IB · E · extended response
The rare deflections that changed the atomic model
Nuclear and quantum physics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Illustrative alpha-scattering observations
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Counts per minute
Horizontal axis (x): Scattering angle / °
Read the exact data as a table
| Scattering angle / ° | Counts per minute |
|---|---|
| 10 | 100000 |
| 30 | 10000 |
| 60 | 1000 |
| 90 | 100 |
| 150 | 10 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
A narrow alpha-particle beam strikes a very thin gold foil. Detectors covering equal solid angles record the supplied count rates; values are illustrative rather than a full scattering-law fit. Most alpha particles pass close to the forward direction. Gold-197 has atomic number 79, and an alpha particle is a helium-4 nucleus of charge +2e. Supplied data (Scattering angle / °; Counts per minute): (10, 100000), (30, 10000), (60, 1000), (90, 100), (150, 10).
- (a)
Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge.
3 marks - (b)
Determine proton, neutron and electron numbers in neutral gold-197 and in a gold-197 ion of charge +3e. Describe an isotope and balance alpha emission from a parent with A = 201, Z = 81.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Most projectiles cross the foil with little change in direction, so atoms are not solid dense obstacles across their full dimensions.
- 2
A rare close encounter with concentrated positive nuclear charge can strongly repel an alpha particle. Very large deflections require a strong local interaction and a sufficiently massive scattering centre.
- 3
A diffuse positive-charge model would spread the repulsion over the atomic volume and struggles to explain the observed rare strong deflections. The evidence supports a compact nucleus, while this count table alone does not determine an exact nuclear radius.
(b)
Determine proton, neutron and electron numbers in neutral gold-197 and in a gold-197 ion of charge +3e. Describe an isotope and balance alpha emission from a parent with A = 201, Z = 81.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Atomic number gives 79 protons. Neutrons are A−Z = 197−79 = 118; neutrality requires 79 electrons.
- 2
A +3e ion has lost three electrons and has 76; its nucleus is unchanged. An isotope has the same proton number but a different neutron number.
- 3
Alpha emission removes A = 4 and Z = 2: ²⁰¹₈₁X → ¹⁹⁷₇₉Au + ⁴₂He. Both nucleon and charge numbers balance.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Most projectiles cross the foil with little change in direction, so atoms are not solid dense obstacles across their full dimensions.
- A rare close encounter with concentrated positive nuclear charge can strongly repel an alpha particle. Very large deflections require a strong local interaction and a sufficiently massive scattering centre.
- A diffuse positive-charge model would spread the repulsion over the atomic volume and struggles to explain the observed rare strong deflections. The evidence supports a compact nucleus, while this count table alone does not determine an exact nuclear radius.
- Atomic number gives 79 protons. Neutrons are A−Z = 197−79 = 118; neutrality requires 79 electrons.
- A +3e ion has lost three electrons and has 76; its nucleus is unchanged. An isotope has the same proton number but a different neutron number.
- Alpha emission removes A = 4 and Z = 2: ²⁰¹₈₁X → ¹⁹⁷₇₉Au + ⁴₂He. Both nucleon and charge numbers balance.
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