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OxfordAQA International GCSE Physics: Atomic structure, isotopes and types of ionising radiation (9203) – Practice Questions

Original practice questions with worked answers on atoms, isotopes, nuclear equations and alpha, beta and gamma radiation for OxfordAQA IGCSE Physics.

Subject
Physics
Level
IGCSE
Topic
Atomic structure, isotopes and types of ionising radiation
Updated

Aligned to OxfordAQA IGCSE Physics (9203), For exams May/June 2018 onwards. Official specification .

Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Physics.

Syllabus points this page covers

9203

  • 3.7.1 Atomic structure
  • 3.7.2 Ionizing radiation from the nucleus

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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 follow the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, sections 3.7.1 a to f (atomic structure) and 3.7.2 a to g (ionizing radiation from the nucleus). The qualification is untiered. Take the radius of a nucleus to be about 10⁻¹⁵ m where a question does not give it.

Links: study guide, revision notes, course hub, printable checklist and the 10-minute diagnostics.

Questions

1. Give the approximate radius of an atom, then describe the simple nuclear model of the atom, saying where almost all of the mass is found. [3]

2. Copy and complete the table for the relative mass and relative charge of the three particles. [3]

Particle Relative mass Relative charge
Proton
Neutron
Electron

3. A neutral atom of lithium is written ⁷₃Li.

(a) Give the numbers of protons, neutrons and electrons it contains. [3] (b) Another lithium atom has 3 protons and 3 neutrons. Explain why the two atoms are isotopes of lithium. [2]

4. (a) An iron atom ⁵⁶₂₆Fe has lost three electrons. Calculate how many electrons it now has and state its overall charge. [2] (b) A sulfur atom ³²₁₆S gains electrons and forms an ion with a charge of –2. State how many electrons the ion has and show your working. [2]

5. Alpha particles are fired at thin metal foil. Describe the three main observations, and for each one say what it shows about the structure of the atom. [6]

6. An atom has a radius of 1.0 × 10⁻¹⁰ m. Its nucleus has a radius of 4.0 × 10⁻¹⁵ m.

(a) Find the factor by which the atom’s radius exceeds the nucleus’s radius. [2] (b) In a scale model the nucleus is a sphere of radius 2.0 mm. Calculate the radius the atom would have in this model. [2]

7. A student holds a radioactive sample in a flame and says that its nuclei will now decay more quickly.

(a) Explain what is meant by saying that radioactive decay is random. [2] (b) Explain whether the student is correct. [2]

8. (a) State two sources of background radiation. [2] (b) A detector with no source nearby records 138 counts in 15 minutes. With a source placed close to it, the detector records 2520 counts in 4 minutes. Calculate the count rate caused by the source alone, in counts per minute, to 3 significant figures. [3]

9. Complete the missing mass number and atomic number in each nuclear equation.

(a) ²³⁸₉₂U → ᴬ_Z X + ⁴₂He [2] (b) ²¹⁴₈₃Bi → ᴬ_Z X + ⁰₋₁e [2] (c) ᴬ_Z X → ²⁰⁶₈₂Pb + ⁰₋₁e [2]

10. (a) Put alpha, beta and gamma in order of ionising power, starting with the greatest. [1] (b) State which of the three has the longest range in air. [1] (c) Explain why the most strongly ionising radiation has the shortest range. [2] (d) State what gamma radiation is. [1]

11. A detector is placed in front of two sources, R and S. The background count rate is 20 counts per minute. Absorbers are placed between each source and the detector.

Absorber Count rate for R (per minute) Count rate for S (per minute)
None 640 905
Paper 26 903
3 mm aluminium 22 902
Thick lead 21 330

(a) Calculate the corrected count rate for R with no absorber. [1] (b) Identify the radiation from R and give a reason. [2] (c) Identify the radiation from S, and explain why it is neither alpha nor beta. [3] (d) A student says that thick lead stopped the radiation from S. Use the data to comment. [2]

12. A small source, T, is placed at different distances from a detector in air. The background count rate has been subtracted from every reading.

Distance (cm) 2 4 6 8
Count rate (per minute) 1210 1195 3 0

(a) State the range of distances in which the radiation from T stops reaching the detector. [1] (b) Identify the type of radiation and give a reason. [2] (c) Explain, in terms of ionisation, why this radiation has such a short range in air. [3] (d) A sheet of paper is placed between T and the detector at a distance of 2 cm. Predict the corrected count rate and explain your prediction. [2]

Answers

1. Radius about 10⁻¹⁰ m [1]. At the centre is a small, positively charged nucleus of protons and neutrons, with electrons around it [1]. The nucleus is far smaller than the atom but holds almost all of the mass [1]. [3] Examiner insight: Cover size, structure and mass location with a statement for each. “Mostly empty space” describes the atom, not the nucleus.

2. Proton: 1, +1 [1]. Neutron: 1, 0 [1]. Electron: very small, –1 [1]. [3] Examiner insight: Write the sign of the charge. “1” and “+1” are not the same thing for a proton, and the electron’s mass should be described as very small, not as zero.

3. (a) Protons = atomic number = 3 [1]. Neutrons = 7 – 3 = 4 [1]. Electrons = protons in a neutral atom = 3 [1]. (b) Both have 3 protons, so they are the same element [1]. They have different numbers of neutrons (4 and 3) [1]. [5] Examiner insight: An isotope answer should compare both numbers: same protons and different neutrons. Giving only one of them is incomplete.

4. (a) Electrons = 26 – 3 = 23 [1]. Charge = 26 – 23 = +3 [1]. (b) The nucleus has 16 protons, and a charge of –2 means two more electrons than protons [1], so 16 + 2 = 18 electrons [1]. [4] Examiner insight: Show the subtraction of protons and electrons in the charge. A negative ion has more electrons than protons, which is a common slip in the other direction.

5. Most alpha particles carry straight on [1], so most of an atom is empty space [1]. Some are deflected through small angles [1], so they are repelled by a positive charge (the nucleus is positive) [1]. Very few bounce back through large angles [1], so a tiny nucleus holds the positive charge and most of the mass [1]. [6] Examiner insight: Match every observation to a conclusion of its own. Three observations followed by a single general statement tend to earn less than three linked pairs.

6. (a) Factor = 1.0 × 10⁻¹⁰ / 4.0 × 10⁻¹⁵ [1] = 2.5 × 10⁴ (25 000) [1]. (b) Atom radius = 2.0 mm × 2.5 × 10⁴ [1] = 5.0 × 10⁴ mm = 50 m [1]. [4] Examiner insight: Convert units at the end, not halfway. 5.0 × 10⁴ mm is 50 m, and an answer given only in millimetres is easy to misread.

7. (a) There is no way to predict the moment at which any individual nucleus decays [1], and which nucleus in the sample goes next is equally unpredictable [1]. (b) The student is wrong [1]. Decay is a change in the nucleus, and physical processes such as heating do not affect it [1]. [4] Examiner insight: Give a decision and a reason. “No, because it is random” is weaker than saying that heating does not alter the nucleus.

8. (a) Any two of these three: radioactive substances in the environment, for example radon from rocks; radiation from space; devices such as hospital X-ray machines [1] [1]. (b) Background rate = 138 / 15 = 9.2 per minute [1]. Rate with source = 2520 / 4 = 630 per minute [1]. Corrected rate = 630 – 9.2 = 620.8, so 621 counts per minute [1]. [5] Examiner insight: Convert both counts to rates before subtracting. Subtracting 138 from 2520 mixes counts taken over different times.

9. (a) A = 238 – 4 = 234 [1]; Z = 92 – 2 = 90 [1]. (b) A = 214 [1]; Z = 83 + 1 = 84 [1]. (c) A = 206 [1]; Z = 82 + 1 = 81 [1]. [6] Examiner insight: In part (c) the unknown is on the left, so the parent’s atomic number is one less than the product’s. Check by adding the right-hand side: 206 + 0 = 206 and 82 + (–1) = 81.

10. (a) Alpha, beta, gamma [1]. (b) Gamma [1]. (c) A strongly ionising radiation transfers energy to atoms as it ionises them [1], so it loses its energy over a short distance and is soon stopped [1]. (d) Electromagnetic radiation from the nucleus [1]. [5] Examiner insight: Link the ideas in part (c): ionisation uses up energy, so range is short. Repeating “alpha is easily stopped” does not explain why.

11. (a) 640 – 20 = 620 counts per minute [1]. (b) Alpha [1], because paper reduces the count rate to close to background (26 against 20) [1]. (c) Gamma [1]. Not alpha, because paper makes almost no difference (905 to 903) [1]. Not beta, because 3 mm of aluminium makes almost no difference (902) [1]. (d) Thick lead reduced the corrected rate from 885 to 310 [1], so about a third still gets through, and lead did not stop it completely [1]. [8] Examiner insight: Quote the figures from the table. Small differences of a few counts are within random variation, so do not call them absorption.

12. (a) Between 4 cm and 6 cm [1]. (b) Alpha [1], because the count rate falls to almost nothing within a few centimetres of air (the range is short) [1]. (c) Alpha particles are strongly ionising, so they knock electrons off many air molecules along their path [1]. Each ionisation transfers energy from the alpha particle [1], so it quickly runs out of energy and stops after a few centimetres [1]. (d) About 0 counts per minute (close to the background) [1], because paper stops alpha particles [1]. [8] Examiner insight: In part (a) give the interval, not a single value, because the readings do not show exactly where the count rate reaches zero. For part (c), name ionisation and energy loss in that order.

Where marks are usually lost

  • Describing the atom as “mostly empty space” without saying that the nucleus is tiny and holds the mass.
  • Listing alpha scattering observations without a matching conclusion.
  • Using the atomic number instead of mass number minus atomic number for neutrons.
  • Getting the charge of an ion the wrong way round: protons minus electrons.
  • Not converting counts to rates before subtracting background.
  • Adding 1 to the atomic number for alpha decay, or subtracting 4 from it.
  • Writing that heating a sample changes its decay.
  • Saying “lead stops gamma” when the data show a count rate that is reduced but not removed.
  • Leaving out the unit (counts per minute, metres) in a final answer.

Next steps

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.7.1 a to f and 3.7.2 a to g.

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