5.1.1Tools and Inquiry — IB Physics skills
Rutherford’s Gold Foil Experiment — IB Physics SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 33 of 33 Unit quiz
Part 1 of 4Observations and Conclusions
Each result gave a clue about the structure of the atom.
1
Most alpha particles passed straight through the foil with little or no deflection, so most of the atom is empty space.
2
A small fraction were deflected through large angles, which requires a strong repulsive force from a concentrated positive charge.
3
Very few, roughly one in many thousands, were scattered back through more than 90°, so the central charge must be very small and massive.
4
The nucleus is therefore tiny compared with the atom, positively charged, and contains almost all of the mass.
5
Negatively charged electrons orbit at large distances, which explains why the atom is mostly empty.
Part 2 of 4Size of the Nucleus
Size of the Nucleus
Closest approach
- An alpha particle fired straight at a nucleus stops when its kinetic energy has become electric potential energy.
Energy equation
- ½mv² = kq₁q₂/r gives an upper limit for the size of the nucleus.
Scale
- Atoms have radii of about 10⁻¹⁰ m while nuclei are about 10⁻¹⁵ m to 10⁻¹⁴ m across.
Replacing a model
- The plum pudding model could not explain large-angle scattering, so it was replaced by the nuclear atom.
| Observation | Conclusion |
|---|---|
| Most pass undeviated | Atom is mostly empty space |
| Some large deflections | Nucleus is positive and concentrated |
| Very few bounce back | Nucleus is small and dense |
Part 3 of 4Worked Example
Question
- An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus (Z = 79) head-on.
- Estimate the distance of closest approach.
- (k = 8.99 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C)
- At closest approach all the kinetic energy is potential energy: kq₁q₂/r = E.
- q₁ = 2e and q₂ = 79e, so q₁q₂ = 158 × (1.6 × 10⁻¹⁹)² = 4.04 × 10⁻³⁶ C².
- E = 5.0 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J.
- r = kq₁q₂ / E = (8.99 × 10⁹ × 4.04 × 10⁻³⁶) / 8.0 × 10⁻¹³ = 4.5 × 10⁻¹⁴ m.
Answer The distance of closest approach is about 4.5 × 10⁻¹⁴ m, so the gold nucleus is smaller than this.
Part 4 of 4Exam Checklist & Tips
5.1.1 – Key Points for the Exam
- Alpha particle A helium nucleus consisting of two protons and two neutrons, with charge +2e.
- Nucleus The small, dense, positively charged centre of an atom.
- Scattering The change in direction of particles as they interact with matter.
- Plum pudding model An early model with positive charge spread through the atom, with electrons embedded in it.
Common Mistakes
- Saying that the nucleus was found because alpha particles were absorbed, rather than deflected.
- Stating that most alpha particles were reflected back, when only a very few were.
- Forgetting that alpha particles and the nucleus are both positive, so the force is repulsive.
- Mixing up the size of the atom with the size of the nucleus, which differ by a factor of about 10⁴ to 10⁵.
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