Questions: Rutherford Scattering and the Nuclear Atom

Question 1

According to the Rutherford formula, how does the scattering rate per unit solid angle at θ = 60° compare with the rate at θ = 180°?

• A. The same
• B. 4 times larger at 60°
• C. 16 times larger at 60°
• D. 16 times smaller at 60°

Answer: (C)

dσ/dΩ ∝ 1/sin⁴(θ/2). At 60°, sin 30° = 0.5 and sin⁴ = 1/16. At 180°, sin 90° = 1. So the 60° rate is 16 times larger.

Question 2

What feature of the Geiger-Marsden results ruled out Thomson's plum-pudding model?

• A. Most alpha particles passed straight through
• B. Some alpha particles were deflected through very large angles
• C. The foil became positively charged
• D. The alpha particles lost all their energy in the foil

Answer: (B)

Charge spread through a whole atom cannot produce a field strong enough to reverse a fast alpha particle. Large-angle scattering requires the positive charge to be concentrated in a very small region.

Question 3

If the alpha particles' kinetic energy is doubled, the Rutherford cross-section at a fixed angle falls to one quarter of its original value.

• T. True
• F. False

Answer: True

dσ/dΩ ∝ 1/K². Doubling K divides the cross-section by 4. Faster particles are deflected less at a given impact parameter.

Question 4

Using e²/(4πε₀) = 1.44 MeV·fm, find the distance of closest approach for a 5.0 MeV alpha particle (z = 2) aimed head-on at a gold nucleus (Z = 79). How does it compare with the gold nuclear radius of about 7 fm?

Model answer: About 45 fm, roughly six times the nuclear radius.

At closest approach all kinetic energy becomes potential energy: K = zZe²/(4πε₀d). So d = 2 × 79 × 1.44 MeV·fm / 5.0 MeV ≈ 45 fm. The alpha particle turns around well outside the nucleus.