infer from the results of the α-particle scattering experiment the existence and small size of the nucleus

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Atoms, Nuclei and Radiation

Objective

To infer from the results of the α‑particle scattering experiment the existence and extremely small size of the atomic nucleus.

1. Historical Background

Before 1911 the prevailing model of the atom was the Thomson “plum‑pudding” model, in which positive charge was spread uniformly throughout the atom with electrons embedded like raisins in a pudding.

2. The Rutherford (Geiger–Marsden) Experiment

2.1 Experimental Arrangement

  • A source of α‑particles (He²⁺ nuclei) is placed in a vacuum chamber.
  • The α‑particles are collimated into a narrow beam that strikes a thin metal foil (typically gold, thickness ≈ 10⁻⁸ m).
  • A circular fluorescent screen surrounds the foil. Scintillations on the screen are observed through a microscope.
  • The angular distribution of scattered α‑particles is recorded by counting scintillations at various angles θ.

Suggested diagram: Schematic of the α‑particle scattering apparatus showing source, collimator, thin foil, and detector screen.

2.2 Key Observations

  1. ~ 98 % of α‑particles passed through the foil with little or no deflection (θ ≈ 0°).
  2. A small fraction were scattered at large angles; a few were reflected backward (θ ≈ 180°).
  3. The number of particles scattered at a given angle decreased sharply with increasing θ.

3. Interpretation of the Results

3.1 Why the Thomson Model Fails

If positive charge were spread uniformly, the electric field inside the atom would be weak and the α‑particles would experience only small, gradual deflections. The observation of large‑angle scattering is incompatible with this picture.

3.2 Rutherford’s Nuclear Model

Rutherford proposed that:

  • The atom consists of a tiny, massive, positively charged nucleus.
  • All of the atomic mass and charge are concentrated in the nucleus.
  • Electrons occupy the surrounding space, contributing negligibly to the scattering of heavy α‑particles.

3.3 Coulomb Scattering Formula

The differential cross‑section for scattering of a particle of charge \$Z1e\$ by a nucleus of charge \$Z2e\$ is given by the Rutherford formula:

\$\frac{d\sigma}{d\Omega} = \left(\frac{Z1 Z2 e^2}{16\pi\varepsilon_0 E}\right)^2 \frac{1}{\sin^4(\theta/2)}\$

where \$E\$ is the kinetic energy of the α‑particle and \$\theta\$ is the scattering angle.

4. Determining the Size of the Nucleus

4.1 Concept of Impact Parameter

The impact parameter \$b\$ is the perpendicular distance between the initial trajectory of the α‑particle and the centre of the nucleus. Large‑angle scattering requires \$b\$ to be comparable to the nuclear radius \$R\$.

4.2 Estimating \$R\$ from the Largest Observed Deflection

For a back‑scattered α‑particle (\$\theta \approx 180^\circ\$) the closest approach \$r_{\min}\$ satisfies the energy‑conservation condition:

\$\frac{1}{4\pi\varepsilon0}\frac{Z1 Z2 e^2}{r{\min}} = E\$

Solving for \$r_{\min}\$ gives an upper bound on the nuclear radius:

\$R \lesssim r{\min} = \frac{Z1 Z2 e^2}{4\pi\varepsilon0 E}\$

For typical α‑particles (\$E \approx 5\,\$MeV, \$Z1=2\$, \$Z2=79\$ for gold):

\$R \lesssim \frac{(2)(79)(1.60\times10^{-19}\,\text{C})^2}{4\pi(8.85\times10^{-12}\,\text{F m}^{-1})(5\times10^{6}\,\text{eV})(1.60\times10^{-19}\,\text{J eV}^{-1})} \approx 1.5\times10^{-14}\,\text{m}\$

This is roughly two orders of magnitude smaller than the atomic radius (\$\sim10^{-10}\,\$m), confirming that the nucleus is extremely compact.

4.3 Empirical Formula for Nuclear Radius

Experimental data across many nuclei lead to the empirical relationship:

\$R = r_0 A^{1/3}\$

where \$A\$ is the mass number and \$r_0 \approx 1.2\,\$fm (\$1\,\$fm = \$10^{-15}\,\$m). This agrees with the order‑of‑magnitude estimate from Rutherford scattering.

5. Summary Table

QuantitySymbolTypical \cdot alue (Gold foil)Units
α‑particle kinetic energy\$E\$5 MeVeV
Charge of α‑particle\$Z_1\$2
Charge of gold nucleus\$Z_2\$79
Estimated nuclear radius\$R\$\$\sim1.5\times10^{-14}\$m
Atomic radius (order of magnitude)\$a\$\$\sim1\times10^{-10}\$m

6. Conclusions

  • The α‑particle scattering experiment provides direct evidence that atoms contain a tiny, dense, positively charged nucleus.
  • Most α‑particles pass straight through because the nucleus occupies only a minute fraction of the atomic volume.
  • The few particles that are deflected at large angles indicate that the nuclear charge is concentrated in a region of radius \$10^{-14}\$\$10^{-15}\$ m, i.e. about 10 000 times smaller than the atom itself.
  • These findings laid the foundation for the modern nuclear model of the atom and for subsequent developments in quantum mechanics and particle physics.