Alpha-Particle Trajectory

The figure shows the trajectory of alpha-particles in the Coulomb field of a target nucleus, along with the impact parameter bb and scattering angle θ.

The trajectory traced by an alpha particle depends on the impact parameter b of the collision. The impact parameter is the perpendicular distance of the initial velocity vector of the alpha particle from the centre of the nucleus.

A given beam of alpha-particles has a distribution of impact parameters, so the beam is scattered in various directions with different probabilities. In a beam, all particles have nearly the same kinetic energy.

• An alpha particle close to the nucleus, that is, with a small impact parameter, suffers large scattering.
• In a head-on collision, the impact parameter is minimum, and the alpha particle rebounds.
• In this case, the scattering angle is approximately equal to ππ.
• For a large impact parameter, the alpha particle goes nearly undeviated.
• In that case, the deflection is small, and the scattering angle is approximately equal to 0.

Only a small fraction of the incident particles rebound. This indicates that the number of alpha-particles undergoing head-on collision is small.

This further implies that the atom's mass and positive charge are concentrated in a small volume. Therefore, Rutherford scattering is a powerful way to determine an upper limit to the size of the nucleus.

Question idea

In Rutherford’s nuclear model, the nucleus, with a radius of about 10⁻¹⁵ m, is analogous to the sun, and the electrons move in orbits with radii of about 10⁻¹⁰ m, like the Earth around the sun. The example asks whether, if the solar system had the same proportions as the atom, the Earth would be closer to or farther from the Sun than it actually is.

Given values

• Radius of nucleus: about 10⁻¹⁵ m.
• Radius of electron orbit: about 10⁻¹⁰ m.
• Radius of Earth’s orbit: about 1.5 × 10¹¹ m.
• Radius of the Sun: 7 × 10⁸ m.

Solution steps

• Ratio of radius of electron orbit to radius of nucleus:
  \[\frac{10^{-10}}{10^{-15}}=10^5\]

This means the electron’s orbital radius is 10⁵ times the nucleus's radius.

• If the radius of the Earth’s orbit around the Sun were 10⁵ times larger than the radius of the Sun, then:
  10⁵ × 7 × 10⁸ m = 7 × 10¹³ m

This is more than 100 times greater than the actual orbital radius of the Earth.

Result

The Earth would be much farther away from the Sun. This implies that an atom contains a much greater fraction of empty space than our solar system does.

Question

In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV alpha particle before it momentarily comes to rest and reverses direction?

Key idea

Throughout the scattering process, the total mechanical energy of the system consisting of an alpha particle and a gold nucleus is conserved. The initial mechanical energy is the kinetic energy K of the incoming alpha particle. The final mechanical energy, when the alpha-particle momentarily stops, is the electric potential energy U of the system.

Let d be the centre-to-centre distance between the alpha-particle and the gold nucleus at the alpha-particle's stopping point.

Energy relation

Thus, the distance of closest approach is:

These relations are directly stated in the source material.

Numerical values used

• Maximum kinetic energy of alpha-particles of natural origin: 7.7 MeV or 1.2 × 10⁻¹² J.

• \[\frac {1}{4πε_0}\] = 9.0 × 10⁹ N m²/C².

• e = 1.6 × 10⁻¹⁹ C.

Substituting these values gives:

For gold, Z = 79, so:

Also, 1 fm (fermi) = 10⁻¹⁵ m.

Result and conclusion

The radius of the gold nucleus is therefore less than 3.0 × 10⁻¹⁴ m. The source then states that this is not in good agreement with the observed result, as the actual radius of the gold nucleus is 6 fm.

The cause of the discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the alpha particle. Thus, the alpha particle reverses its motion without ever actually touching the gold nucleus.