OCR GCSE Physics A: Atomic model, atomic size and density (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements P1.1a to P1.1f of the OCR GCSE (9-1) Physics A (Gateway Science) specification (J249), Version 5.0 (July 2026). None of these statements is Higher Tier only, so every question is suitable for both tiers. Read the study guide first if a topic is new, or use the revision notes to refresh it.

Course links: OCR GCSE Physics hub, printable checklist and the free 10-minute diagnostics. More density practice is in the Matter practice questions.

Questions

1. State two features of the plum pudding model of the atom. [2]

2. Alpha particles were fired at thin gold foil. Describe two observations and state what each observation shows about the structure of the atom. [4]

3. The atomic model changed again after Rutherford’s nuclear model.

(a) Describe what Bohr changed in the model. [2] (b) State two differences between Rutherford’s nuclear model and the plum pudding model. [2]

4. Describe the atom, referring to charge, mass and size. [3]

5. An atom is drawn 80 m across for a scale model. The nucleus is about 10 000 times smaller across than the atom.

(a) Calculate the diameter of the nucleus in this model, in millimetres. [2] (b) An atom has a diameter of about 1 × 10⁻¹⁰ m. Calculate how many atoms fit side by side along a line 2.0 mm long. [2]

6. A metal block measures 6.0 cm × 5.0 cm × 2.0 cm and has mass 420 g.

(a) Define density. [1] (b) Calculate the density of the metal in g/cm³ and in kg/m³. [3]

7. (a) Mercury has a density of 13.6 g/cm³. Calculate the mass in kilograms of 250 cm³ of mercury. [3] (b) Ethanol has a density of 790 kg/m³. Calculate the volume of 0.46 kg of ethanol, in m³ to two significant figures. [3]

8. A student finds the density of a small irregular metal object. The mass is 133.5 g. The water level in a measuring cylinder rises from 40.0 cm³ to 55.0 cm³ when the object is lowered in.

(a) Calculate the volume of the object. [1] (b) Calculate the density in kg/m³. [2] (c) Reference densities are: copper 8900 kg/m³, brass 8500 kg/m³, iron 7900 kg/m³. Identify the most likely metal. [1] (d) Suggest two reasons the measured value might not match a reference value exactly. [2]

9. Water has density 1000 kg/m³. Air has density 1.2 kg/m³.

(a) Explain, in terms of particles, why air is much less dense than water. [3] (b) Calculate how many times denser water is than air, to two significant figures. [2]

10. A syringe holds 0.0150 m³ of a gas of density 0.80 kg/m³. The nozzle is blocked and the gas is compressed to 0.0040 m³.

(a) Calculate the mass of the gas. [2] (b) State why the mass is the same after compression. [1] (c) Calculate the density of the gas after compression. [2]

11. A block of ice has volume 400 cm³ and density 920 kg/m³. It melts completely to water of density 1000 kg/m³.

(a) Calculate the mass of the ice, in kg. [2] (b) Calculate the volume of water formed, in cm³. [2] (c) Calculate the percentage decrease in volume. [2]

12. A sheet of gold leaf has area 0.010 m² and thickness 1.0 × 10⁻⁷ m. The density of gold is 19 300 kg/m³.

(a) Calculate the volume of the leaf. [2] (b) Calculate the mass of the leaf, giving your answer in milligrams to three significant figures. [2] (c) Taking a gold atom as 1 × 10⁻¹⁰ m across, estimate how many atoms thick the leaf is. [2] (d) Use the nuclear model to explain why most alpha particles can pass through a foil that is many atoms thick. [2]

Answers

1. The atom is a ball of positive charge [1]; electrons are embedded in / spread through it [1]. [2] Examiner insight: Give two separate features. Saying “positive ball with electrons” is incomplete unless you also state where the electrons are.

2. Observation 1: most alpha particles went straight through [1], so most of the atom is empty space [1]. Observation 2: a small number were deflected through large angles (a very few bounced almost straight back) [1], so the atom has a small, dense, positively charged nucleus that repels the alpha particles [1]. [4] Examiner insight: Each observation needs its own conclusion. A list of observations with a single conclusion at the end tends to gain fewer points than pairing them.

3. (a) Bohr placed electrons in orbits (energy levels) at fixed distances from the nucleus [1], which explained why atoms are stable / give out light only at particular wavelengths [1]. [2] (b) In the nuclear model the positive charge and nearly all the mass are in a small central nucleus [1]; in the plum pudding model they are spread through the whole atom [1]. [2] Examiner insight: Make each difference a comparison of the two models, not a description of one model alone.

4. The atom has a positively charged nucleus in the centre with negatively charged electrons surrounding it [1]. Almost all the mass is in the nucleus [1]. The nuclear radius is much smaller than the radius of the atom [1]. [3] Examiner insight: Cover each of charge, mass and size in turn. Calling the nucleus “the biggest part” costs the size point.

5. (a) Diameter = 80 m ÷ 10 000 [1] = 0.0080 m = 8.0 mm [1]. [2] (b) Number = 2.0 × 10⁻³ ÷ 1 × 10⁻¹⁰ [1] = 2 × 10⁷ atoms [1]. [2] Examiner insight: Convert millimetres to metres before dividing. Dividing 2.0 by 1 × 10⁻¹⁰ gives a power of ten that is 1000 times too large.

6. (a) Density is mass per unit volume. [1] (b) Volume = 6.0 × 5.0 × 2.0 = 60 cm³ [1]. Density = 420 ÷ 60 = 7.0 g/cm³ [1] = 7000 kg/m³ [1]. [3] Examiner insight: Write the unit with each answer. If a question asks for two units, show the conversion factor (× 1000) rather than just the new number.

7. (a) Mass = 13.6 × 250 [1] = 3400 g [1] = 3.4 kg [1]. [3] (b) V = m ÷ ρ = 0.46 ÷ 790 [1] = 5.82 × 10⁻⁴ [1] = 5.8 × 10⁻⁴ m³ [1]. [3] Examiner insight: Rearranging is shown by writing the new form (V = m ÷ ρ). The conversion from grams to kilograms is the step most often left out in part (a).

8. (a) Volume = 55.0 − 40.0 = 15.0 cm³ [1]. (b) Density = 133.5 ÷ 15.0 = 8.9 g/cm³ [1] = 8900 kg/m³ [1]. [2] (c) Copper, because 8900 kg/m³ matches the copper value. [1] (d) Any two from: air bubbles on the object, making the volume read too large [1]; reading the meniscus inaccurately [1]. [2] An alloy rather than a pure metal would also change the density. Examiner insight: For error sources, say which way the result is affected. “Human error” on its own is too vague to gain credit.

9. (a) In a gas the particles are far apart [1], while in a liquid they are close together / touching [1]. So a given volume of air contains far fewer particles and much less mass [1]. [3] (b) Ratio = 1000 ÷ 1.2 [1] = 833, so 830 times (2 s.f.) [1]. [2] Examiner insight: Explain density as a difference in spacing, not in particle size. Statements such as “the particles in a gas are lighter” do not explain the difference.

10. (a) Mass = density × volume = 0.80 × 0.0150 [1] = 0.012 kg [1]. [2] (b) No gas has entered or left, so the mass is conserved. [1] (c) Density = 0.012 ÷ 0.0040 [1] = 3.0 kg/m³ [1]. [2] Examiner insight: Use the mass found in part (a) in part (c), and show the substitution so your method is clear even if an earlier value is slightly off.

11. (a) Volume = 400 cm³ = 4.0 × 10⁻⁴ m³ [1]. Mass = 920 × 4.0 × 10⁻⁴ = 0.368 kg [1]. [2] (b) Volume of water = 0.368 ÷ 1000 = 3.68 × 10⁻⁴ m³ [1] = 368 cm³ [1]. [2] (c) Decrease = 400 − 368 = 32 cm³ [1]. Percentage = 32 ÷ 400 × 100 = 8.0% [1]. [2] Examiner insight: The percentage change is taken relative to the original volume (the ice), not the final volume. Dividing by 368 gives a different value.

12. (a) Volume = area × thickness = 0.010 × 1.0 × 10⁻⁷ [1] = 1.0 × 10⁻⁹ m³ [1]. [2] (b) Mass = 19 300 × 1.0 × 10⁻⁹ = 1.93 × 10⁻⁵ kg [1] = 19.3 mg [1]. [2] (c) Number = 1.0 × 10⁻⁷ ÷ 1 × 10⁻¹⁰ [1] = about 1000 atoms [1]. [2] (d) The atom is mostly empty space [1], and the nucleus is tiny compared with the atom, so most alpha particles pass between the nuclei without being deflected [1]. [2] Examiner insight: In part (b) the unit conversion (1 kg = 1 000 000 mg) is needed to reach milligrams. In part (d), name both the empty space and the small size of the nucleus.

Where marks are usually lost

• Stating alpha scattering observations without saying what each shows about the atom.
• Mixing up the models, for example giving the plum pudding model a nucleus.
• Calling the nucleus the biggest part of the atom rather than the heaviest.
• Leaving standard form powers uncorrected when converting millimetres to metres.
• Not converting grams to kilograms, or cm³ to m³, before using the density equation.
• Treating 1 g/cm³ as 100 kg/m³.
• Using the final volume instead of the original as the base for a percentage change.
• Explaining low gas density using lighter or smaller particles instead of larger spacing.
• Giving an error source with no direction of effect on the result.

Next steps

• Review with the revision notes or the study guide.
• Browse the OCR GCSE Physics hub and tick off the printable checklist.
• Find gaps with the all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Physics A (Gateway Science) specification (J249), Version 5.0 (July 2026), for first assessment in 2018, published by OCR. Statements P1.1a to P1.1f.