IGCSE Physics Paper 4 (Theory / Extended)

Topic 5: Nuclear Physics — Mock Exam 1

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0625 / 0972

Instructions

Answer all questions in the spaces provided.
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 × 10&sup8; 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]

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]

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)

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]

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)

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]

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)

Question 2: Isotopes, Nuclide Notation & Nuclear Equations

11 marks

(a) [2]

Define the term isotope.

Model Answer — 2(a)

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]

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.

Hint: Carbon has atomic number 6.

Model Answer — 2(b)

Carbon-12: ¹²⁄&sub6;C — 6 protons, 6 neutrons, 6 electrons [2]

¹²⁄&sub6;C → protons = 6, neutrons = 12 − 6 = 6, electrons = 6

Carbon-14: ¹&sup4;⁄&sub6;C — 6 protons, 8 neutrons, 6 electrons [2]

¹&sup4;⁄&sub6;C → protons = 6, neutrons = 14 − 6 = 8, electrons = 6

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)

Radium-226 undergoes alpha decay. Complete the nuclear equation below by identifying the daughter nuclide.

²²&sup6;⁄&sub8;&sub8;Ra → ? + &sup4;⁄&sub2;He

Model Answer — 2(c)(i)

Daughter nuclide is Radon (Rn) [1]

Mass number = 226 − 4 = 222, Atomic number = 88 − 2 = 86 [1]

²²&sup6;⁄&sub8;&sub8;Ra → ²²²⁄&sub8;&sub6;Rn + &sup4;⁄&sub2;He

Mark 19 -- Correct daughter element radon Rn identified (1 mark)

Mark 20 -- Correct mass number 222 and atomic number 86 (1 mark)

Carbon-14 undergoes beta decay to form nitrogen-14. Identify the missing particle in the equation below.

¹&sup4;⁄&sub6;C → ¹&sup4;⁄&sub7;N + ?

Model Answer — 2(c)(ii)

The missing particle is a beta particle (electron): &sup0;⁄−¹e (or &sup0;⁄−¹β) [1]

¹&sup4;⁄&sub6;C → ¹&sup4;⁄&sub7;N + &sup0;⁄−¹e

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]

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]

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]

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)

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]

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]

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 (²³&sup8;⁄&sub9;&sub2;U) 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]

²³&sup8;⁄&sub9;&sub2;U → ²³&sup4;⁄&sub9;&sub0;Th + &sup4;⁄&sub2;He

Check: mass 238 = 234 + 4 ✓ atomic 92 = 90 + 2 ✓

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 (&sup9;&sup0;⁄&sub3;&sub8;Sr) 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]

&sup9;&sup0;⁄&sub3;&sub8;Sr → &sup9;&sup0;⁄&sub3;&sub9;Y + &sup0;⁄−¹e

Check: mass 90 = 90 + 0 ✓ atomic 38 = 39 + (−1) ✓

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)

An unknown nuclide X undergoes alpha decay to form lead-208 (²&sup0;&sup8;⁄&sub8;&sub2;Pb).

X → ²&sup0;&sup8;⁄&sub8;&sub2;Pb + &sup4;⁄&sub2;He

(i) Determine the mass number of X.
(ii) Determine the atomic number of X.
(iii) Identify element X.
(iv) Write the complete balanced equation.

Model Answer — 4(c)

Mass number of X = 208 + 4 = 212 [1]

Atomic number of X = 82 + 2 = 84 [1]

Element X is Polonium (Po) [1]

Complete equation: ²¹²⁄&sub8;&sub4;Po → ²&sup0;&sup8;⁄&sub8;&sub2;Pb + &sup4;⁄&sub2;He [1]

²¹²⁄&sub8;&sub4;Po → ²&sup0;&sup8;⁄&sub8;&sub2;Pb + &sup4;⁄&sub2;He

Check: mass 212 = 208 + 4 ✓ atomic 84 = 82 + 2 ✓

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]

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.

Model Answer — 5(b)(i)

Corrected count rate = measured count rate − background count rate = 400 − 30 = 370 cpm [1]

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

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)

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)

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]

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.

Model Answer — 5(d)

Starting corrected count rate = 370 cpm [1]

Successive halving: 370 → 185 (1 half-life) → 92.5 (2 half-lives) → 46.25 (3 half-lives) → 23.125 (4 half-lives) [1]

After 4 half-lives the corrected count rate is 23.125 cpm which is below 25 cpm. Answer: 4 half-lives [1]

370 ÷ 2&sup4; = 370 ÷ 16 = 23.125 cpm < 25 cpm

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]

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]

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)

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]

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]

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]

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]

Mark 71 -- Joining or combining of two light or small nuclei (1 mark)

Mark 72 -- To form a heavier nucleus releasing energy (1 mark)

State where nuclear fusion occurs naturally.

Model Answer — 7(c)

Nuclear fusion occurs naturally in stars (including the Sun) [1]

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]

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]

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 radiation intensity decreases with distance (inverse square law for gamma). 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]

Mark 79 -- Time distance and shielding all three named correctly (1 mark)

Mark 80 -- Explanation of how each principle reduces radiation exposure (1 mark)

Score Summary

Total Score

0 / 80

Percentage