Show all working for calculation questions — marks are awarded for method.
Useful constants: g = 10 N/kg (or 10 m/s²), speed of light c = 3.0 × 10⁸ m/s.
After completing all questions, click Submit Exam at the end of Question 7 to enter marking mode.
After submitting, answers are auto-graded using keyword matching. Review each model answer to verify your marks.
Question 1: Atomic Structure & Alpha Scattering
12 marks
(a)[3]
Draw a labelled diagram of an atom. Your diagram should show the nucleus containing protons and neutrons, and the electrons orbiting the nucleus.
Model Answer — 1(a)
Nucleus drawn at the centre of the atom containing protons and neutrons [1]
Electrons shown orbiting the nucleus in shells/orbits [1]
All three particles correctly labelled (protons, neutrons, electrons) [1]
⚠ If you missed marks here: Electrons drawn inside the nucleus, or a nucleus that fills most of the circle, loses the first mark — the nucleus must be a tiny dot at the centre with the electrons in orbits well outside it. The third mark is for writing, not drawing: a perfect sketch with nothing written on it scores 2 of 3, because “protons”, “neutrons” and “electrons” all have to be labelled.
Mark 1 -- Nucleus drawn at centre containing protons and neutrons (1 mark)
Mark 2 -- Electrons shown orbiting nucleus in shells (1 mark)
Mark 3 -- All three particles correctly labelled (1 mark)
(b)[3]
State the relative charge and relative mass of each of the three subatomic particles: proton, neutron, and electron.
Model Answer — 1(b)
Proton: relative charge = +1, relative mass = 1 [1]
Neutron: relative charge = 0, relative mass = 1 [1]
Electron: relative charge = −1, relative mass = 1/1836 (or negligible / approximately zero) [1]
⚠ If you missed marks here: “Charge = −1.6 × 10−19 C” is a real value, not a RELATIVE one, and scores nothing here — the answer wanted is −1. The other frequent loss is giving the electron a relative mass of 1: it is 1/1836, or “negligible”. The neutron row (charge 0, mass 1) is the one most often left blank.
Mark 4 -- Proton charge +1 and mass 1 stated correctly (1 mark)
Mark 5 -- Neutron charge 0 and mass 1 stated correctly (1 mark)
Mark 6 -- Electron charge -1 and mass negligible stated correctly (1 mark)
(c) (i)[3]
Describe the experimental setup of the Geiger-Marsden (Rutherford) alpha-particle scattering experiment.
Model Answer — 1(c)(i)
A beam of alpha particles is directed at a very thin gold foil [1]
A movable detector (or fluorescent screen) is placed around the foil to detect scattered alpha particles [1]
The experiment is carried out in a vacuum to prevent alpha particles being absorbed by air [1]
⚠ If you missed marks here: “Fire alpha particles at gold foil” on its own is one mark of three. The foil has to be described as VERY THIN and the detector as movable, or as a screen placed all the way round — a single detector behind the foil cannot see the particles that come back. The mark thrown away most is the vacuum: in air the alpha particles are absorbed before they ever reach the foil.
Mark 7 -- Alpha particles directed at thin gold foil (1 mark)
Mark 8 -- Movable detector or fluorescent screen around the foil (1 mark)
Mark 9 -- Experiment in vacuum to prevent absorption by air (1 mark)
(c) (ii)[3]
State three observations from the alpha scattering experiment, and for each observation, state what it tells us about the structure of the atom.
Model Answer — 1(c)(ii)
Most alpha particles pass straight through the foil → the atom is mostly empty space [1]
Some alpha particles are deflected through small angles → the nucleus is positively charged (repelling the positive alpha particles) [1]
A very small number of alpha particles bounce back (deflected more than 90°) → the nucleus is very small, dense and contains most of the mass [1]
⚠ If you missed marks here: The question asks for an observation AND what it shows, so three bare observations score at most half. Watch the scale words: “most bounced back” is wrong — most passed straight through and only a very small number came back. Blaming the deflection on the electrons scores nothing; it is the positive nucleus repelling the positive alpha particle.
Mark 10 -- Most pass through so atom is mostly empty space (1 mark)
Mark 11 -- Some deflected so nucleus is positively charged (1 mark)
Mark 12 -- Very few bounce back so nucleus is small dense and massive (1 mark)
Isotopes are atoms of the same element with the same number of protons (same atomic number) [1]
but different numbers of neutrons (different mass number) [1]
⚠ If you missed marks here: “Atoms with different masses” is not enough for either mark. Mark 1 is specifically SAME NUMBER OF PROTONS (same proton number); mark 2 is DIFFERENT NUMBER OF NEUTRONS. Reversing them — different proton number, same mass number — describes two different elements, not isotopes.
Mark 13 -- Same element same number of protons (1 mark)
Mark 14 -- Different numbers of neutrons or different mass number (1 mark)
(b)[4]
Carbon-12 and Carbon-14 are isotopes of carbon. For each isotope, write the nuclide notation and state the number of protons, neutrons, and electrons in a neutral atom.
⚠ If you missed marks here: The neutron count is the trap: carbon-14 has 8 neutrons, not 14, because neutrons = nucleon number − proton number = 14 − 6. Keep the two numbers on the correct corners as well; putting the 14 bottom-left writes an atom with 14 protons, which is silicon. Electrons equal protons here because the atoms are neutral, so 6 for both isotopes.
Mark 15 -- Carbon-12 nuclide notation correct with mass 12 and atomic number 6 (1 mark)
Mark 16 -- Carbon-12 has 6 protons 6 neutrons 6 electrons (1 mark)
Mark 17 -- Carbon-14 nuclide notation correct with mass 14 and atomic number 6 (1 mark)
Mark 18 -- Carbon-14 has 6 protons 8 neutrons 6 electrons (1 mark)
(c) (i)[2]
Radium-226 undergoes alpha decay. Complete the nuclear equation below by identifying the daughter nuclide.
²²⁶⁄₈₈Ra → ? + ⁴⁄₂He
Model Answer — 2(c)(i)
Daughter nuclide is Radon (Rn) [1]
Mass number = 226 − 4 = 222, Atomic number = 88 − 2 = 86 [1]
²²⁶⁄₈₈Ra → ²²²⁄₈₆Rn + ⁴⁄₂He
⚠ If you missed marks here: Keeping the element as radium is the commonest error — losing two protons changes WHICH element it is, so the daughter must be radon. The other slip is subtracting the wrong way round: taking 4 off the proton number and 2 off the nucleon number gives 224 and 84 instead of 222 and 86. Alpha removes 4 from the top number and 2 from the bottom one.
Mark 19 -- Correct daughter element radon Rn identified (1 mark)
Mark 20 -- Correct mass number 222 and atomic number 86 (1 mark)
(c) (ii)[1]
Carbon-14 undergoes beta decay to form nitrogen-14. Identify the missing particle in the equation below.
¹⁴⁄₆C → ¹⁴⁄₇N + ?
Model Answer — 2(c)(ii)
The missing particle is a beta particle (electron): ⁰⁄₋₁e (or ⁰⁄₋₁β) [1]
¹⁴⁄₆C → ¹⁴⁄₇N + ⁰⁄₋₁e
⚠ If you missed marks here: Writing “an electron” with no notation loses the mark — the equation needs it as a nuclide, nucleon number 0 and proton number −1. Putting +1 on the bottom describes a positron, which is not on this syllabus and does not balance: 6 has to equal 7 + (−1).
Mark 21 -- Beta particle or electron identified with correct notation (1 mark)
(d)[2]
Explain why the mass number changes in alpha decay but does not change in beta decay.
Model Answer — 2(d)
In alpha decay, an alpha particle (2 protons + 2 neutrons, mass number 4) is emitted from the nucleus, so the mass number decreases by 4 [1]
In beta decay, a neutron in the nucleus changes into a proton and an electron; the electron (beta particle) is emitted but has mass number 0, so the total mass number remains unchanged [1]
⚠ If you missed marks here: “The beta particle is too light to count” does not score. The mark is for the NUMBER: a beta particle has nucleon number 0, so the total is unchanged, while an alpha particle carries away 2 protons and 2 neutrons, nucleon number 4. Say where the beta particle comes from too — a neutron in the nucleus turning into a proton, not an electron leaving a shell.
Mark 22 -- Alpha particle has mass number 4 so mass number decreases by 4 (1 mark)
Mark 23 -- Beta particle has mass number 0 so mass number unchanged neutron converts to proton (1 mark)
Question 3: Three Types of Radiation — Properties & Comparison
12 marks
(a)[6]
Complete the table below comparing the three types of nuclear radiation: alpha (α), beta (β) and gamma (γ). For each type, state the nature, relative charge, relative mass, ionising ability, penetrating power, and what stops it.
Model Answer — 3(a)
Alpha: helium nucleus (2p + 2n), charge +2, mass 4, strongly ionising, least penetrating, stopped by paper or few cm of air [2]
Beta: high-speed electron, charge −1, mass 1/1836 (negligible), moderately ionising, moderate penetrating, stopped by a few mm of aluminium [2]
Gamma: electromagnetic radiation (photon), charge 0, mass 0, weakly ionising, most penetrating, reduced by several cm of lead or thick concrete [2]
⚠ If you missed marks here: The mark lost most here is “gamma is stopped by lead”. Gamma is never fully stopped, only REDUCED — several centimetres of lead or thick concrete cut the count rate down. Check the alpha row as well: relative charge +2 and relative mass 4, not +1 and 2; and beta needs a few millimetres of aluminium, not paper.
Mark 24 -- Alpha is helium nucleus with charge +2 and mass 4 (1 mark)
Mark 25 -- Alpha strongly ionising least penetrating stopped by paper (1 mark)
Mark 26 -- Beta is high speed electron with charge -1 (1 mark)
Mark 27 -- Beta moderately ionising stopped by aluminium (1 mark)
Mark 28 -- Gamma is electromagnetic radiation with no charge and no mass (1 mark)
Mark 29 -- Gamma weakly ionising most penetrating stopped by thick lead (1 mark)
(b)[3]
Describe an experiment to identify the type of radiation emitted by a source, using absorbers (paper, aluminium and lead) and a Geiger-Müller (GM) tube.
Model Answer — 3(b)
Measure the count rate from the source with no absorber using a GM tube, then subtract the background count rate [1]
Place a sheet of paper between the source and the GM tube: if the count rate drops significantly, the source emits alpha radiation [1]
If paper does not stop it, place a few mm of aluminium: if count rate drops, the source emits beta. If aluminium does not stop it, the source emits gamma (only thick lead significantly reduces the count rate) [1]
⚠ If you missed marks here: Skipping the background count throws away mark 1 — every reading has to have the background subtracted, and a reading with no absorber is needed as the control. When you reach lead, do not write “lead stops the gamma”: say the count rate falls but stays well above background. A count rate unchanged by paper rules alpha out; one unchanged by a few millimetres of aluminium rules beta out.
Mark 30 -- Measure count rate with GM tube and subtract background count (1 mark)
Mark 31 -- Paper absorber test for alpha radiation (1 mark)
Mark 32 -- Aluminium absorber test for beta and lead for gamma (1 mark)
(c)[3]
Explain why alpha particles are the most ionising but the least penetrating type of radiation.
Model Answer — 3(c)
Alpha particles are relatively large and have a charge of +2, so they interact strongly with atoms they pass close to [1]
Each ionisation event removes energy from the alpha particle [1]
Because alpha particles cause so many ionisations in a short distance, they lose all their kinetic energy quickly and are stopped after only a few centimetres of air (or a sheet of paper) [1]
⚠ If you missed marks here: “Alpha is too big to get through paper” scores nothing — what is wanted is an energy argument. Alpha carries charge +2 and is massive, so it strips electrons from the atoms it passes, and EVERY ionisation costs it some kinetic energy. Thousands of ionisations within a couple of centimetres leave it with none left, which is why it stops. Strong ionisation and poor penetration are the same fact, not two.
Mark 33 -- Alpha particles are large and highly charged so interact strongly with atoms (1 mark)
Mark 34 -- Each ionisation removes energy from the alpha particle (1 mark)
Mark 35 -- Many ionisations so energy lost quickly and stopped in short distance (1 mark)
Question 4: Radioactive Decay Equations
10 marks
(a)[2]
State what is meant by radioactive decay being described as spontaneous and random.
Model Answer — 4(a)
Spontaneous means the decay is not caused by any external factors and cannot be influenced by changes in physical conditions (e.g. temperature, pressure) [1]
Random means it is impossible to predict which particular nucleus will decay next, or when a particular nucleus will decay [1]
⚠ If you missed marks here: These two get swapped constantly. Spontaneous means nothing sets the decay off and nothing you do to the sample — heating it, compressing it, reacting it chemically — alters the rate. Random means you cannot say WHICH nucleus goes next or WHEN. “It happens by itself at random times” is one idea covering both marks and will collect only one.
Mark 36 -- Spontaneous means not caused by external factors cannot be influenced (1 mark)
Mark 37 -- Random means cannot predict which nucleus will decay next (1 mark)
(b) (i)[2]
Uranium-238 (²³⁸⁄₉₂U) undergoes alpha decay. Write the balanced nuclear equation for this decay.
Model Answer — 4(b)(i)
Correct daughter nuclide: Thorium-234 with atomic number 90 [1]
Balanced equation with alpha particle emitted [1]
²³⁸⁄₉₂U → ²³⁴⁄₉₀Th + ⁴⁄₂He
Check: mass 238 = 234 + 4 ✓ atomic 92 = 90 + 2 ✓
⚠ If you missed marks here: Thorium-234 with proton number 88 is the usual wrong answer — alpha decay removes 2 protons from 92, giving 90, not 4. Put the balance check in your working (238 = 234 + 4 and 92 = 90 + 2). An equation without the alpha particle written as nucleon number 4, proton number 2 loses the second mark even when the daughter is right.
Mark 38 -- Thorium-234 identified as daughter with atomic number 90 (1 mark)
Mark 39 -- Equation balanced with alpha particle mass 4 charge 2 (1 mark)
(b) (ii)[2]
Strontium-90 (⁹⁰⁄₃₈Sr) undergoes beta decay. Write the balanced nuclear equation for this decay.
Model Answer — 4(b)(ii)
Correct daughter nuclide: Yttrium-90 with atomic number 39 [1]
Balanced equation with beta particle emitted [1]
⁹⁰⁄₃₈Sr → ⁹⁰⁄₃₉Y + ⁰⁄₋₁e
Check: mass 90 = 90 + 0 ✓ atomic 38 = 39 + (−1) ✓
⚠ If you missed marks here: Beta decay pushes the proton number UP, so strontium 38 becomes yttrium 39; writing 37 and calling it rubidium is the standard slip, made by assuming every decay makes the nucleus smaller. The nucleon number stays at 90, so an answer of 89 means you treated the beta particle as though it had mass.
Mark 40 -- Yttrium-90 identified as daughter with atomic number 39 (1 mark)
Mark 41 -- Equation balanced with beta particle mass 0 charge -1 (1 mark)
(c)[4]
An unknown nuclide X undergoes alpha decay to form lead-208 (²⁰⁸⁄₈₂Pb).
X → ²⁰⁸⁄₈₂Pb + ⁴⁄₂He
(i) Determine the mass number of X.
(ii) Determine the atomic number of X.
(iii) Identify element X.
(iv) Write the complete balanced equation.
⚠ If you missed marks here: Work BACKWARDS: X existed before the decay, so add rather than subtract. 208 + 4 = 212 and 82 + 2 = 84, which is polonium-212. Subtracting gives 204 and 80, mercury — the answer almost everyone who rushes this writes. Part (iv) also wants the whole equation written out: three correct numbers with no equation still drops a mark.
Mark 42 -- Mass number of X is 212 (1 mark)
Mark 43 -- Atomic number of X is 84 (1 mark)
Mark 44 -- Element X identified as polonium Po (1 mark)
Mark 45 -- Complete balanced equation written correctly (1 mark)
Question 5: Half-life Calculations from Data Table
12 marks
(a)[2]
Define half-life of a radioactive isotope.
Model Answer — 5(a)
The half-life is the time taken for half of the radioactive nuclei (in a sample) to decay [1]
OR the time taken for the activity / count rate of a source to fall to half its initial value [1]
⚠ If you missed marks here: “The time for the source to decay” or “the time for half the source to disappear” is too loose. The mark needs half of WHAT: half the radioactive nuclei in the sample, or the count rate falling to half its starting value. Nuclei do not disappear either — they change into a different nuclide.
Mark 46 -- Time taken for half the radioactive nuclei to decay (1 mark)
Mark 47 -- Or time for count rate or activity to halve (1 mark)
(b)[4]
A student measures the count rate from a radioactive source at regular intervals. The results are shown in the table below. The background count rate is 30 counts per minute (cpm).
(b)(i) Calculate the corrected count rate at time = 0.
⚠ If you missed marks here: Background is subtracted, never added: 400 + 30 = 430 is the wrong-direction answer, and 400 on its own ignores the correction altogether. The 400 cpm already contains the 30 cpm coming from rocks, cosmic rays and the air, so the source itself is producing 370 cpm.
Mark 48 -- Corrected count rate at time 0 is 370 cpm (1 mark)
(b)(ii) Using the corrected count rates, determine the half-life of the source. Show your working.
Model Answer — 5(b)(ii)
Half of 370 cpm = 185 cpm [1]
From the table/graph: 185 cpm falls between t = 20 min (210 cpm) and t = 30 min (160 cpm) [1]
By interpolation: half-life ≈ 25 minutes (accept 24–28 minutes) [1]
Alternative method: 370 → 185 cpm, reading from graph gives t½ ≈ 25 min
⚠ If you missed marks here: The half-life must come from the CORRECTED figures. Halving the raw 400 to 200 gives roughly 22 minutes instead of 25, and it is wrong in principle because a real count can never fall below the 30 cpm background. Write 185 cpm down explicitly and name the two rows it lies between — “25 minutes” with no working scores 1 of 3.
Mark 49 -- Half of 370 is 185 cpm calculated correctly (1 mark)
Mark 50 -- Identified that 185 falls between 20 and 30 minutes (1 mark)
Mark 51 -- Half-life stated as approximately 25 minutes accept 24 to 28 (1 mark)
(c)[3]
On the axes below, sketch the corrected decay curve using the data from the table. Label the axes and plot the points clearly.
Model Answer — 5(c)
Your diagram should look like this:
Correct axes with labels (time on x-axis, corrected count rate on y-axis) and sensible scales [1]
All corrected data points plotted accurately (370, 280, 210, 160, 120, 90, 65) [1]
Smooth exponential decay curve drawn through the points (not straight lines between points) [1]
⚠ If you missed marks here: Plot the corrected values (370, 280, 210 …), not the measured ones; a curve starting at 400 cannot earn mark 2. Join them with one smooth curve, because ruled lines from point to point is the mark most often lost on any decay graph. Each axis needs a quantity and a unit, so “time” on its own is not a label.
Mark 52 -- Axes correctly labelled with sensible scales (1 mark)
Mark 53 -- All corrected data points plotted accurately (1 mark)
Mark 54 -- Smooth exponential decay curve drawn (1 mark)
(d)[3]
After how many half-lives will the corrected count rate fall below 25 cpm? Show your working.
After 4 half-lives the corrected count rate is 23.125 cpm which is below 25 cpm. Answer: 4 half-lives [1]
370 ÷ 2⁴ = 370 ÷ 16 = 23.125 cpm < 25 cpm
⚠ If you missed marks here: Three half-lives leaves 46.25 cpm, still above 25, so stopping at 3 is the near-miss answer — you want the first value UNDER 25, which is 23.125 after 4. Start from 370 rather than 400: halving the uncorrected value four times lands on exactly 25 cpm and makes the answer ambiguous.
Mark 55 -- Starting value 370 cpm identified (1 mark)
Mark 56 -- Successive halving shown 370 to 185 to 92.5 to 46.25 to 23.125 (1 mark)
Mark 57 -- Answer 4 half-lives with correct reasoning below 25 cpm (1 mark)
Question 6: Applications of Radioactivity
11 marks
(a)[3]
Explain how a smoke alarm uses a radioactive source. State the type of radiation used and explain why that type is chosen.
Model Answer — 6(a)
An alpha source (e.g. americium-241) is placed inside the smoke alarm; it ionises the air between two plates, creating a small current [1]
When smoke enters the alarm, the smoke particles absorb the alpha particles, reducing the ionisation and the current drops, triggering the alarm [1]
Alpha radiation is used because it is strongly ionising (so it effectively ionises air) and has low penetrating power (so it is safe and does not escape the detector casing) [1]
⚠ If you missed marks here: Naming gamma loses all three marks, since gamma would pass through smoke unabsorbed and nothing would change. Say what alpha actually does: it ionises the air between the plates so a small current flows, then smoke ABSORBS the alpha, the current FALLS and the alarm sounds. “The smoke sets the alarm off” with no mention of ionisation or current scores nothing.
Mark 58 -- Alpha source ionises air creating current between plates (1 mark)
Mark 59 -- Smoke absorbs alpha particles reducing current triggering alarm (1 mark)
Mark 60 -- Alpha used because strongly ionising and low penetrating power safe (1 mark)
(b)[3]
Explain how radioactive tracers are used in medicine. State the type of radiation used and explain why it is suitable.
Model Answer — 6(b)
A radioactive tracer (e.g. technetium-99m) is injected into or swallowed by the patient; it travels through the body and concentrates in the organ being investigated [1]
A gamma camera outside the body detects the gamma radiation emitted by the tracer and produces an image of the organ [1]
Gamma radiation is used because it can penetrate out of the body to reach the detector, and a tracer with a short half-life is chosen to minimise radiation exposure to the patient [1]
⚠ If you missed marks here: Alpha or beta as the tracer scores nothing — both would be absorbed inside the patient and the camera outside would see nothing at all. Two marks hang on details students skip: the tracer goes IN (injected or swallowed) and collects in the organ being investigated, and its half-life must be SHORT so the activity dies away instead of irradiating the patient for weeks.
Mark 61 -- Tracer injected or swallowed and concentrates in target organ (1 mark)
Mark 62 -- Gamma camera detects radiation from outside the body produces image (1 mark)
Mark 63 -- Gamma used because penetrates body and short half-life reduces exposure (1 mark)
(c)[3]
Explain how radiation is used to control the thickness of paper in a factory. Include the type of radiation used and how the system works.
Model Answer — 6(c)
A beta source is placed on one side of the paper and a detector (GM tube) on the other side [1]
If the paper is too thick, fewer beta particles pass through and the count rate decreases; if too thin, more pass through and the count rate increases [1]
Beta radiation is used because it is partially absorbed by paper (alpha would be fully stopped, gamma would pass straight through regardless of thickness changes) [1]
⚠ If you missed marks here: Gamma is the wrong choice even though it sounds the most powerful: it passes through paper of any thickness, so the count rate would not respond to thickness at all. Alpha fails the opposite way, since paper stops it completely. Get the direction right too — thicker paper means MORE absorption and a LOWER count rate.
Mark 64 -- Beta source on one side and detector on the other side of paper (1 mark)
Mark 65 -- Thickness change affects count rate too thick less detected too thin more detected (1 mark)
Mark 66 -- Beta used because partially absorbed by paper alpha stopped gamma not affected (1 mark)
(d)[2]
State one use of gamma radiation in medicine (other than tracers) and explain why gamma is suitable for this purpose.
Model Answer — 6(d)
Gamma radiation is used to sterilise medical equipment OR to treat cancer (radiotherapy) [1]
Gamma is suitable because it is highly penetrating so it can pass through packaging to kill bacteria on equipment, or it can be focused on a tumour deep inside the body to destroy cancer cells [1]
⚠ If you missed marks here: Naming tracers again scores nothing, because the question rules them out. Take sterilising equipment or radiotherapy and explain through PENETRATION: gamma passes through sealed packaging to kill the bacteria inside, or reaches a tumour deep in the body. “Gamma is dangerous so it kills things” is not an explanation.
Mark 67 -- Sterilising equipment or treating cancer radiotherapy (1 mark)
Mark 68 -- Gamma suitable because highly penetrating passes through packaging or reaches tumour (1 mark)
Question 7: Nuclear Fission, Fusion & Safety
12 marks
(a)[2]
Define nuclear fission.
Model Answer — 7(a)
Nuclear fission is the splitting of a large / heavy nucleus [1]
into two smaller nuclei (of roughly equal size), releasing a large amount of energy (and usually neutrons) [1]
⚠ If you missed marks here: “Splitting an atom” is the phrasing that costs a mark — it is the NUCLEUS that splits, and it has to be a large or heavy one. The second mark needs the products: two smaller nuclei of roughly equal size, plus a large release of energy and neutrons. Describing neutrons going on to hit other nuclei is the chain reaction, which this part does not ask for.
Mark 69 -- Splitting of a large or heavy nucleus (1 mark)
Mark 70 -- Into two smaller nuclei releasing energy and neutrons (1 mark)
(b)[2]
Define nuclear fusion.
Model Answer — 7(b)
Nuclear fusion is the joining / combining of two light / small nuclei [1]
to form a single heavier nucleus, releasing a large amount of energy [1]
⚠ If you missed marks here: Fusion and fission get written the wrong way round under time pressure — fusion JOINS two light nuclei into one heavier nucleus. “Two atoms join” is loose, because atoms carry electrons and it is bare nuclei that fuse. Do not say energy is needed overall either: energy is released, even though high temperature and pressure are required to start it.
Mark 71 -- Joining or combining of two light or small nuclei (1 mark)
Mark 72 -- To form a heavier nucleus releasing energy (1 mark)
(c)[1]
State where nuclear fusion occurs naturally.
Model Answer — 7(c)
Nuclear fusion occurs naturally in stars (including the Sun) [1]
⚠ If you missed marks here: This single mark is for stars, or the Sun. “In a nuclear power station” describes fission, not fusion, and a fusion reactor on Earth is not natural. One word is enough, so never leave this one blank.
Mark 73 -- In stars or the Sun (1 mark)
(d)[3]
Explain three safety precautions that should be taken when handling radioactive sources in a school laboratory.
Model Answer — 7(d)
Always handle the source with tongs (at arm's length) and never touch it directly — this increases the distance between the source and the body, reducing exposure [1]
Point the source away from the body (and other people) and keep exposure time as short as possible [1]
Store the source in a lead-lined container when not in use to shield against radiation, and never eat or drink near radioactive sources [1]
⚠ If you missed marks here: A list with no reasons is how this part ends up on 1 or 2 marks. Each precaution has to say what it reduces: tongs increase the DISTANCE, taking the source out only briefly reduces the TIME of exposure, and the lead-lined box provides SHIELDING. “Wear gloves” and “be careful” earn nothing, since gloves stop nothing more penetrating than alpha.
Mark 74 -- Handle with tongs at arms length never touch directly (1 mark)
Mark 75 -- Point away from body and minimise exposure time (1 mark)
Mark 76 -- Store in lead-lined container and no eating or drinking nearby (1 mark)
(e)[2]
Describe how ionising radiation can damage living cells and state two possible consequences of this damage.
Model Answer — 7(e)
Ionising radiation can damage or destroy the DNA in living cells by ionising atoms within the cell [1]
Two consequences: (1) cells may mutate, leading to cancer, and (2) cells may die, leading to radiation sickness / burns (accept: cell death, tissue damage, genetic mutations passed to offspring) [1]
⚠ If you missed marks here: “It burns you” or “it makes you ill” is not a mechanism. Mark 1 is for the radiation IONISING atoms inside the cell and so damaging or destroying DNA. Then give two DIFFERENT consequences — mutation leading to cancer, and cell death causing radiation sickness or burns. Writing “cancer” twice in two wordings is one consequence.
Mark 77 -- Radiation damages DNA by ionising atoms in cells (1 mark)
Mark 78 -- Two consequences such as cancer mutations and cell death radiation sickness (1 mark)
(f)[2]
State and explain the three principles of radiation protection: time, distance and shielding.
Model Answer — 7(f)
Time: minimise the time spent near a radioactive source to reduce the total dose of radiation received. Distance: maximise the distance from the source, because the radiation spreads out and its intensity falls rapidly as you move away (and alpha and beta are absorbed by a few centimetres or a few metres of air). Shielding: place absorbing materials (e.g. lead, concrete) between the source and people to absorb the radiation [1]
All three principles correctly stated with at least a brief explanation of each [1]
⚠ If you missed marks here: Naming time, distance and shielding without saying what each one does is worth one mark, not two. Time: less time near the source means a smaller total dose. Distance: the radiation spreads out and the air absorbs alpha and beta, so the intensity falls quickly as you step back. Shielding: lead or concrete placed between you and the source absorbs the radiation.
Mark 79 -- Time distance and shielding all three named correctly (1 mark)
Mark 80 -- Explanation of how each principle reduces radiation exposure (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Exam Submitted — Marking Mode Active
Click "Show Model Answer" on each question to check your work.