E.1 · IB Physics
Rutherford scattering
Alpha particles are fired at a single nucleus and feel only the Coulomb repulsion, integrated step by step. Change the impact parameter b, the energy or the target, or fire a whole beam and count how many bounce back.
Predict first
Set Impact parameter b to 0 (head-on), then raise Alpha kinetic energy from 5.0 to 10.0 MeV. What happens to the closest approach?
At closest approach all of Eₖ has become electric potential energy kQq ÷ d, so d = kQq ÷ Eₖ. Doubling Eₖ halves d, from 45.5 fm to 22.8 fm.
Check it in the model: Set the Impact parameter, b slider to 0, then move Alpha kinetic energy, Eₖ from 5.0 to 10.0 MeV and read Closest approach.
path turned back through more than 90° · nucleus drawn to scale
Head-on: d = kQq ÷ Ek = 2Z ke² ÷ Ek
tan(θ/2) = d ÷ (2b)
Nuclear radius ≈ 1.2 A1/3 fm
Try this
- Press 'Head-on', then raise Ek from 5.0 to 10.0 MeV and fire again. How does the closest approach compare with d = kQq ÷ Ek?
- With gold at 5.0 MeV, reduce b from 100 fm to 30 fm to 0 fm, firing each time. Compare the angle with tan(θ/2) = d ÷ (2b).
- Press 'Beam on gold' and read how many of the 400 particles are turned back through more than 90°. Why so few?
Common error: drawing the alpha bouncing off a hard surface. The repulsion acts along the whole path, so the track is a smooth curve that bends away before reaching the nucleus, and the closest approach from d = kQq ÷ Ek is only an upper limit for the nuclear radius.