An atom consists of a small, dense nucleus surrounded by electrons arranged in shells.
(a)State the radius of a typical atom and the radius of a typical nucleus, both in standard form, in metres.(2)
(b)Complete the table to give the relative charge of a proton, a neutron and an electron.(3)
(c)State the overall charge of a neutral atom and explain why it has this charge.(2)
(Total for Question 1 is 7 marks)
2
An atom of sodium can be written as Na-23, with atomic number 11.
(a)Define what is meant by 'atomic number'.(1)
(b)Define what is meant by 'mass number'.(1)
(c)Calculate the number of neutrons in an atom of Na-23.(2)
(d)Explain what is meant by the term 'isotope', using two isotopes of an element other than sodium as an example.(2)
(Total for Question 2 is 6 marks)
3
The nucleus of an unstable atom can change to become more stable by emitting radiation.
(a)State what is meant by 'radioactive decay'.(2)
(b)State why radioactive decay is described as a random process.(1)
(c)Name the three main types of ionising nuclear radiation that can be emitted by unstable nuclei.(3)
(Total for Question 3 is 6 marks)
4
Scientific understanding of the atom has changed significantly over time.
(a)State the name of the scientist credited with discovering the neutron.(1)
(b)The 'plum pudding model' pictured the atom as a ball of positive charge with electrons embedded in it. State one observation from the α particle scattering experiment that this model could not explain.(2)
(Total for Question 4 is 3 marks)
5
Radioactive isotopes have important uses in medicine. Gamma-emitting isotopes such as technetium-99m are used as tracers to diagnose disease, and γ or β emitters can be used in radiotherapy to treat cancer by destroying tumour cells. However, radioactive sources also carry risks because ionising radiation can damage or kill living cells. Evaluate the use of radioactive sources in medicine, weighing up the benefits against the risks.
(Total for Question 5 is 6 marks)
6
Alpha, β and γ radiation have different penetrating powers and ionising abilities.
(a)State which material is needed to reduce each type of radiation to a negligible level: (i) α (ii) β (iii) γ.(3)
(b)State what type of particle is emitted as β radiation, and where in the atom it originates.(1)
(c)An α particle consists of two protons and two neutrons. Which of the following gives the relative charge of an α particle?(1)
A) +1
B) +2
C) -1
D) 0
(Total for Question 6 is 5 marks)
7
Nuclear equations show how the mass number and atomic number of a nucleus change during radioactive decay. Total mass number and total atomic number must both be conserved (balanced) on each side of the equation.
(a)Americium-241 (241/95 Am) decays by α emission to form neptunium (Np). Determine the mass number and atomic number of the neptunium produced, and write the balanced nuclear equation.(2)
(b)Carbon-14 (14/6 C) decays by β emission to form nitrogen (N). Write a balanced nuclear equation for this decay, including mass numbers, atomic numbers and the symbol for a β particle.(3)
(Total for Question 7 is 5 marks)
8
The half-life of a radioactive isotope is the time taken for the number of unstable nuclei in a sample (or its activity) to halve.
(a)State what is meant by 'half-life'.(2)
(b)A sample of a radioactive isotope has an initial activity of 800 Bq. Its half-life is 6 hours. Calculate the activity of the sample after 24 hours.(3)
(Total for Question 8 is 5 marks)
9
A student measures the total count rate from a radioactive source using a Geiger-Muller tube, then subtracts the background count rate to find the corrected count rate. The initial corrected count rate from the source is 960 counts per minute (cpm).
(a)The corrected count rate falls to 60 cpm. Calculate the number of half-lives that have passed.(2)
(b)The half-life of the isotope is 5 minutes. Calculate the total time that has passed.(1)
(c)At the end of this time, the measured background count rate was 15 cpm and the total (uncorrected) count rate recorded was 75 cpm. Explain whether this is consistent with your answer to part (a).(1)
(Total for Question 9 is 4 marks)
10
Background radiation is the low-level radiation that is around us all the time.
(a)Give two natural sources of background radiation.(2)
(b)State one man-made (artificial) source of background radiation.(1)
(c)Explain, in terms of atoms, why some rocks (such as granite) emit background radiation.(1)
(Total for Question 10 is 4 marks)
11
A radiographer stands behind a lead screen while a patient has an X-ray taken.
(a)Explain the difference between radioactive contamination and irradiation.(2)
(b)Explain why standing behind the lead screen reduces the radiographer's exposure to X-rays.(2)
(c)State one other precaution taken by hospital workers who regularly handle radioactive sources, to reduce their risk of contamination.(1)
(Total for Question 11 is 5 marks)
12
A machine controls the thickness of aluminium foil as it is produced by passing a beam of radiation through the foil and measuring the count rate with a detector on the other side.
(a)Explain why a β source, rather than an α or a γ source, is suitable for this application.(3)
(b)Suggest what would happen to the detected count rate if the foil became too thick, and how this reading could be used to control the machine.(2)
(Total for Question 12 is 5 marks)
13
A student carried out the required practical investigation into the absorption of radiation from a radioactive source, using a Geiger-Muller (GM) tube and counter, and sheets of paper, aluminium and lead as absorbers.
(a)State the name of the piece of apparatus used to detect and count the radiation in this experiment.(1)
(b)Before starting, the student measured the background count rate for several minutes with no source present. Explain why this step is necessary.(2)
(c)Suggest two variables that should be controlled during this investigation to make it a fair test.(2)
(d)The student's results, already corrected for background, are shown below. Absorber: none, corrected count rate = 340 cpm. Absorber: paper, corrected count rate = 335 cpm. Absorber: aluminium (5 mm), corrected count rate = 210 cpm. Absorber: lead (5 cm), corrected count rate = 8 cpm. Using these results, identify the type(s) of radiation emitted by the source, justifying your answer.(3)
(e)Calculate the percentage of the count rate that is absorbed when the aluminium sheet is added, compared with no absorber.(2)
(Total for Question 13 is 10 marks)
14
In nuclear fission, a large, unstable nucleus (such as uranium-235) absorbs a neutron and splits into two smaller nuclei, releasing energy and further neutrons.
(a)State what is meant by 'nuclear fission'.(2)
(b)A nucleus of uranium-235 absorbs a neutron and splits into a nucleus of krypton-92, a nucleus of barium-141, and some neutrons. Use conservation of mass number to calculate how many neutrons are released.(3)
(c)Explain how the neutrons released in fission reactions inside a nuclear reactor are controlled, to keep the chain reaction steady rather than allowing it to increase uncontrollably.(2)
(Total for Question 14 is 7 marks)
15
Nuclear fusion is the process that powers stars, including the Sun.
(a)State what is meant by 'nuclear fusion'.(2)
(b)Explain why very high temperatures and pressures are needed for nuclear fusion to take place.(2)
(Total for Question 15 is 4 marks)
16
The activity of a radioactive source halves every half-life. Show that, after 3 half-lives have passed, the activity of a source has fallen to approximately 12.5% of its initial value.
(Total for Question 16 is 3 marks)
17
A radioactive source that emits α radiation is considered much more hazardous if it is swallowed or inhaled than if it is held outside the body at a safe distance.
(a)Explain why an α-emitting source is more hazardous inside the body but less hazardous outside the body, compared with a γ-emitting source.(3)
(b)State one way a radiographer can reduce their own exposure when taking a patient's X-ray image.(1)
(Total for Question 17 is 4 marks)
18
Ionisation smoke detectors contain a small source of americium-241, an α-emitting isotope.
(a)Explain how an ionisation smoke detector containing an α-emitting source works.(3)
(b)Suggest why an α emitter, rather than a β or γ emitter, is used in a smoke detector.(2)
(Total for Question 18 is 5 marks)
19
A student used a GM tube to measure the corrected count rate of a radioactive source at 10-minute intervals, as part of a required practical to determine its half-life. Time (minutes): 0, 10, 20, 30, 40. Corrected count rate (cpm): 640, 320, 160, 80, 40.
(a)Explain how these results show the half-life of the source, without needing to plot a graph.(2)
(b)State the half-life of the source shown by this data.(1)
(c)Suggest one improvement the student could make to increase the reliability of the half-life value obtained.(2)
(d)Explain why radioactive decay is described as a random process, and why this means a single count rate reading might not be completely reliable.(2)
(Total for Question 19 is 7 marks)
20
Radon-220 (220/86 Rn) decays by α emission to polonium-216, which itself decays by α emission to lead-212.
(a)Write a balanced nuclear equation for the decay of radon-220 to polonium.(2)
(b)Polonium-216 (216/84 Po) then decays by α emission to lead. Write a balanced nuclear equation for this second decay.(2)
(c)A pure sample of radon-220 has an initial activity of 2.4 x 106 Bq. Its half-life is 55 seconds. Calculate the activity of the sample after 275 seconds, giving your answer in standard form.(3)
(d)Suggest why the activity measured experimentally after 275 seconds might differ slightly from your answer to part (c).(1)
(Total for Question 20 is 8 marks)
Mark scheme · P4 Atomic Structure
Question 1
(a) B1 radius of atom = 1 x 10-10 m, oe (allow 10-10 m)
(a) B1 radius of nucleus = 1 x 10-14 m (or up to 1 x 10-15 m), oe
(a) Answer: Atom: about 1 x 10-10 m. Nucleus: about 1 x 10-14 m.
(c) B1 because the number of protons (positive) equals the number of electrons (negative), so their charges cancel out, oe
(c) Answer: Overall charge = 0, because the number of protons equals the number of electrons.
Question 2
(a) B1 the number of protons in the nucleus of an atom, oe
(a) Answer: The number of protons in the nucleus of an atom.
(b) B1 the total number of protons and neutrons in the nucleus of an atom, oe
(b) Answer: The total number of protons and neutrons in the nucleus of an atom.
(c) M1 correct method: neutrons = mass number - atomic number, i.e. 23 - 11
(c) A1 12 neutrons, cao
(c) Answer: 12 neutrons
(d) B1 isotopes are atoms of the same element (same number of protons/same atomic number) with different numbers of neutrons (different mass numbers), oe
(d) B1 correct named example of two isotopes of the same element, e.g. carbon-12 and carbon-14, oe
(d) Answer: Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, e.g. carbon-12 and carbon-14.
Question 3
(a) B1 the nucleus of an unstable atom breaks down/changes
(a) B1 and emits radiation (in the form of particles and/or energy) to become more stable, oe
(a) Answer: The nucleus of an unstable atom breaks down and emits radiation to become more stable.
(b) B1 it is impossible to predict which nucleus in a sample will decay next, or exactly when any particular nucleus will decay, oe
(b) Answer: It is impossible to predict which nucleus will decay, or when.
(c) B1 α (particle/radiation)
(c) B1 β (particle/radiation)
(c) B1 γ (radiation/rays)
(c) Answer: Alpha, β and γ radiation.
Question 4
(a) B1 (James) Chadwick
(a) Answer: James Chadwick.
(b) B1 a small number of α particles were deflected through large angles (or bounced almost straight back), oe
(b) B1 this could not be explained by the plum pudding model because its charge and mass were thought to be spread evenly through the atom, so nothing was expected to be dense/charged enough to repel an α particle strongly, oe
(b) Answer: A small number of α particles were deflected through large angles, which the plum pudding model could not explain since it had no small, dense, charged region to cause such large deflections.
Question 5
Level 1 (1-2): Simple, largely unlinked statements are made about a benefit or a risk of radioactive sources in medicine, with little reference to specific uses or the ionising properties of radiation.
Level 2 (3-4): Some relevant benefits (e.g. diagnosis, destroying cancer cells) and risks (e.g. damage to healthy cells, risk of causing cancer) are given, with partial links between the properties of the radiation used and its medical purpose, but the evaluation is not fully balanced or justified.
Level 3 (5-6): A clear, well-linked evaluation is given that explains why gamma radiation is suitable as a tracer (penetrates the body to be detected externally, does not need to be intensely ionising) and why radiotherapy deliberately uses ionising radiation to kill tumour cells while accepting some damage to nearby healthy cells, and weighs these benefits against risks such as radiation exposure to patients and staff, damage to healthy tissue and the small increased risk of secondary cancers, reaching a justified overall conclusion (e.g. that the benefits generally outweigh the risks when doses are carefully controlled).
Indicative content:
Gamma-emitting tracers are used because gamma radiation penetrates body tissue and can be detected by a scanner outside the body, allowing doctors to see how an organ is functioning without surgery
A tracer's half-life should be short enough to minimise the time the patient is exposed to radiation, but long enough to allow the scan to be completed
Radiotherapy uses carefully targeted, ionising radiation (beam or implanted source) to kill cancer cells by damaging their DNA
Risk: ionising radiation can also damage healthy cells near the tumour, causing side effects, and can increase the small risk of the patient developing a different cancer later
Risk: staff and other patients need protection (e.g. lead shielding, distance, limited exposure time) to reduce their own radiation dose
Benefit: early and accurate diagnosis, or successful treatment of cancer, can save lives, and the doses used are carefully calculated and monitored to keep risk as low as reasonably possible
An overall judgement that, for most patients, the medical benefit outweighs the risk because doses are controlled and monitored, though this must be balanced individually for each patient and use
Question 6
(a) B1 α: a sheet of paper (or a few cm of air)
(a) B1 β: a few mm of aluminium
(a) B1 γ: several cm (a thick layer) of lead, or metres of concrete
(a) Answer: Alpha: paper. Beta: a few mm of aluminium. Gamma: thick lead (or thick concrete).
(b) B1 a fast-moving electron, emitted from the nucleus when a neutron changes into a proton, oe
(b) Answer: A fast-moving electron, emitted from the nucleus when a neutron changes into a proton.
(c) B1 B) +2, cao
(c) Answer: B) +2
Question 7
(a) M1 mass number of Np = 241 - 4 = 237; atomic number of Np = 95 - 2 = 93
(a) A1 correct balanced equation: 241/95 Am -> 237/93 Np + 4/2 He, cao
(a) Answer: 237/93 Np; equation: 241/95 Am -> 237/93 Np + 4/2 He
(b) M1 correct mass and atomic numbers for nitrogen: mass number stays 14; atomic number increases by 1 to 7
(b) M1 correct β particle notation: 0/-1 e
(b) A1 fully correct balanced equation: 14/6 C -> 14/7 N + 0/-1 e, cao
(b) Answer: 14/6 C -> 14/7 N + 0/-1 e
Question 8
(a) B1 the time taken for the number of unstable/radioactive nuclei in a sample to halve
(a) B1 or, equivalently, the time taken for the activity (count rate) of a sample to halve, oe
(a) Answer: The time taken for the number of unstable nuclei (or the activity) of a sample to halve.
(c) B1 yes, consistent: corrected count rate = 75 - 15 = 60 cpm, which matches the value used in part (a), ft
(c) Answer: Yes: 75 - 15 = 60 cpm, which matches the corrected count rate used above.
Question 10
(a) B1 any one valid natural source, e.g. radon gas (from rocks/ground), cosmic rays (from space), rocks and soil (e.g. granite), food and drink
(a) B1 any second different valid natural source from the list above
(a) Answer: Any two of: radon gas, cosmic rays, rocks/soil, food and drink.
(b) B1 any valid man-made source, e.g. medical X-rays/scans, nuclear industry/power stations, fallout from nuclear weapons testing
(b) Answer: Medical X-rays or scans (or the nuclear industry).
(c) B1 they contain small amounts of naturally occurring radioactive/unstable isotopes, which decay and emit radiation, oe
(c) Answer: They contain naturally occurring unstable (radioactive) isotopes that decay and emit radiation.
Question 11
(a) B1 contamination is the unwanted presence of radioactive material (particles) on or inside an object or person, which continues to expose it/them to radiation until removed, oe
(a) B1 irradiation is being exposed to radiation from a source without being in contact with it, and exposure stops once the person moves away from (or the source is removed from) the radiation, oe
(a) Answer: Contamination is unwanted radioactive material on or in an object/person, which keeps exposing it. Irradiation is exposure to radiation from a source without contact, which stops when the source is removed or the person moves away.
(b) B1 lead is a dense material that strongly absorbs (ionising) X-ray/γ-type radiation, oe
(b) B1 so most of the radiation is absorbed by the screen and does not reach the radiographer, reducing their dose, oe
(b) Answer: Lead strongly absorbs X-rays, so the screen blocks most of the radiation from reaching the radiographer, reducing their exposure.
(c) B1 any valid precaution, e.g. wearing gloves and protective clothing, using remote handling tools/tongs, washing hands thoroughly after handling, storing sources in sealed/shielded containers
(c) Answer: Wearing gloves and protective clothing, or using tongs/remote handling tools.
Question 12
(a) B1 an α source would be completely absorbed by the foil (or even the air gap), so no change in count rate could be detected as thickness varies
(a) B1 a γ source would barely be absorbed by such thin foil, so the count rate would change very little as thickness varies
(a) B1 a β source is partially absorbed by the foil, so small changes in thickness produce a measurable/detectable change in the count rate, oe
(a) Answer: Alpha would be completely absorbed by the foil (no useful signal). Gamma would barely be absorbed (little change with thickness). Beta is partially absorbed, so its count rate changes measurably with foil thickness.
(b) B1 the count rate detected would decrease, because more β radiation is absorbed by the thicker foil, oe
(b) B1 this drop in count rate can be used to automatically adjust the rollers (move them closer together) to reduce the foil thickness back to the correct value, oe
(b) Answer: The count rate would decrease (more radiation absorbed), which can trigger the rollers to move closer together and reduce the thickness back to the correct value.
Question 13
(a) B1 Geiger-Muller (GM) tube (and counter)
(a) Answer: A Geiger-Muller (GM) tube connected to a counter.
(b) B1 background radiation is always present and would otherwise be included in (add to) every reading taken with the source
(b) B1 measuring it separately allows it to be subtracted from later readings, giving the corrected count rate due to the source alone, oe
(b) Answer: Background radiation would otherwise add to every reading; measuring it first allows it to be subtracted so the count rate due to the source alone is known.
(c) B1 the distance between the source and the GM tube (kept the same for each absorber)
(c) B1 the time over which each count rate reading is taken (kept the same for each absorber), oe
(c) Answer: The distance between source and GM tube, and the time each reading is taken over.
(d) B1 the count rate barely changes with paper (340 to 335), so the source emits little or no α radiation
(d) B1 the count rate falls significantly with aluminium (340 to 210), showing the source emits β radiation, which aluminium mostly absorbs
(d) B1 the count rate falls to close to zero with lead (8 cpm, near background), showing any γ radiation present is almost fully absorbed, so the source is best described as a β (and very weak or no γ) source, oe
(d) Answer: The source emits mainly β radiation (little or no α, and little or no significant γ), based on the pattern of absorption by paper, aluminium and lead.
(e) M1 correct method: (340 - 210) / 340 x 100
(e) A1 38.2% (or awrt 38%), cao
(e) Answer: 38.2% (to 3 s.f.)
Question 14
(a) B1 the splitting of a large, unstable nucleus (usually after absorbing a neutron)
(a) B1 into two smaller nuclei (plus neutrons), releasing energy, oe
(a) Answer: The splitting of a large, unstable nucleus into two smaller nuclei, releasing energy (and further neutrons).
(b) M1 total mass number before fission = 235 + 1 (absorbed neutron) = 236
(b) M1 total mass number of the two fragments = 92 + 141 = 233
(b) A1 number of neutrons released = 236 - 233 = 3, cao
(b) Answer: 3 neutrons
(c) B1 control rods (e.g. made of boron) are inserted into or withdrawn from the reactor core
(c) B1 the control rods absorb some of the released neutrons, reducing the number available to cause further fission, keeping the chain reaction steady (rather than increasing), oe
(c) Answer: Control rods absorb some of the released neutrons, limiting how many go on to cause further fission, so the chain reaction stays steady rather than increasing uncontrollably.
Question 15
(a) B1 two light (small) nuclei join/combine to form a single larger nucleus
(a) B1 releasing energy in the process, oe
(a) Answer: Two light nuclei join together to form a larger nucleus, releasing energy.
(b) B1 both nuclei are positively charged, so they repel each other electrostatically as they approach, oe
(b) B1 very high temperature and pressure give the nuclei enough kinetic energy/force them close enough together to overcome this repulsion, so the strong nuclear force can bind them together, oe
(b) Answer: Both nuclei are positively charged and repel each other; extreme temperature and pressure give them enough energy to get close enough for the strong nuclear force to overcome this repulsion and fuse them.
Question 16
M1 recognises the activity halves each half-life: after 1 half-life, activity = 50% of initial
M1 after 2 half-lives, activity = 50% of 50% = 25% of initial
A1 after 3 half-lives, activity = 50% of 25% = 12.5% of initial, cso (answer given in question, full working must be shown)
Answer: 12.5% of the initial activity (shown).
Question 17
(a) B1 α radiation is strongly ionising but has a very short range, so outside the body it is easily absorbed by the outer layer of dead skin (or a short distance of air) before reaching living cells, oe
(a) B1 if swallowed or inhaled, an α source is very close to living cells and all of its (strongly ionising) energy is absorbed by nearby tissue, causing significant damage, oe
(a) B1 γ radiation is weakly ionising and penetrates the body from outside, so it is comparatively more hazardous than α when the source is external, oe
(a) Answer: Alpha is strongly ionising but very short-range, so it is stopped before reaching living cells from outside the body, but if swallowed/inhaled all its ionising energy is absorbed by nearby living cells, causing serious damage; γ is weakly ionising but penetrates the body from outside, making it relatively more hazardous externally.
(b) B1 standing behind a lead screen, or leaving the room/standing further away, or wearing a lead apron
(b) Answer: Standing behind a lead screen (or leaving the room).
Question 18
(a) B1 the α particles ionise the air between two electrodes inside the detector
(a) B1 this allows a small, steady electric current to flow between the electrodes
(a) B1 if smoke enters the detector, it absorbs/blocks the α particles, reducing the ionisation and the current, which triggers the alarm, oe
(a) Answer: Alpha particles ionise the air between two electrodes, allowing a small current to flow; smoke absorbs the α particles, reducing the current and triggering the alarm.
(b) B1 α radiation has a very short range and is strongly ionising, so it effectively ionises the air over the short distance inside the detector
(b) B1 its short range also means it does not travel far enough to escape the casing and reach the user, making it relatively safe to have in the home; a β or γ source would either ionise the air less effectively or be more penetrating/hazardous, oe
(b) Answer: Alpha strongly ionises air over the short distance inside the detector, and its very short range keeps it (and any hazard) contained within the casing, unlike a more penetrating β or γ source.
Question 19
(a) B1 the count rate halves every 10 minutes throughout the table (640 to 320 to 160 to 80 to 40), oe
(a) B1 so the half-life can be read directly as the fixed time interval over which the count rate always halves, without needing to plot and interpret a graph
(a) Answer: The count rate halves every 10 minutes, so the half-life is 10 minutes.
(b) B1 10 minutes, ft from 19a
(b) Answer: 10 minutes
(c) B1 valid improvement identified, e.g. repeat the count rate reading at each time interval and take a mean, or take each count over a longer time period
(c) B1 explanation of how this improves reliability, e.g. reduces the effect of the random/statistical fluctuations in the count rate on the result, oe
(c) Answer: Repeat each reading (or count for longer at each interval) and take a mean, to reduce the effect of random statistical fluctuation on the result.
(d) B1 it is impossible to predict which individual nucleus will decay, or exactly when, so the number of decays in a given time interval varies naturally from one measurement to the next, oe
(d) B1 a single reading may therefore be higher or lower than the true average count rate purely by chance, so it may not be a reliable representation of the source's activity, oe
(d) Answer: Decay is random, so the exact number of counts recorded in a fixed time varies by chance each time; a single reading might be unusually high or low and not reliably represent the true count rate.
Question 20
(a) M1 correct mass and atomic numbers for polonium: mass number = 220 - 4 = 216; atomic number = 86 - 2 = 84
(a) A1 fully correct balanced equation: 220/86 Rn -> 216/84 Po + 4/2 He, cao
(a) Answer: 220/86 Rn -> 216/84 Po + 4/2 He
(b) M1 correct mass and atomic numbers for lead: mass number = 216 - 4 = 212; atomic number = 84 - 2 = 82
(b) A1 fully correct balanced equation: 216/84 Po -> 212/82 Pb + 4/2 He, cao
(b) Answer: 216/84 Po -> 212/82 Pb + 4/2 He
(c) M1 number of half-lives = 275 / 55 = 5
(c) M1 correct method: activity = 2.4 x 106 / 25 = 2.4 x 106 / 32
(c) A1 7.5 x 104 Bq, cao
(c) Answer: 7.5 x 104 Bq
(d) B1 radioactive decay is random, so the measured activity fluctuates statistically around the calculated (theoretical) value, oe; or background radiation/detector efficiency was not fully accounted for
(d) Answer: Decay is random, so the measured activity naturally fluctuates around the calculated value (or background radiation/detector limitations were not fully accounted for).