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In marking mode, compare your answers with the model answers and tick the marks you earned.
Question 1 — The Nucleus: Composition & Isotopes
Total: 11 marks
(a)[2]
State the composition of the nucleus of an atom.
Model Answer — 1(a)
The nucleus contains protons [1]
The nucleus contains neutrons [1]
⚠ If you missed marks here: Answering “protons and electrons” costs a mark — electrons sit OUTSIDE the nucleus, in shells. There are two separate marks on offer, one for protons and one for neutrons, so a single word such as “nucleons” will not collect both.
Mark 1 -- Protons stated as part of nucleus composition (1 mark)
Mark 2 -- Neutrons stated as part of nucleus composition (1 mark)
(b) (i)[1]
Define the term proton number (atomic number).
Model Answer — 1(b)(i)
The proton number (atomic number) is the number of protons in the nucleus of an atom [1]
⚠ If you missed marks here: Defining it as “the number of protons and neutrons” gives the nucleon number instead and scores zero. “The bottom number” says where it is written rather than what it is — the mark wants the number of protons in the nucleus of an atom.
Mark 1 -- Number of protons in the nucleus correctly defined (1 mark)
(b) (ii)[1]
Define the term nucleon number (mass number).
Model Answer — 1(b)(ii)
The nucleon number (mass number) is the total number of protons and neutrons in the nucleus [1]
⚠ If you missed marks here: The word TOTAL is doing the work: nucleon number is protons PLUS neutrons, so “the number of neutrons” gets nothing. It is also not a mass in kilograms or grams; it is a count of particles, which is why it carries no unit.
Mark 1 -- Total number of protons and neutrons in nucleus correctly defined (1 mark)
(c)[3]
The nuclide notation for a sodium atom is 2311Na.
State the number of:
(i) protons
(ii) neutrons
(iii) electrons in a neutral atom of sodium.
Model Answer — 1(c)
(i) Number of protons = 11 [1]
(ii) Number of neutrons = 23 − 11 = 12 [1]
(iii) Number of electrons = 11 (same as protons in a neutral atom) [1]
⚠ If you missed marks here: Neutrons = 23 − 11 = 12; writing 23, by reading the top number as the neutron count, is the error that appears most often. Electrons are 11 because the question specifies a NEUTRAL atom — had it said sodium ion, the answer would have been 10.
Mark 1 -- Protons stated as 11 (1 mark)
Mark 2 -- Neutrons stated as 12 (1 mark)
Mark 3 -- Electrons stated as 11 (1 mark)
(d)[2]
Chlorine has two isotopes: 3517Cl and 3717Cl.
Explain why these are described as isotopes of the same element.
Model Answer — 1(d)
Isotopes are atoms of the same element with the same proton number (same number of protons) [1]
But different nucleon numbers (different numbers of neutrons) [1]
⚠ If you missed marks here: One mark comes from each half and most answers supply only one: same proton number (17 for both) AND different nucleon number, meaning 18 neutrons against 20. “They have different masses” names the difference but never states what is the same, so it cannot score twice.
Mark 1 -- Same proton number or same number of protons stated (1 mark)
Mark 2 -- Different nucleon number or different number of neutrons stated (1 mark)
(e)[2]
Explain why isotopes of the same element have the same chemical properties.
Model Answer — 1(e)
Chemical properties depend on the number of electrons (or electron arrangement) [1]
Isotopes have the same number of electrons because they have the same proton number [1]
⚠ If you missed marks here: “They are both chlorine” restates the question. Chemical behaviour is set by the ELECTRONS — how many there are and how they are arranged — and both isotopes have 17 protons, so both have 17 electrons and react identically. The extra neutrons sit in the nucleus and take no part in bonding.
Mark 1 -- Chemical properties depend on electrons or electron arrangement (1 mark)
Mark 2 -- Isotopes have same number of electrons because same proton number (1 mark)
Question 2 — Alpha Scattering Experiment
Total: 12 marks
(a)[3]
Draw a labelled diagram of the Geiger-Marsden alpha scattering experiment. Your diagram should show the alpha source, collimator, gold foil, and fluorescent detector screen.
Model Answer — 2(a)
Your diagram should look like this:
Alpha source shown in lead container with collimator to produce narrow beam [1]
Very thin gold foil in the path of the beam [1]
Fluorescent screen or movable detector surrounding the foil to detect scattered particles [1]
⚠ If you missed marks here: Two features get left out of this drawing: the LEAD block or collimator that turns the source into a narrow beam, and a screen or detector that can be moved right round the foil. A detector drawn only behind the foil cannot record the particles that come back, which is the entire point of the experiment. The foil must be labelled as thin gold.
Mark 1 -- Alpha source with lead container and collimator shown (1 mark)
Mark 2 -- Thin gold foil shown in beam path (1 mark)
Mark 3 -- Fluorescent screen or detector surrounding the foil (1 mark)
(b)[3]
State three observations from the alpha scattering experiment.
Model Answer — 2(b)
Most alpha particles passed straight through the gold foil undeflected [1]
Some alpha particles were deflected through small angles [1]
A very small number of alpha particles bounced back (deflected through angles greater than 90 degrees) [1]
⚠ If you missed marks here: This part wants what was SEEN, not what it means, so “the atom is mostly empty space” is a conclusion and belongs in part (c). Three separate observations are needed and the proportions must be right: most straight through, some deflected slightly, a very few back through more than 90°.
Mark 1 -- Most alpha particles pass straight through undeflected (1 mark)
Mark 2 -- Some deflected through small angles (1 mark)
Mark 3 -- Very few bounce back or deflected more than 90 degrees (1 mark)
(c)[3]
Explain what each of the three observations in part (b) tells us about the structure of the atom.
Model Answer — 2(c)
Most pass through: the atom is mostly empty space [1]
Some deflected: there is a concentration of positive charge in the atom (the nucleus) which repels the positive alpha particles [1]
Very few bounce back: the nucleus is very small, very dense, and has a large positive charge [1]
⚠ If you missed marks here: Here the traffic runs the other way — repeating the observations from part (b) scores nothing, because each one has to be turned into a statement about the atom. Pair them up one to one: empty space, a concentrated positive charge, and a nucleus that is tiny yet holds nearly all the mass. Saying “this proves the nucleus exists” three times is one idea, not three.
Mark 1 -- Most pass through because the atom is mostly empty space (1 mark)
Mark 2 -- Some deflected because nucleus has positive charge that repels alpha particles (1 mark)
Mark 3 -- Very few bounce back because nucleus is small dense and has large positive charge (1 mark)
(d)[3]
Before the alpha scattering experiment, the “plum pudding” model of the atom was widely accepted. In this model, the atom was thought to be a uniform sphere of positive charge with electrons embedded in it.
Explain how the results of the alpha scattering experiment disproved the plum pudding model.
Model Answer — 2(d)
In the plum pudding model, the positive charge is spread out evenly across the whole atom, so alpha particles should pass through with only very slight deflections [1]
The observation that some alpha particles were deflected at large angles shows the positive charge must be concentrated in a very small region (the nucleus), not spread out [1]
The observation that most particles passed straight through shows the atom is mostly empty space, contradicting the idea of a solid sphere of charge [1]
⚠ If you missed marks here: A model cannot be disproved until you state what it PREDICTED. Say that a positive charge spread evenly through the atom could only nudge an alpha particle slightly, then set that against the large-angle deflections actually observed. “The plum pudding model was wrong because Rutherford proved it” scores nothing.
Mark 1 -- Plum pudding model has positive charge spread evenly so only slight deflections expected (1 mark)
Mark 2 -- Large angle deflections show positive charge concentrated in small nucleus (1 mark)
Mark 3 -- Most passing through shows atom is mostly empty space not a solid sphere (1 mark)
State the nature of each type of nuclear radiation:
(i) alpha (α) radiation
(ii) beta (β) radiation
(iii) gamma (γ) radiation
Model Answer — 3(a)
(i) Alpha radiation consists of 2 protons and 2 neutrons (a helium nucleus) [1]
(ii) Beta radiation consists of a high-speed electron emitted from the nucleus [1]
(iii) Gamma radiation is a high-frequency electromagnetic wave (photon) [1]
⚠ If you missed marks here: Alpha is a helium NUCLEUS — calling it a helium atom implies electrons are attached and loses the mark. Beta has to be a fast electron emitted FROM THE NUCLEUS, not one knocked off a shell, and gamma is an electromagnetic wave or photon rather than a particle.
Mark 1 -- Alpha is 2 protons and 2 neutrons or helium nucleus (1 mark)
Mark 2 -- Beta is high-speed electron from nucleus (1 mark)
Mark 3 -- Gamma is electromagnetic wave or photon (1 mark)
(b)[3]
The diagram below shows the three types of radiation passing between two electrically charged plates. Describe and explain the path of each type of radiation in the electric field.
Model Answer — 3(b)
Alpha (α): deflected toward the negative plate because it has a positive charge (+2); deflection is relatively small because of its large mass [1]
Beta (β): deflected toward the positive plate because it has a negative charge (−1); deflection is larger than alpha because of its much smaller mass [1]
Gamma (γ): passes straight through undeflected because it is uncharged (electromagnetic radiation) [1]
⚠ If you missed marks here: Sending alpha towards the positive plate is the single most common error. Alpha is positive, so it is pulled towards the NEGATIVE plate; beta is negative, so it moves to the positive one. Half of each mark is the SIZE of the bend: beta curves far more than alpha because its mass is thousands of times smaller. Gamma is uncharged and carries straight on.
Mark 1 -- Alpha deflected toward negative plate because positively charged (1 mark)
Mark 2 -- Beta deflected toward positive plate because negatively charged with larger deflection (1 mark)
Mark 3 -- Gamma undeflected because uncharged electromagnetic radiation (1 mark)
(c)[2]
Explain why alpha particles cause the greatest ionisation of air molecules compared to beta particles and gamma rays.
Model Answer — 3(c)
Alpha particles have a larger charge (+2) compared to beta (−1) and gamma (0), so they exert a stronger electric force on air molecules [1]
Alpha particles are large and slow-moving, so they interact with more air molecules per unit distance (higher ionisation density) and lose energy more quickly [1]
⚠ If you missed marks here: “Alpha is the most dangerous” is not physics. Two things do the work here: a charge of +2 against beta’s −1 and gamma’s 0, giving a stronger pull on the electrons in air molecules; and its large mass and low speed, so it lingers near each molecule and ionises far more per centimetre travelled.
Mark 1 -- Alpha has larger charge so exerts stronger force on molecules (1 mark)
Mark 2 -- Alpha is larger and slower so interacts with more molecules per unit distance (1 mark)
(d)[3]
A radioactive source is placed in front of a Geiger-Müller detector. Different absorbers are placed between the source and the detector:
• With no absorber: the corrected count rate is 450 counts per minute.
• With a sheet of paper: the count rate drops to 280 counts per minute.
• With 5 mm of aluminium: the count rate drops to 35 counts per minute (close to background).
What type(s) of radiation does this source emit? Explain your reasoning.
Model Answer — 3(d)
The source emits alpha and beta radiation [1]
Paper reduces the count rate significantly: this means some of the radiation is alpha (stopped by paper) [1]
Aluminium reduces the count rate to near background: the remaining radiation after paper must be beta, which is stopped by a few mm of aluminium. If gamma were present, it would not be stopped by 5 mm aluminium [1]
⚠ If you missed marks here: “Alpha only” ignores the second drop, and “gamma is present” contradicts the final reading. Paper cutting 450 to 280 shows alpha; aluminium then bringing it to 35, which is about background, shows the remainder was beta and that NO gamma is there, since 5 mm of aluminium would barely dent a gamma count. Quote those numbers as your evidence.
Mark 1 -- Source emits alpha and beta radiation (1 mark)
Mark 2 -- Paper stops alpha causing the first drop in count rate (1 mark)
Mark 3 -- Aluminium stops beta and no gamma present because count drops to background (1 mark)
Question 4 — Decay Curves & Half-life
Total: 12 marks
(a)[2]
Define the term half-life of a radioactive isotope.
Model Answer — 4(a)
The half-life is the time taken for the activity (or count rate) of a radioactive isotope to decrease to half its original value [1]
OR the time taken for half the radioactive nuclei in a sample to decay [1]
⚠ If you missed marks here: State half of WHAT. “The time for the source to halve” or “to run out” scores nothing; the mark needs the activity or count rate falling to half its original value, or half the undecayed nuclei having decayed. Half-life is not half the time the source lasts.
Mark 1 -- Time taken for activity or count rate to halve (1 mark)
Mark 2 -- Or time for half the nuclei to decay (1 mark)
(b) (i)[3]
The table below shows how the corrected count rate of a radioactive source changes with time.
Time / hours
0
2
4
6
8
10
12
Corrected count rate / Bq
1600
1130
800
566
400
283
200
Plot these values on a decay curve. Label the axes clearly.
Model Answer — 4(b)(i)
Your diagram should look like this:
Axes correctly labelled with quantities and units (time/hours on x-axis, corrected count rate/Bq on y-axis) [1]
All data points plotted correctly at the right positions [1]
Smooth curve of best fit drawn through the points (not straight lines between points) [1]
⚠ If you missed marks here: Ruling straight lines from point to point is the mark lost most often — decay is smooth, so draw one curve of best fit through all seven points. Both axes need a quantity AND a unit (time/hours, corrected count rate/Bq), and pick a scale that fills most of the grid instead of squashing the data into a corner.
Mark 1 -- Axes labelled with correct quantities and units (1 mark)
Mark 2 -- All data points plotted accurately (1 mark)
Mark 3 -- Smooth curve of best fit drawn through points (1 mark)
(b) (ii)[2]
Use your graph to determine the half-life of this radioactive source. Show your working clearly.
Model Answer — 4(b)(ii)
Method shown: reading from 1600 Bq down to 800 Bq on the y-axis and reading across to the x-axis [1]
1600 → 800 Bq at t = 4 hours
Half-life = 4 hours (confirmed by 800 → 400 Bq from t = 4 to t = 8 hours) [1]
⚠ If you missed marks here: Two hours is the trap answer, lifted from the first row of the table — but 1600 falls only to 1130 after 2 hours, which is not a halving. Read 1600 down to 800 and across to t = 4 hours. Mark 1 is for showing that construction on the graph, so a bare “4 hours” with no lines drawn takes half the marks away.
Mark 1 -- Correct method shown reading from graph 1600 to 800 (1 mark)
Mark 2 -- Half-life correctly stated as 4 hours (1 mark)
(b) (iii)[1]
State the corrected count rate after 5 half-lives.
Model Answer — 4(b)(iii)
1600 → 800 → 400 → 200 → 100 → 50 Bq
After 5 half-lives the corrected count rate is 50 Bq [1]
⚠ If you missed marks here: Halve five times; do not divide by five. The sequence runs 1600, 800, 400, 200, 100, 50 Bq. Dividing 1600 by 5 gives 320 and stopping one halving early gives 100, and both are collected every year. Count the ARROWS between the numbers, not the numbers themselves.
Mark 1 -- Count rate after 5 half-lives correctly stated as 50 Bq (1 mark)
(c)[2]
A radioactive sample initially has an activity of 4800 Bq. After 3 hours, the activity has fallen to 600 Bq.
Calculate the half-life of this sample. Show your working.
Model Answer — 4(c)
4800 → 2400 → 1200 → 600 Bq = 3 half-lives
Three half-lives occur in 3 hours [1]
Half-life = 3 hours ÷ 3 = 1 hour
Half-life = 1 hour [1]
⚠ If you missed marks here: Do not answer “3 hours”; that is the time given, not the half-life. Count the halvings first — 4800, 2400, 1200, 600 is three of them — then divide: 3 hours ÷ 3 = 1 hour. Counting four halvings, which would end at 300 Bq, produces 45 minutes and loses both marks.
Mark 1 -- Working shown: 4800 to 2400 to 1200 to 600 equals 3 half-lives (1 mark)
Mark 2 -- Half-life correctly calculated as 1 hour (1 mark)
(d)[2]
Explain why radioactive decay is described as a “random” process.
Model Answer — 4(d)
It is impossible to predict which particular nucleus will decay next [1]
It is impossible to predict when a particular nucleus will decay; the decay is not affected by external conditions such as temperature or pressure [1]
⚠ If you missed marks here: Random and spontaneous mean different things and this part wants random: you cannot predict which nucleus decays next, nor when any particular one will go. “We do not have the equation yet” or “scientists have not worked it out” makes the unpredictability sound like a gap in knowledge, which is wrong — it is a property of the decay itself.
Mark 1 -- Cannot predict which nucleus will decay next (1 mark)
Mark 2 -- Cannot predict when a nucleus will decay or decay is unaffected by external conditions (1 mark)
Question 5 — Writing Nuclear Decay Equations
Total: 10 marks
(a) (i)[2]
Write the balanced nuclear equation for the alpha decay of radon-222 (22286Rn).
In alpha decay, the nucleus emits an alpha particle (42He).
Model Answer — 5(a)(i)
22286Rn → 21884Po + 42He
Correct daughter nucleus: polonium-218 with nucleon number 218 and proton number 84 [1]
Equation balanced: nucleon numbers (222 = 218 + 4) and proton numbers (86 = 84 + 2) both balance [1]
⚠ If you missed marks here: Alpha decay takes 4 off the top and 2 off the bottom. Leaving the proton number at 86, writing radon-218, is the standard error; the daughter has to be polonium-218, a different element. The alpha particle must appear on the right as nucleon number 4, proton number 2, or the equation does not balance.
Mark 1 -- Correct daughter product polonium-218 with Z=84 and A=218 (1 mark)
Mark 2 -- Equation balanced with nucleon and proton numbers correct on both sides (1 mark)
(a) (ii)[2]
Write the balanced nuclear equation for the alpha decay of thorium-232 (23290Th).
Model Answer — 5(a)(ii)
23290Th → 22888Ra + 42He
Correct daughter nucleus: radium-228 with nucleon number 228 and proton number 88 [1]
Equation balanced: nucleon numbers (232 = 228 + 4) and proton numbers (90 = 88 + 2) both balance [1]
⚠ If you missed marks here: Thorium 90 minus 2 is 88, which is radium — not radon (86), the element used in the part above. The nucleon number goes 232 to 228. Copying the pattern down from part (i) is a genuine risk here, so recompute both numbers rather than matching the previous line.
Mark 1 -- Correct daughter product radium-228 with Z=88 and A=228 (1 mark)
Mark 2 -- Equation balanced with nucleon and proton numbers correct on both sides (1 mark)
(b) (i)[2]
Write the balanced nuclear equation for the beta decay of iodine-131 (13153I).
In beta decay, a neutron in the nucleus changes into a proton and emits a beta particle (0−1e).
Model Answer — 5(b)(i)
13153I → 13154Xe + 0−1e
Correct daughter nucleus: xenon-131 with proton number 54 and nucleon number 131 [1]
Equation balanced: nucleon numbers (131 = 131 + 0) and proton numbers (53 = 54 + (−1)) both balance [1]
⚠ If you missed marks here: Beta decay RAISES the proton number, so iodine 53 becomes 54, which is xenon; dropping to 52 (tellurium) is the usual mistake. The nucleon number does not move at all — it stays at 131, because the beta particle has nucleon number 0.
Mark 1 -- Correct daughter product xenon-131 with Z=54 and A=131 (1 mark)
Mark 2 -- Equation balanced with nucleon and proton numbers correct on both sides (1 mark)
(b) (ii)[2]
Write the balanced nuclear equation for the beta decay of phosphorus-32 (3215P).
Model Answer — 5(b)(ii)
3215P → 3216S + 0−1e
Correct daughter nucleus: sulfur-32 with proton number 16 and nucleon number 32 [1]
Equation balanced: nucleon numbers (32 = 32 + 0) and proton numbers (15 = 16 + (−1)) both balance [1]
⚠ If you missed marks here: Phosphorus 15 goes UP to 16, giving sulfur-32; writing silicon-32 with proton number 14 means you subtracted. Hold the nucleon number at 32, since changing it to 31 as though the electron carried mass away is the other frequent slip. Test it with 15 = 16 + (−1) before moving on.
Mark 1 -- Correct daughter product sulfur-32 with Z=16 and A=32 (1 mark)
Mark 2 -- Equation balanced with nucleon and proton numbers correct on both sides (1 mark)
(c)[2]
Explain why the atomic number (proton number) increases by 1 during beta decay, even though the nucleon number stays the same.
Model Answer — 5(c)
In beta decay, a neutron in the nucleus is converted into a proton and an electron (beta particle) [1]
The number of protons increases by 1 (so the proton number goes up by 1), but the total number of nucleons stays the same because one neutron has been replaced by one proton [1]
⚠ If you missed marks here: “The nucleus loses a negative charge so it becomes more positive” sounds plausible but is not what happens, and it does not earn mark 1. A NEUTRON converts into a proton and an electron, and the electron is ejected as the beta particle. That is why the proton count rises by one while the nucleon count holds steady — a neutron has been swapped for a proton, and both are nucleons.
Mark 1 -- A neutron changes into a proton and an electron (1 mark)
Mark 2 -- Proton number increases by 1 but nucleon number unchanged because neutron replaced by proton (1 mark)
Question 6 — Choosing Radioisotopes for Applications
Total: 12 marks
(a) (i)[2]
A factory produces aluminium sheets of a specific thickness. A radioactive source and detector are used to monitor the thickness during production.
Which type of radiation (alpha, beta, or gamma) should be used? Explain your choice.
Model Answer — 6(a)(i)
Beta radiation should be used [1]
Beta particles are partially absorbed by aluminium, so changes in sheet thickness will cause measurable changes in the count rate at the detector [1]
⚠ If you missed marks here: Beta is the answer. Gamma is tempting because it sounds strong, but it goes through the sheet almost unaffected, so the reading would not change when the thickness did. Say WHY beta works: it is partly absorbed by aluminium, so the count rate rises and falls as the sheet thins and thickens.
Mark 1 -- Beta radiation correctly chosen (1 mark)
Mark 2 -- Beta partially absorbed by aluminium so thickness changes affect count rate (1 mark)
(a) (ii)[2]
Explain why the other two types of radiation would be unsuitable for monitoring the thickness of aluminium sheet.
Model Answer — 6(a)(ii)
Alpha particles would be completely absorbed by even a thin sheet of aluminium, so no radiation would reach the detector regardless of thickness [1]
Gamma rays would pass through the aluminium sheet with very little absorption, so changes in thickness would not produce a noticeable change in count rate [1]
⚠ If you missed marks here: Both marks need a specific outcome at the detector, not a verdict on strength. Alpha is absorbed completely by aluminium, so the count sits at background whatever the thickness. Gamma passes through with almost no absorption, so the count hardly moves. “Alpha is too weak and gamma is too strong” scores nothing.
Mark 1 -- Alpha completely absorbed by aluminium so cannot detect thickness changes (1 mark)
Mark 2 -- Gamma passes through with little absorption so thickness changes not detected (1 mark)
(a) (iii)[2]
Should the radioactive source used in the factory have a long half-life or a short half-life? Explain your answer.
Model Answer — 6(a)(iii)
The source should have a long half-life [1]
So that the activity remains approximately constant over a long period and the source does not need to be replaced frequently, ensuring consistent and reliable readings [1]
⚠ If you missed marks here: Short half-life is wrong here even though it is right for a medical tracer — the correct choice depends on the job. A factory source needs a LONG half-life so its activity is effectively steady; otherwise the count rate would drift downwards on its own and be misread as the sheet getting thicker.
Mark 1 -- Long half-life correctly chosen (1 mark)
Mark 2 -- Activity stays constant so source lasts a long time and readings remain reliable (1 mark)
(b) (i)[2]
A doctor wants to use a radioactive tracer to investigate a blockage inside a patient’s body. The tracer is injected into the patient and its progress is tracked using a detector outside the body.
Which type of radiation should the tracer emit? Explain your choice.
Model Answer — 6(b)(i)
The tracer should emit gamma radiation [1]
Gamma rays can pass through the body and be detected outside; alpha and beta would be absorbed by body tissue before reaching the detector [1]
⚠ If you missed marks here: Gamma, and the reason has to be about escaping the body: alpha and beta are absorbed by tissue within centimetres, so a detector held outside would register nothing. Choosing beta because it is “less harmful than gamma” misses the point, since radiation that cannot get out is useless as a tracer.
Mark 1 -- Gamma radiation correctly chosen (1 mark)
Mark 2 -- Gamma passes through body tissue to be detected outside (1 mark)
(b) (ii)[2]
Should the tracer have a long half-life or a short half-life? Explain your answer.
Model Answer — 6(b)(ii)
The tracer should have a short half-life [1]
So that the radioactive source decays quickly and does not remain active in the patient’s body for a long time, reducing the radiation dose and harm to the patient [1]
⚠ If you missed marks here: Short, and the reason concerns the patient, not the equipment. The activity has to fall away quickly so the tracer stops irradiating the body once the scan is finished; a long half-life would leave the source active inside the patient for weeks. Avoid “so the dose halves” — that is the definition of half-life, not a reason.
Mark 1 -- Short half-life correctly chosen (1 mark)
Mark 2 -- Decays quickly so patient receives minimal radiation dose (1 mark)
(c)[2]
Explain how radioactive sources are used in the sterilisation of medical equipment.
Model Answer — 6(c)
Gamma rays from a strong radioactive source (such as cobalt-60) are directed at the sealed medical equipment [1]
The gamma radiation kills bacteria and other microorganisms on the equipment without damaging it or requiring high temperatures, and the equipment does not become radioactive [1]
⚠ If you missed marks here: The mark is for GAMMA doing the killing; describing heat sterilisation is a different method and earns nothing here. Note that the equipment can be sealed in its packaging first, because gamma passes through the wrapping to reach the bacteria inside. Add that the equipment does not itself become radioactive — that is the reassurance this question is fishing for.
Mark 1 -- Gamma rays directed at sealed medical equipment (1 mark)
Mark 2 -- Kills bacteria without damaging equipment or making it radioactive (1 mark)
Question 7 — Nuclear Power & Safety Precautions
Total: 12 marks
(a)[3]
Describe the process of nuclear fission, including the role of neutrons.
Model Answer — 7(a)
Nuclear fission is the splitting of a large, heavy nucleus (such as uranium-235) into two smaller daughter nuclei of roughly equal size [1]
Fission is initiated when a slow-moving (thermal) neutron is absorbed by the heavy nucleus [1]
The fission process releases a large amount of energy and also releases two or three additional neutrons [1]
⚠ If you missed marks here: Three marks means three separate points, and the one missed most is the START: a slow-moving neutron is ABSORBED by the heavy nucleus. It then splits into two smaller nuclei of similar size, releasing energy and two or three further neutrons. “Splitting an atom” drops the first mark, because it is the nucleus that splits.
Mark 1 -- Fission is splitting of a large heavy nucleus into two smaller nuclei (1 mark)
Mark 2 -- Initiated by absorption of a neutron (1 mark)
Mark 3 -- Releases energy and two or three additional neutrons (1 mark)
(b)[2]
State what is meant by a chain reaction in nuclear fission.
Model Answer — 7(b)
The neutrons released from one fission event can go on to cause further fission reactions in other nuclei [1]
This process is self-sustaining: each fission produces more neutrons that cause more fissions, leading to a chain reaction [1]
⚠ If you missed marks here: “It keeps going on its own” is half an answer. Name the neutrons: the ones released by a fission are absorbed by other nuclei and cause further fissions, so the number of fissions grows and the reaction sustains itself. A chain reaction is not the same as fission, which is what many answers end up describing instead.
Mark 1 -- Neutrons from one fission cause further fission in other nuclei (1 mark)
Mark 2 -- Process is self-sustaining as each fission produces more neutrons (1 mark)
(c)[3]
Describe the process of nuclear fusion and state where it occurs naturally.
Model Answer — 7(c)
Nuclear fusion is the joining (combining) of two small, light nuclei to form a single larger nucleus [1]
Fusion releases a very large amount of energy; it requires extremely high temperatures and pressures to overcome the electrostatic repulsion between the positively charged nuclei [1]
Fusion occurs naturally in stars, including the Sun, where hydrogen nuclei fuse to form helium [1]
⚠ If you missed marks here: The mark for high temperature and pressure is the one that vanishes. Fusion needs both because the two nuclei are POSITIVE and repel one another, and they must be forced close enough to join. State where as well — stars and the Sun, hydrogen fusing into helium; a nuclear power station runs on fission.
Mark 1 -- Fusion is joining of two small light nuclei to form a larger nucleus (1 mark)
Mark 2 -- Releases large energy and requires very high temperature and pressure (1 mark)
Mark 3 -- Occurs naturally in stars or the Sun (1 mark)
(d)[4]
State two safety precautions that should be taken when handling radioactive sources in a school laboratory. For each precaution, explain why it is necessary.
Model Answer — 7(d)
Precaution 1: Hold the source with long tongs or forceps (never with bare hands) [1]
Reason: This increases the distance between the source and the handler’s body; the radiation spreads out and is absorbed by the air, so the dose received falls sharply with distance, and there is no direct skin contact [1]
Precaution 2: Keep the source in a lead-lined container when not in use and limit the time of exposure [1]
Reason: Lead absorbs most types of radiation, preventing unnecessary exposure; limiting time reduces the total radiation dose received by the handler [1]
⚠ If you missed marks here: Two precautions each with a reason is four marks, so listing four precautions and no reasons caps you at two. Pair them explicitly: tongs, because greater distance from the source cuts the exposure; and a lead-lined store or a short handling time, because lead absorbs the radiation and less time means a smaller total dose. Never let a reason read “because it is safer”.
Mark 1 -- Precaution 1: hold with tongs or forceps at a distance (1 mark)
Mark 2 -- Reason: increases distance and reduces radiation exposure to the body (1 mark)
Mark 3 -- Precaution 2: store in lead-lined container or limit exposure time (1 mark)
Mark 4 -- Reason: lead absorbs radiation and limiting time reduces total dose received (1 mark)
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