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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

📚 IB Physics SL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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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
  • Closest approach An alpha particle fired straight at a nucleus stops when its kinetic energy has become electric potential energy.
Energy equation
  • Energy equation ½mv² = kq₁q₂/r gives an upper limit for the size of the nucleus.
Scale
  • Scale Atoms have radii of about 10⁻¹⁰ m while nuclei are about 10⁻¹⁵ m to 10⁻¹⁴ m across.
Replacing a model
  • Replacing a model The plum pudding model could not explain large-angle scattering, so it was replaced by the nuclear atom.
ObservationConclusion
Most pass undeviatedAtom is mostly empty space
Some large deflectionsNucleus is positive and concentrated
Very few bounce backNucleus 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)

  1. At closest approach all the kinetic energy is potential energy: kq₁q₂/r = E.
  2. q₁ = 2e and q₂ = 79e, so q₁q₂ = 158 × (1.6 × 10⁻¹⁹)² = 4.04 × 10⁻³⁶ C².
  3. E = 5.0 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J.
  4. 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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