach heavy nuclei at typical source energies. High-energy electrons\textbf{electrons} are better — they are not affected by the strong force, and at high energy their de Broglie wavelength λ=h/p\lambda = h/p is comparable to nuclear dimensions, so they diffract around nuclei and reveal their size.

The same approach, at still higher energies, showed that electrons scatter from three point-like charges inside each proton — the first evidence for quarks.

λ=hp\lambda = \dfrac{h}{p}
de Broglie wavelength of a probe particle.

Equation recap

Ek=kQ1Q2/rE_{k} = kQ_{1}Q_{2}/r
Closest approach.
N=A−ZN = A - Z
Neutron number.
λ=h/p\lambda = h/p
Probe wavelength.

Common mistakes to avoid

Saying most alpha particles bounced back.
Most passed straight through; only a tiny fraction were back-scattered.
Using Q=eQ = e for the alpha particle.
An alpha particle carries +2e+2e.
Forgetting to convert MeV to joules.
Multiply by 1.60×10−131.60\times10^{-13} J per MeV.
Confusing AA and ZZ.
ZZ is protons; AA is protons plus neutrons.

Key takeaways

  • Most alphas passed through: the atom is mostly empty space.
  • Large deflections: a concentrated positive charge.
  • Rare back-scattering: mass and charge are in a tiny nucleus.
  • Closest approach, from Ek=kQ1Q2/rE_{k} = kQ_{1}Q_{2}/r, gives an upper limit on nuclear size.
  • High-energy electrons diffract around nuclei and measure their size more accurately.

Test yourself

Answer a few questions, then review anything you missed.

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