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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]
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]
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]
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]
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]
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]
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)
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]
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]
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]
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]
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]
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]
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]
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]
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]
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)
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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, reducing the intensity of radiation exposure (inverse square law) and preventing direct skin contact which could cause burns or tissue damage [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]
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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