Full chapter
Radioactivity
All 7 topics and the revision summary on one page.
01
Atoms, nuclei and isotopes
Make a guess. You are not marked.
Carbon-12 and carbon-14 are isotopes of carbon. What is different about their nuclei?
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The number of neutrons: carbon-14 has two more.
Both are carbon, so both have 6 protons. The nucleon numbers 12 and 14 differ by two, so carbon-14 has 8 neutrons and carbon-12 has 6.
An atom has a small positive nucleus, with negatively charged electrons outside it. The nucleus contains protons and usually neutrons.
- Proton
- A positively charged particle in the nucleus. Its charge is +1 when measured in units of the proton's charge.
- Neutron
- An uncharged particle in the nucleus. Protons and neutrons are collectively called nucleons.
- Electron
- A negatively charged particle outside the nucleus. Its charge is -1 in the same units. A neutral atom has equal numbers of protons and electrons.
The nucleus is much smaller than the whole atom. A drawing that makes its particles visible cannot also show their sizes and separations to one realistic scale. Ordinary hydrogen has one proton and no neutron in its nucleus, so a neutron is not present in every atom.
Read a nuclide symbol
A nuclide is a type of nucleus specified by its proton and neutron numbers. Its notation puts the nucleon number A at the upper left and the proton number Z at the lower left of the element symbol.
means sodium-23, with 23 nucleons and 11 protons.
- Proton number, Z
- The number of protons in a nucleus. It is also called atomic number and identifies the element.
- Nucleon number, A
- The total number of protons and neutrons. It is also called mass number. It is a count, not the atom's mass in grams.
Particles, nuclide notation and isotope comparisons
A neutral helium-4 atom
The nucleus is greatly enlarged. Positions are schematic; the drawing is not to scale and does not show fixed electron paths.
Read sodium-23's left-hand numbers
Compare neutral atoms
Same element, different neutron number
For sodium-23, the neutron count is 23 - 11 = 12. Sodium-24 still has 11 protons, but has 24 - 11 = 13 neutrons. They are isotopes: atoms of the same element with the same proton number but different neutron numbers.
Magnesium-24 has 12 protons and 12 neutrons. It shares nucleon number 24 with sodium-24, but its different proton number makes it a different element. Equal nucleon numbers alone do not identify isotopes.
Removing or adding electrons changes an atom's charge, producing an ion. It does not change its proton or neutron numbers, so it does not create another isotope. Also, the existence of isotopes does not mean that every isotope of an element is radioactive.
Check your understanding
Three neutral atoms have these nucleon/proton numbers: P has A = 23, Z = 11; Q has A = 24, Z = 11; R has A = 24, Z = 12. Which pair consists of isotopes of the same element?
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P and Q, because their proton numbers match while their neutron numbers, and so nucleon numbers, differ.
Both have Z = 11. P has 23 - 11 = 12 neutrons; Q has 24 - 11 = 13. Isotopes have the same proton number and different neutron numbers.
02
Radioactive decay and radiation
Make a guess. You are not marked.
A radiation passes through a sheet of paper but is stopped by a few millimetres of aluminium. Which radiation is it?
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Beta, which is more penetrating than alpha but less than gamma.
Alpha is stopped by paper. Gamma passes through aluminium and needs thick lead or concrete to reduce it greatly. Beta lies between: through paper, stopped by a few millimetres of aluminium.
During radioactive decay, an unstable nucleus loses energy by emitting radiation. The material contains the radioactive nuclei; the radiation is what leaves them.
A nucleus contains protons and usually neutrons. Nuclear decay changes a nucleus, rather than simply removing an electron from an atom's outer region.
Random events, a predictable average pattern
- Random
- The exact time when a particular unstable nucleus will decay cannot be predicted. Identical nuclei do not all decay after one fixed waiting time.
- Spontaneous
- The decay does not require an external trigger. An unstable nucleus can undergo the change on its own.
For a large collection of the same radioactive nuclide, the average rate follows a reproducible pattern. This is compatible with uncertainty about individual nuclei. The half-life describes that population pattern.
The nucleus remaining after an emission is called the daughter nucleus. It may itself be radioactive; one emission does not guarantee a stable final nucleus.
Check your understanding
A large sample of one radioactive nuclide has a reproducible average half-life. What does this tell us about an individual undecayed nucleus?
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Its decay time cannot be predicted, although a large sample of such nuclei has a predictable average pattern.
Random individual events and a reproducible population trend are compatible. The half-life does not identify which nuclei will decay during a particular interval.
Ionisation changes atoms along the radiation's path
Ionisation can occur when radiation removes an electron from an atom or molecule, leaving it charged. This can happen in material that the radiation enters, away from the original radioactive nucleus.
Alpha, beta and gamma can all cause ionisation. Compare what each emission is, how densely it ionises along its path and how readily it penetrates a stated material.
- Alpha, α
An alpha particle is a helium nucleus: two protons and two neutrons, with charge +2 in proton-charge units. It is not a whole neutral helium atom.
It typically produces strong, dense ionisation over a short path. Its penetration is low: paper or the outer layer of intact skin can stop typical alpha particles.
- Beta-minus, β-
A beta-minus particle is a fast electron emitted in a nuclear process, with charge -1. It is produced during that change, rather than being an existing electron ejected from an atomic shell.
Compared with typical alpha radiation, it produces less dense ionisation and penetrates farther. A suitable thin metal absorber can substantially reduce it. The energy and absorber determine how much passes through.
- Gamma, γ
Gamma radiation is electromagnetic radiation from a nucleus. It has no electric charge and no rest mass. Its nature connects to the electromagnetic spectrum.
It is highly penetrating. Thick lead or concrete can reduce its intensity, but shielding does not necessarily remove every gamma ray. It can cause ionisation even though its ionisation is less dense in the usual comparison.
Emission type and shielding
Only selected emissions travelling to the right are drawn. Arrows show their direction, not an exact particle count or transmitted fraction. These are typical comparisons: radiation energy, material and thickness matter.
Alpha: low penetration
Paper can stop typical alpha particles. Low penetration does not make alpha-emitting material harmless, especially if it is inside the body.
Beta-minus: greater penetration than typical alpha
Beta can pass through paper. A suitable thin metal layer substantially reduces it; the amount transmitted depends on the beta energies and the absorber.
Gamma: high penetration, reduced by dense shielding
Thick lead or concrete attenuates gamma radiation. Some can still pass through; neither the drawn thickness nor the transmitted fraction is a universal value.
Passing through paper does not, by itself, identify an emission as gamma: beta can also pass paper. Use the full set of observations under the stated absorber and source conditions.
Penetration and ionisation describe different properties. A short range does not mean harmlessness, especially if radioactive material is close to living cells. The uses and hazards depend on the source's location and the exposure conditions.
03
Nuclear decay equations
A nuclear equation records the daughter nucleus and emitted radiation. Check both the total nucleon number and the total charge number.
In nuclide notation, A counts nucleons and Z counts protons. An alpha particle contains two protons and two neutrons; beta-minus is an electron; gamma carries away nuclear energy.
Balance the upper and lower numbers separately
- Upper numbers: the total number of nucleons is the same before and after the emission.
- Lower numbers: the total charge number is the same. For a nucleus, this is its proton number; for an emitted electron, it is -1.
- Identify the daughter: its new proton number determines the element. Use the supplied element information when needed.
The lower -1 on a beta-minus electron does not mean that it contains negative protons. Its upper 0 means zero nucleons, not zero electron mass.
Balance nucleon and charge numbers
For a nucleus, the lower number is its proton number. For the emitted electron, -1 is its charge number; its upper 0 means no nucleons, not zero mass. The nucleus boxes give counts, not a drawing of every nucleon.
Alpha: four nucleons leave together
Both nucleon number and total charge balance. Two protons and two neutrons leave the parent nucleus together.
Beta-minus: a neutron changes into a proton
The beta electron is produced during the nuclear change; it is not ejected from an existing electron shell or stored inside the nucleus. An antineutrino is also emitted; the displayed school-level bookkeeping omits it. Antineutrino recall is not needed here.
Gamma: the same nuclide at lower nuclear energy
The star marks extra nuclear energy. Removing it changes the energy state, while the proton and neutron counts remain the same.
- Alpha: A falls by 4; Z falls by 2
- The daughter loses the four nucleons carried away in the helium nucleus.
- Beta-minus: A is unchanged; Z rises by 1
- One neutron becomes a proton, so the total nucleon count stays the same. The emitted negative charge balances the daughter's increased positive charge.
- Gamma: A and Z are unchanged
- The daughter has the same composition but lower nuclear energy. A star on the initial symbol marks a higher-energy nuclear state; it does not mean an extra proton or neutron.
Two emissions in order
Start with A = 214 and Z = 84
The nucleus emits an alpha particle, then a beta-minus particle.
- After alpha: A = 214 - 4 = 210; Z = 84 - 2 = 82.
- After beta-minus: A stays 210; Z = 82 + 1 = 83.
- Count the final neutrons: A - Z = 210 - 83 = 127.
The second step acts on the daughter from the first step. Do not apply both emissions separately to the original nucleus and discard one of the changes.
Try it step by step
Work it out in steps: Nucleon number after the decay, then Proton number after the decay.
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Problem 1.
Nucleon number after the decay: A = 226 - 4 = 222
Proton number after the decay: Z = 88 - 2 = 86 (radon)
Problem 2.
Nucleon number after the decay: A = 14 - 0 = 14
Proton number after the decay: Z = 6 - (-1) = 7 (nitrogen)
Problem 3.
Nucleon number after the decay: A = 238 - 4 = 234
Proton number after the decay: Z = 92 - 2 = 90 (thorium)
A nucleus losing energy need not lose nucleons: gamma emission is the example. Conversely, the beta electron's negative charge does not mean the daughter's proton number decreases.
Check your understanding
A nucleus with A = 214 and Z = 84 emits an alpha particle, then a beta-minus particle. What are its final nucleon, proton and neutron numbers?
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A = 210, Z = 83, with 127 neutrons.
Alpha emission subtracts 4 from A and 2 from Z, giving 210 and 82. Beta-minus emission changes one neutron into a proton, leaving A at 210 and increasing Z to 83. Neutrons = 210 - 83 = 127.
04
Background radiation and activity
A detector can register radiation even when the particular source being investigated is absent. Separate this background contribution from the source reading.
Radioactive decay is random, so counts fluctuate. A rate means an amount per unit time: equal raw counts collected over different durations need not mean equal rates.
Background is present without the study source
Natural contributions include cosmic radiation, radioactive materials in rocks and soil, radon from the ground, and radionuclides in food and body materials. Artificial sources can also contribute. Background varies with location and conditions.
Removing the study source therefore does not normally make the detector reading zero. Repeated equal-time measurements also need not give exactly the same count.
Convert counts to matching rates
Five equal one-minute background observations give 18, 21, 19, 22 and 20 counts. Their total is 100 counts in 5 min, so the mean background rate is 100/5 = 20 counts/min.
Compare rates over the correct counting intervals
These are supplied readings. A counter records detected events; its display does not directly give the source's activity. Teal arrows show selected radiation directions, not a count of every emission.
Background only: 100 counts in 5 min
Source plus background: 300 counts in 2 min
Net source rate = 150 - 20 = 130 counts/min. Subtract the rates, because the two raw totals were collected over different durations.
Subtract rates, not unmatched counts
300 counts in two minutes with the source present
- Total rate: 300/2 = 150 counts/min.
- Mean background rate: 100/5 = 20 counts/min.
- Estimated source rate: 150 - 20 = 130 counts/min.
Subtracting 100 directly from 300 would compare counts collected over different durations. For this example, take the source activity as approximately constant over the collection intervals.
Try it step by step
Work it out in steps: Background rate in counts/min, then Rate with the source present in counts/min, then Corrected source rate in counts/min.
Show the answer
Problem 1.
Background rate in counts/min: Background = 150 / 5 = 30 counts/min
Rate with the source present in counts/min: With source = 420 / 2 = 210 counts/min
Corrected source rate in counts/min: Source alone = 210 - 30 = 180 counts/min
Problem 2.
Background rate in counts/min: Background = 120 / 4 = 30 counts/min
Rate with the source present in counts/min: With source = 900 / 3 = 300 counts/min
Corrected source rate in counts/min: Source alone = 300 - 30 = 270 counts/min
Problem 3.
Background rate in counts/min: Background = 200 / 10 = 20 counts/min
Rate with the source present in counts/min: With source = 630 / 3 = 210 counts/min
Corrected source rate in counts/min: Source alone = 210 - 20 = 190 counts/min
Check your understanding
A background measurement records 100 counts in 5.0 min. With the source present, the same setup records 300 counts in 2.0 min. What is the estimated net source count rate?
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130 counts/min
The source-plus-background rate is 150 counts/min. Subtract the mean background rate of 20 counts/min to estimate 130 counts/min from the source. A separate calibration is needed to turn this detector rate into an activity estimate.
Beyond the syllabus: activity in becquerels (not examined)
Activity counts decays, not detector events
The activity A of a radioactive source is the number of nuclear decays per second. Its unit is the becquerel, Bq: 1 Bq means one decay per second.
Here A names a rate. In a nuclide symbol, A instead names nucleon number, which has no unit. Use the quantity and context to distinguish them.
A detector generally registers only a fraction of the source's emissions. Direction, distance, absorbers and detector response matter. A reading of 80 counts/s is therefore not automatically an activity of 80 Bq.
Use a supplied calibration
One recorded source count for every four decays, on average
Suppose a calibration for the long-lived source and unchanged arrangement above gives this response. The net source count rate is 130 counts/min.
- Estimate the decay rate: 130 x 4 = 520 decays/min.
- Convert to seconds: 520/60 = 8.67 decays/s.
- State the activity estimate: about 8.7 Bq.
This result depends on the supplied average calibration. It is not an exact activity obtained merely by changing the label on a counter.
Keep the measurement conditions matched
- Detector and range: use a suitable detector and counter with a usable rate range and response for the emission. At high rates, some detectors miss events arriving close together.
- Background: measure it with the study source absent, keeping the detector settings, location and surrounding conditions comparable.
- Geometry: keep source-detector distance and alignment fixed when comparing a source over time. Changed separation or an added absorber can change counts without any change in source activity.
- Timing: record the actual collection duration. Longer counting or repeated observations can reduce relative random variation in a steady or background rate.
- Calibration: the one-in-four factor above applies only to its stated emission, geometry, shielding and settings. A change can invalidate that estimate.
Longer counting is not always better for following a decaying source: a long interval averages over its change. Use intervals short compared with the decay timescale when assigning a rate to a particular time. Repetition cannot correct an inappropriate calibration or changed geometry.
05
Half-life from numbers and graphs
Make a guess. You are not marked.
A source has a half-life of 10 minutes. After 20 minutes, what fraction of the original number of undecayed nuclei is left?
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One quarter, after halving twice.
Twenty minutes is two half-lives. After the first, half remains. After the second, half of that half remains: one quarter.
Half-life is the time taken for half the undecayed radioactive nuclei in a large sample of one nuclide to decay. Over that time, its activity falls to half.
Use the background-corrected count rate. If the detector stays in the same place with the same settings, the corrected count rate halves whenever the activity halves.
The symbol is t1/2. It is a time, with SI unit s. A problem may use minutes, hours or days consistently; compare the elapsed time and half-life in matching units.
Halve what remains each time
Repeated halving
A corrected count rate of 800 counts/min, with a half-life of 6 h
- After 6 h: 800/2 = 400 counts/min.
- After 12 h: 400/2 = 200 counts/min.
- After 18 h: 200/2 = 100 counts/min.
Eighteen hours contains three half-lives. The remaining fraction is (1/2) x (1/2) x (1/2) = 1/8; the decayed fraction is 7/8. The same 6 h half-life is 6 x 3600 = 21600 s.
Try it step by step
Work it out in steps: Number of half-lives, then Corrected count rate remaining in counts/min.
Show the answer
Problem 1.
Number of half-lives: 24 / 8 = 3 half-lives
Corrected count rate remaining in counts/min: 1200 -> 600 -> 300 -> 150 counts/min
Problem 2.
Number of half-lives: 20 / 5 = 4 half-lives
Corrected count rate remaining in counts/min: 640 -> 320 -> 160 -> 80 -> 40 counts/min
Problem 3.
Number of half-lives: 12 / 3 = 4 half-lives
Corrected count rate remaining in counts/min: 2000 -> 1000 -> 500 -> 250 -> 125 counts/min
Half of the radioactive parent nuclei remain after one half-life on average. Half of the whole sample has not vanished: daughter material is still present. After the next half-life, half of the remaining parents decay.
Find half-life from corrected data
This separate supplied model follows one radioactive nuclide. Background has a constant mean of 20 counts/min; daughter contributions are neglected. The rates represent the labelled instants, with detection conditions unchanged.
| Time / min | Total / (counts/min) | Source / (counts/min) |
|---|---|---|
| 0 | 180 | 160 |
| 4 | 100 | 80 |
| 8 | 60 | 40 |
| 12 | 40 | 20 |
| 16 | 30 | 10 |
Remove the background before finding the half-life
Supplied smooth model: one radioactive nuclide, unchanged detection conditions, constant mean background of 20 counts/min and negligible daughter contribution. Real readings fluctuate around such a trend.
Total count rate includes the background
The dashed line is the 20 counts/min background. The total curve approaches that level, rather than zero.
Net source rate: subtract 20 counts/min
160 to 80 and 40 to 20 counts/min each take 4 min. The net rate approaches zero; it is still 10 counts/min at 16 min.
The axes use the same scales in both panels. Each marked dot matches a supplied table value; the curve represents the ideal average pattern between them.
- Correct the first pair: 180 - 20 = 160 and 100 - 20 = 80 counts/min.
- Read the time interval: 160 to 80 takes 4 - 0 = 4 min.
- Check another pair: the net rate falls from 40 at 8 min to 20 at 12 min, again taking 4 min.
A fall from 160 to 20 counts/min is three halvings: 160, 80, 40, 20. It therefore takes 3 x 4 = 12 min. It is not one half-life simply because the final reading is small.
Halving the uncorrected 180 to 90 counts/min would also halve the background contribution. The background does not follow the source's decay, so that would give the wrong interval.
Read a curve and interpret real measurements
Label elapsed time horizontally and the stated rate vertically. On a corrected curve, choose a source rate, find half that rate, and read the difference between their times. Use another well-separated pair as a check.
The source loses the same fraction over equal half-life intervals, not the same amount. The average curve bends as the rate falls; it does not become zero after two half-lives. Individual measurements scatter around this trend because decay and detection are random.
For a logged count series, retain the timestamps and counting durations. Calculate rates, subtract a comparable background estimate and keep source-detector geometry fixed. A short counting window can give an approximate rate at its labelled time, but an excessively long window blurs a rapid decay. Do not mistake a changed detector position or setting for a new half-life.
Watch out for this
A student says: "After two half-lives, none of the radioactive nuclei are left." What is wrong with this?
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Each half-life halves the number that remain, not the original number. After two half-lives, one quarter of the original nuclei are still undecayed.
Graph reader
Find a half-life from measured counts
A detector records the total count rate from a source plus background. Move the time cursor to read the graph.
Check your understanding
A single-nuclide model has background 20 counts/min. Its total rates are 180 counts/min at 0 min, 100 at 4 min and 60 at 8 min. What half-life follows from these readings?
Show the answer
4 min: the corrected rate falls from 160 to 80, then from 80 to 40 counts/min.
Both corrected halving intervals last 4 min under unchanged detection conditions. Equal time intervals halve the remaining source contribution; they do not subtract an equal number of counts each time.
06
Uses and hazards
Make a guess. You are not marked.
A paper mill checks paper thickness with a source on one side of the sheet and a detector on the other. Which radiation should it use?
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Beta, because paper partly absorbs it, so the count changes with thickness.
Beta passes through paper but is partly absorbed, so a thicker sheet gives a lower count. The change in count can control the rollers.
Radioactivity is useful when its emissions can be detected or transfer energy where needed. Those same interactions can damage living tissue.
Alpha is strongly ionising with low penetration; gamma is highly penetrating. Beta lies between them in the usual comparison. Half-life describes how rapidly a source's activity falls, not the range of its radiation.
Follow the source, material and detector
- A gamma-emitting medical tracer
Radioactive material inside the body emits radiation that can reach a detector outside. The detected radiation gives information about the tracer's distribution.
The source is inside, and radiation travels outwards towards the detector. This differs from an external X-ray source sending radiation through the body. Suitable penetration is needed for the emitted signal to reach the detector.
- A beta thickness gauge
A source and detector sit on opposite sides of a moving sheet. Under matched material and geometry conditions, a thicker sheet absorbs more, giving a lower detected rate. A thinner sheet gives greater transmission.
The chosen radiation must be partly transmitted through the relevant sheet. Alpha may be absorbed too completely, while very penetrating radiation may show too little change. Beta is useful for suitable paper, plastic or light-metal sheets, but is not the correct choice for every material and thickness.
Keep the source output and detector conditions accounted for: a falling count rate caused by source decay or changed alignment would not, by itself, show that the sheet became thicker.
- Gamma radiation in cancer treatment
Ionisation can damage or kill tumour cells. It can also damage healthy cells, so the useful effect requires controlled exposure. Gamma rays do not automatically recognise and target cancer cells.
Follow the radiation and locate the radioactive material
Filled brown dots mark radioactive material. Teal arrows show selected radiation paths; blue arrows show sheet movement. These schematic paths and thicknesses are not numerical transmission measurements.
Thinner sheet: a larger detector rate
Thicker sheet: a smaller detector rate
Compare the same sheet material with the source, detector and other relevant conditions fixed. The useful comparison has partial transmission: some radiation is absorbed and some reaches the detector.
A tracer is an internal source
The source is inside and the detector is outside. Some emitted gamma radiation travels out to the detector.
Irradiation: radiation reaches the object
Contamination: radioactive material is present
Irradiation concerns radiation reaching an object. Contamination concerns radioactive material on or inside it. The two descriptions refer to different things and can occur together.
Distinguish irradiation from contamination
- Irradiation or exposure
- Radiation reaches an object or person. Exposure to an external source does not, by itself, mean radioactive source material has been transferred.
- Radioactive contamination
- Radioactive material is present where it is not wanted, such as on a surface or inside a body. It can continue emitting from that location.
Ionisation can damage cells or DNA, kill cells or increase cancer risk. The consequences depend on the radiation, amount and duration of exposure, and where its energy is absorbed.
Typical alpha particles have little penetration from outside, but alpha-emitting material inside the body can damage nearby cells. There may be no protective outer layer between that source and the tissue. Gamma can reach tissue from outside. Penetrating power alone does not rank every exposure's hazard.
A source's activity is its rate of decay. It does not directly give the energy absorbed by a particular person. Equal activities can produce different exposures because the emissions, shielding, separation and source locations can differ.
Match protection to the route of exposure
- Reduce exposure time: less time exposed can reduce the energy received under otherwise unchanged conditions.
- Increase separation: greater distance from an external source can reduce the radiation reaching a person.
- Use suitable shielding: the material and thickness must suit the emission and conditions. Reduced transmission does not mean the source has stopped being radioactive.
- Contain the source material: containment addresses transfer and spread of radioactive material, rather than only shielding radiation arriving from outside.
Check your understanding
Typical alpha particles are stopped by paper or the outer layer of intact skin. Why can an alpha-emitting material inside the body still be hazardous?
Show the answer
It emits strongly ionising particles straight into nearby living cells.
Low penetration means a short range, not an absence of energy transfer. Internal alpha-emitting material can cause intense local ionisation. The source's location and the exposure conditions matter.
Choose the half-life and radiation for the task
A tracer must stay active long enough for the investigation, while a shorter subsequent persistence can limit continued activity afterwards. A continuously used industrial gauge needs suitably sustained output; a long-lived source also leaves storage and disposal considerations.
Penetration must suit the path. Radiation must reach a detector or target where required, while useful partial absorption makes a thickness gauge sensitive to changes. Ionisation explains both useful cell damage and the hazard. A suitable half-life cannot compensate for an emission that is unsuitable for the task.
Use the stated requirements
An invented industrial-tracer comparison
A detector at the far end of a pipe will follow the tracer. Each candidate source gives a corrected count rate of 1600 counts/min at the start and emits a suitable radiation. At least 400 counts/min must remain after a 2 h investigation. The source should not stay active for long afterwards.
- R: half-life 20 min
- Two hours is 120 min, containing six half-lives. The count rate falls to 1600/64 = 25 counts/min, below the required 400 counts/min.
- S: half-life 2 h
- One half-life leaves 800 counts/min. It meets the investigation requirement, then continues halving over the following hours.
- T: half-life 8 days
- It retains ample activity during the short investigation, but stays active much longer afterwards than S.
S best matches the supplied conditions. R decays too quickly for the task, while T remains active much longer. This conclusion depends on the sources starting at the same count rate with the same kind of radiation.
The shortest half-life is not automatically best, and half-life alone does not rank hazard. Initial activity, emission type and energy, exposure time and the location of the material also matter.
Check your understanding
A pipe tracer must still give at least 400 counts/min (corrected) after 2 h. Sources R, S and T each start at 1600 counts/min and emit a suitable radiation. Their half-lives are 20 min, 2 h and 8 days. Which source meets the task without staying active for long afterwards?
Show the answer
S, with a half-life of 2 h.
One half-life leaves 800 counts/min after 2 h, which meets the requirement. It then keeps halving, so within a day it has fallen to a tiny fraction of its starting rate.
07
Fission and fusion
Make a guess. You are not marked.
In a nuclear reactor, what starts the fission of a uranium-235 nucleus?
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A neutron absorbed by the nucleus, often one released by an earlier fission.
When a uranium-235 nucleus absorbs a slow neutron it becomes unstable and splits, releasing energy and two or three more neutrons. Those neutrons can cause further fissions.
Fission splits a heavy nucleus. Fusion joins light nuclei. Suitable reactions involving nuclear fuels release energy.
A nucleus contains protons and usually neutrons. Nuclear changes differ from ordinary chemical burning, which rearranges electrons and chemical bonds.
- Nuclear fission
- A heavy nucleus splits into smaller nuclei. In a suitable fuel reaction, energy and usually neutrons are released. An incoming neutron can initiate a fuel-fission event, though an external neutron is not part of the definition of every possible fission event.
- Nuclear fusion
- Light nuclei join to form a heavier nucleus. Suitable light-nucleus fuel reactions release energy. Fusion changes the nuclei; it is not simply mixing two substances together.
Two ways suitable nuclear fuels release energy
The large shapes represent whole nuclei, without displaying their nucleon counts. Sizes and arrows are schematic. These examples show the processes, not a calculation of reaction energy.
Fission: a heavy nucleus splits
A neutron can initiate a suitable fuel-fission event. The daughter nuclei and emitted particles carry energy away; different fission events can have different products.
Fusion: light nuclei join
This suitable light-nucleus reaction releases energy. The change is in the nuclei; it is not ordinary chemical burning or simply mixing substances.
Follow the released energy
Reaction products can move rapidly and radiation can carry energy away. Energy transferred from those products to surrounding material can increase its internal energy. This is the connection between a nuclear reaction and a useful energy supply.
Both suitable fission and fusion reactions can release energy. The choice of nuclei matters: arbitrary splitting or joining of any chosen nuclei is not guaranteed to release energy.
For electricity-generation context, nuclear fuel as an energy resource connects a heat supply to a turbine and generator. The nuclear reaction is the source of the released energy, while the generator transfers mechanical energy electrically.
Revision summary
Read the nucleus and atom
- Proton number Z
- Number of protons; identifies the element. A neutral atom also has Z electrons.
- Nucleon number A
- Protons plus neutrons. Neutrons = A - Z. A and Z are counts with no unit.
- Nuclide and isotope
- A nuclide has specified proton and neutron numbers. Isotopes have the same Z but different neutron numbers and therefore different A.
Put A at the upper left and Z at the lower left of the element symbol. A change in electrons makes an ion; it does not change the isotope. Equal A alone does not establish the same element.
Explain decay and emissions
An unstable nucleus loses energy by emitting radiation. Decay is random: an individual decay time cannot be predicted. It is spontaneous: no external trigger is needed. A large population can still have a predictable average pattern.
- Alpha: helium nucleus, two protons and two neutrons, charge +2. Typically dense ionisation and low penetration.
- Beta-minus: fast electron produced in a nuclear change, charge -1. Typically less dense ionisation and more penetration than alpha.
- Gamma: electromagnetic radiation from the nucleus, uncharged and highly penetrating. Thick shielding reduces transmission but need not eliminate it.
All three can ionise. Energy and absorber conditions affect their penetration. Radiation leaving a source is different from radioactive material spreading to a new location.
Track a nuclear change
- Alpha: daughter A decreases by 4 and Z by 2.
- Beta-minus: A is unchanged and Z increases by 1.
- Gamma: both A and Z stay unchanged while nuclear energy decreases.
Balance total upper nucleon numbers and lower charge numbers. The electron's lower -1 is a charge number, not a negative proton count; its upper 0 is not a claim of zero mass. Alpha then beta-minus takes A/Z from 214/84 to 210/83, leaving 127 neutrons.
Separate activity from detector rates
Net source rate = total rate - mean background rateUse matching rate units and comparable detector conditions.
Background comes from sources such as cosmic radiation, rocks, radon and radioactive materials in food and the body, with artificial contributions possible. Counts fluctuate; removing the study source does not normally make them zero.
Activity is the number of decays per second in the source. A detector records only some of the emissions, so its count rate is smaller than the activity. The example 300 counts in 2 min minus 100 counts in 5 min gives 150 - 20 = 130 counts/min from the source.
Choose suitable detector response and rate range. Keep geometry, shielding and settings fixed, record durations and estimate background. Longer counting reduces relative random variation for a steady rate, but can blur rapid decay. It cannot repair a wrong calibration.
Use repeated halving
Half-life t1/2 is the time for half the undecayed nuclei of one nuclide in a large sample to decay, or its activity to halve. It is measured in s, with other consistent time units possible.
After one, two and three half-lives, the remaining fractions are 1/2, 1/4 and 1/8. With half-life 6 h, a corrected count rate of 800 counts/min becomes 100 counts/min after 18 h. Daughter material remains; the whole sample does not disappear.
Use background-corrected rates for a decay curve. In the supplied model, 160 to 80 counts/min takes 4 min; 40 to 20 also takes 4 min. The total curve approaches 20 counts/min background, while the source curve approaches zero. Keep the single-nuclide, negligible-daughter and unchanged-detection conditions.
Connect uses and hazards to the radiation
- An internal gamma tracer emits towards a detector outside the body.
- A suitable beta gauge detects lower transmission through a thicker sheet under otherwise matched conditions.
- Gamma ionisation can damage tumour cells and healthy cells.
- Irradiation means radiation reaches a target. Contamination means radioactive material is present on or inside it.
- Internal alpha-emitting material can be hazardous despite low penetration. Activity alone does not give the energy absorbed by a person.
- Time, separation and suitable shielding affect external exposure; containment addresses spread of source material.
Choose half-life to retain useful activity for the task without unnecessarily long persistence. Also match penetration to the detector/target path and explain the useful or harmful ionisation. For the tracer comparison starting at 1600 counts/min, the 2 h half-life leaves 800 counts/min after 2 h; the 20 min half-life leaves only 25 counts/min.
Distinguish fission and fusion
Fission splits a heavy nucleus into smaller nuclei. Fusion joins light nuclei to form a heavier one. Suitable nuclear-fuel reactions in both processes release energy carried by their products and radiation; ordinary chemical burning does not change the nuclei in this way.
Back to atoms and nuclidesPractise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
A sample contains 80 g of a radioactive isotope with a half-life of 6.0 days. Calculate the mass of this isotope that remains undecayed after 18 days. [2]
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Exam-style question
Define half-life. [1]
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Exam-style question
A factory uses a radioactive source and a detector to check the thickness of paper as it is made. Explain why a beta source is used, rather than an alpha or a gamma source. [3]
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Exam-style question
The background count rate in a laboratory is 20 counts/min. A graph of the measured count rate from a source against time shows 420 counts/min at 0 min, 220 counts/min at 8.0 min and 120 counts/min at 16 min. Use these readings to find the half-life of the source. [3]
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Exam-style question
Data-based question: Finding a leak with a radioactive tracer [10]
An engineer wants to find a leak in a water pipe buried 1.0 m underground. She adds a small amount of a radioactive isotope to the water. A day later she walks along the ground above the pipe with a detector and records the count rate. The background count rate is 25 counts per minute.
| isotope | radiation emitted | half-life |
|---|---|---|
| A | alpha | 6 hours |
| B | beta | 15 hours |
| C | gamma | 6 hours |
| D | gamma | 30 years |
| distance along pipe / m | count rate / counts per min |
|---|---|
| 0 | 31 |
| 10 | 29 |
| 20 | 30 |
| 30 | 186 |
| 40 | 33 |
| 50 | 28 |
- (a) Choose the most suitable isotope from Table 1. Explain your choice using both the radiation and the half-life. [3]
- (b) Calculate the corrected count rate at 30 m. [1]
- (c) State where the leak is and explain why the count rate is high there. [2]
- (d) State one assumption you made in (c). [1]
- (e) A technician says the readings at the other positions show that the isotope is present there too. Evaluate this claim. [3]
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Quick recall
Three neutral atoms have these nucleon/proton numbers: P has A = 23, Z = 11; Q has A = 24, Z = 11; R has A = 24, Z = 12. Which pair consists of isotopes of the same element?