K323 / 2027
Radioactivity overview

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?

  1. The number of protons: carbon-14 has two more.
  2. The number of neutrons: carbon-14 has two more.
  3. The number of electrons orbiting each nucleus.
Show the answer

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.
Number of neutrons = A - ZA and Z have no units. For a neutral atom, the electron number is Z as well.

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.

Two protons, two neutrons and two electrons in a neutral helium-4 atomA greatly enlarged central nucleus contains four separately labelled particles: two p-plus protons and two n neutrons. Two e-minus electrons are outside the nucleus, in a shaded region that does not depict their paths. Protons have charge plus one, neutrons zero, and electrons minus one in proton-charge units. The equal proton and electron numbers make the atom neutral. Lengths, positions and particle sizes are schematic.p+nnp+e-e-NucleusElectrons outside the nucleusp+2 protons; charge +1 eachn2 neutrons; charge 0e-2 electrons; charge -1 each

Read sodium-23's left-hand numbers

Sodium-23: nucleon number 23, proton number 11The element symbol Na has 23 at its upper left and 11 at its lower left. The upper number A counts all protons plus neutrons; the lower number Z counts protons. Na means sodium. Therefore this nucleus contains 11 protons and 23 minus 11 equals 12 neutrons. A neutral sodium-23 atom also has 11 electrons. The upper number is a count, not a mass in grams.23 = protons + neutronsNucleon number A2311NaProton number Z11 protonsElement:sodiumNeutrons = 23 - 11 = 12Neutral atom: 11 electrons

Compare neutral atoms

Same proton number identifies the element; different neutron numbers identify its isotopesThree cards use proper upper-left nucleon numbers and lower-left proton numbers. Sodium-23 has 11 protons, 12 neutrons and 11 electrons. Sodium-24 has 11 protons, 13 neutrons and 11 electrons. Magnesium-24 has 12 protons, 12 neutrons and 12 electrons. Both sodium atoms have proton number 11, but different neutron counts, so they are isotopes. Sodium-24 and magnesium-24 share nucleon number 24 but have different proton numbers, so they are different elements.2311NaSodium-2311 protons12 neutrons11 electrons2411NaSodium-2411 protons13 neutrons11 electrons2412MgMagnesium-2412 protons12 neutrons12 electronsThe sodium pair: same Z, different ANa-24 and Mg-24: same A, different Z
The atom schematic makes the particles visible without using a realistic size scale. In a nuclide symbol, the upper number counts nucleons and the lower number counts protons. Compare the two numbers before deciding whether two atoms are isotopes.

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?

  1. P and Q, because their proton numbers match while their neutron numbers, and so nucleon numbers, differ.
  2. P and R, because they each have 12 neutrons.
  3. Q and R, because they have the same nucleon number and so the same mass.
Show the answer

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?

  1. Gamma, which needs thick lead or concrete to absorb most of it.
  2. Beta, which is more penetrating than alpha but less than gamma.
  3. Alpha, which is stopped by aluminium because it is heavy.
Show the answer

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?

  1. Its decay time cannot be predicted, although a large sample of such nuclei has a predictable average pattern.
  2. It must decay exactly one half-life after the sample is prepared, like the other nuclei in it.
  3. It needs an external impact to trigger its decay.
Show the answer

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

Alpha: low penetrationThe nature inset shows exactly two positive protons and two uncharged neutrons: an alpha particle is a helium nucleus, not a neutral helium atom. In the shielding schematic, selected emissions travel right from a source towards paper. None is drawn beyond the paper for this typical-alpha example. The indicated particle count and absorber thickness are not a measured transmission result.p+nnp+Helium nucleus2 protons + 2 neutronsCharge +2SourcePaperNo transmittedalpha in thisexampleSchematic: no fixed absorber thicknessTransmission depends on the conditions

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-minus: greater penetration than typical alphaThe nature inset shows a fast electron produced in a nuclear change, not an electron falling from a shell. Selected beta emissions travel right, through paper and towards a suitable aluminium layer. Most shown paths end there and a smaller transmitted set continues. This illustrates reduction, not a universal stopping thickness or a measured fraction.e-Fast electronProduced during decayCharge -1SourcePaperAluminiumSchematic: no fixed absorber thicknessTransmission depends on the conditions

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

Gamma: high penetration, reduced by dense shieldingThe nature inset represents electromagnetic radiation emitted from a nucleus. Selected gamma emissions travel right through paper and aluminium towards a thick lead shield. Fewer paths continue beyond the lead: gamma is attenuated, not assumed completely stopped. Arrow counts, spacing and shield thicknesses are schematic, not measured transmission probabilities.Electromagneticradiation from nucleusNo charge; no rest massSourcePaperAluminiumLeadSchematic: no fixed absorber thicknessTransmission depends on the conditions

Thick lead or concrete attenuates gamma radiation. Some can still pass through; neither the drawn thickness nor the transmitted fraction is a universal value.

The absorber comparisons are schematic. Material and radiation energy affect transmission. Gamma radiation is attenuated by shielding rather than necessarily being stopped completely.

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

  1. Upper numbers: the total number of nucleons is the same before and after the emission.
  2. Lower numbers: the total charge number is the same. For a nucleus, this is its proton number; for an emitted electron, it is -1.
  3. 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

Alpha: four nucleons leave togetherUranium-238, proton number 92, becomes thorium-234, proton number 90, and a helium-4 nucleus with proton number 2. The parent has 92 protons and 146 neutrons; the daughter has 90 protons and 144 neutrons. The emitted alpha particle contains the remaining two protons and two neutrons. Nucleon totals balance as 238 equals 234 plus 4; lower charge numbers balance as 92 equals 90 plus 2.23892U23490Th42He+Before: nucleusAfter: nucleus92 protons146 neutrons90 protons144 neutronsEmitted alpha particlep+nnp+2 protons + 2 neutronsCharge +2238 = 234 + 492 = 90 + 2

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

Beta-minus: a neutron changes into a protonCarbon-14, proton number 6, becomes nitrogen-14, proton number 7, with a newly emitted electron. The nucleus changes from 6 protons and 8 neutrons to 7 protons and 7 neutrons. The inset shows one neutron changing into a proton with an electron produced during the change; no electron is drawn waiting in the original nucleus. This school bookkeeping omits the antineutrino, which is also emitted. The upper totals balance as 14 equals 14 plus 0; the lower charge totals balance as 6 equals 7 plus negative 1.146C147N0-1e+Before: nucleusAfter: nucleus6 protons8 neutrons7 protons7 neutronsOne neutron changesnp++e-NeutronProtonElectronThe electron is produced during the change.14 = 14 + 06 = 7 + (-1)

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

Gamma: the same nuclide at lower nuclear energyAn excited nucleus X, with nucleon number A and proton number Z, emits a gamma photon and reaches a lower nuclear energy. A star on the parent symbol marks its higher-energy nuclear state. Both nuclei have Z protons and A minus Z neutrons. Neither count changes. The photon carries energy away; its upper and lower bookkeeping numbers are both zero. The starred and unstarred symbols describe the same nuclide, not different isotopes.AZX*AZX00γ+Before: nucleusAfter: nucleusZ protonsA - Z neutronsZ protonsA - Z neutronsHigher energyLower energy* = higher-energy nucleusPhoton carries energy awaySame A and Z; lower nuclear energyA = A + 0Z = Z + 0

The star marks extra nuclear energy. Removing it changes the energy state, while the proton and neutron counts remain the same.

Alpha removes two protons and two neutrons. Beta-minus changes one neutron into a proton while emitting an electron. Gamma lowers nuclear energy without changing either nucleon count. The upper and lower bookkeeping rows must both balance.
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.

  1. After alpha: A = 214 - 4 = 210; Z = 84 - 2 = 82.
  2. After beta-minus: A stays 210; Z = 82 + 1 = 83.
  3. 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.

  1. Problem 1. A radium nucleus with nucleon number 226 and proton number 88 emits an alpha particle (nucleon number 4, proton number 2). Find the nucleon and proton numbers of the new nucleus.
  2. Problem 2. A carbon-14 nucleus (nucleon number 14, proton number 6) emits a beta-minus particle (an electron, nucleon number 0, charge -1). Find the nucleon and proton numbers of the new nucleus.
  3. Problem 3. A uranium nucleus with nucleon number 238 and proton number 92 emits an alpha particle. Find the nucleon and proton numbers of the new nucleus.
Show the answer

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?

  1. A = 214, Z = 85, with 129 neutrons.
  2. A = 210, Z = 83, with 127 neutrons.
  3. A = 210, Z = 81, with 129 neutrons.
Show the answer

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

Count rate = number of detected counts / counting timeExamples of units are counts/s and counts/min. State the interval as well as the count.

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

Background only: 100 counts in 5 minThe study source is absent. Dashed teal arrows represent background radiation reaching a detector. A grey signal cable connects the detector to a counter displaying 100 counts, the total from five one-minute background intervals. On a linear zero-to-five-minute scale below, the shaded counting interval lasts five minutes. Dividing 100 counts by five minutes gives an estimated mean background rate of twenty counts per minute. The displayed total is a number of counts, not a count rate.BackgroundStudy sourceabsentDetectorCounter total100 countsCounting time / min012345100 / 5 = 20 counts/min

Source plus background: 300 counts in 2 min

Source plus background: 300 counts in 2 minThe study source is present to the left of the same detector. Filled brown dots mark radioactive material; solid teal arrows show selected radiation paths towards the detector. Dashed arrows show that background radiation is still present. The counter displays 300 counts during two minutes. The time bar ends at two on the same zero-to-five-minute scale as the background panel. Dividing 300 by two gives 150 total counts per minute. Subtracting the background rate of twenty gives 130 net source counts per minute. Neither the arrow count nor this uncalibrated detector total measures all nuclear decays.BackgroundStudysourceDetectorCounter total300 countsCounting time / min012345300 / 2 = 150 counts/min

Net source rate = 150 - 20 = 130 counts/min. Subtract the rates, because the two raw totals were collected over different durations.

The same detector is used with the study source absent and present. The intervals are different, so convert each count to a rate before subtracting. A calibrated activity estimate is separate from the detector's raw count display.

Subtract rates, not unmatched counts

300 counts in two minutes with the source present

  1. Total rate: 300/2 = 150 counts/min.
  2. Mean background rate: 100/5 = 20 counts/min.
  3. 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.

  1. Problem 1. With no source present, a detector records 150 counts in 5 minutes. With a source in place it records 420 counts in 2 minutes. Find the count rate due to the source alone.
  2. Problem 2. With no source present, a detector records 120 counts in 4 minutes. With a source in place it records 900 counts in 3 minutes. Find the count rate due to the source alone.
  3. Problem 3. With no source present, a detector records 200 counts in 10 minutes. With a source in place it records 630 counts in 3 minutes. Find the count rate due to the source alone.
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?

  1. 130 counts/min
  2. 200 counts/min
  3. 150 counts/min
Show the answer

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.

  1. Estimate the decay rate: 130 x 4 = 520 decays/min.
  2. Convert to seconds: 520/60 = 8.67 decays/s.
  3. 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?

  1. One half, because only one more half-life has passed since the first.
  2. None. Two half-lives make a whole.
  3. One quarter, after halving twice.
Show the answer

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.

  1. Problem 1. A source gives a corrected count rate of 1200 counts/min. Its half-life is 8 minutes. What will the corrected count rate be after 24 minutes?
  2. Problem 2. A source gives a corrected count rate of 640 counts/min. Its half-life is 5 hours. What will the corrected count rate be after 20 hours?
  3. Problem 3. A source gives a corrected count rate of 2000 counts/min. Its half-life is 3 days. What will the corrected count rate be after 12 days?
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.

Subtract the same mean background from each total rate.
Time / minTotal / (counts/min)Source / (counts/min)
0180160
410080
86040
124020
163010

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

Total count rate includes the backgroundAn ideal smooth total-count-rate curve uses time in minutes horizontally and counts per minute vertically. Marked values are 180 at zero minutes, 100 at four, 60 at eight, 40 at twelve, and 30 at sixteen. The dashed horizontal line is the constant mean background of twenty counts per minute. The curve remains above that line and approaches it as the source contribution becomes smaller. It does not approach zero. These are supplied model values, not invented experimental measurements.Count rate / counts per min02040801201602000481216Time / min

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

Net source rate: subtract 20 counts/minOn the same axes and scales, the corrected smooth source-rate curve passes through 160, 80, 40, 20 and 10 counts per minute at zero, four, eight, twelve and sixteen minutes. Two shaded intervals, zero to four minutes and eight to twelve minutes, each last four minutes. In the first, 160 halves to 80; in the second, 40 halves to 20. Brackets below the graph label the equal durations. The corrected curve approaches zero but is still ten counts per minute at sixteen minutes. It is a continuous ideal model for one nuclide with unchanged detection conditions and no daughter contribution.Count rate / counts per min02040801201602000481216Time / min4 min160 to 804 min40 to 20

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.

The total-rate curve approaches the background level of 20 counts/min. The corrected source curve approaches zero. The marked 0-4 min and 8-12 min intervals both halve the source contribution.
  1. Correct the first pair: 180 - 20 = 160 and 100 - 20 = 80 counts/min.
  2. Read the time interval: 160 to 80 takes 4 - 0 = 4 min.
  3. 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?

Show the answer

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.

050100150200250300051015202530time / mincount rate / (counts/min)

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?

  1. It cannot be found until the total rate reaches zero.
  2. 4 min: the corrected rate falls from 160 to 80, then from 80 to 40 counts/min.
  3. Between 4 and 8 min, when the total rate falls from 180 to 90 counts/min.
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?

  1. Gamma, because it passes straight through and is easy to detect.
  2. Alpha, because any sheet of paper stops it before it reaches the detector.
  3. Beta, because paper partly absorbs it, so the count changes with thickness.
Show the answer

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

Thinner sheet: a larger detector rateThe same beta source is on the left and detector on the right, with a relatively thin sheet between them. Five selected incident radiation paths are illustrated: four continue to the detector and one ends inside the sheet to represent absorption. The drawing is qualitative, not a measured four-fifths transmission. A separate blue arrow below the sheet shows its downward movement. Under matched material, source, detector and geometry conditions, the thinner sheet gives the larger corrected detector count rate.BetasourceSheetDetectorSheetmotion

Thicker sheet: a smaller detector rate

Thicker sheet: a smaller detector rateThe source and detector remain in the same places, but the intervening sheet of the same material is thicker. Of the same five illustrated incident paths, two continue to the detector and three end within the sheet. This represents greater absorption with some radiation still transmitted; the arrow count is not a measured efficiency or universal thickness law. The blue downward arrow again means sheet movement. Other relevant conditions are held fixed.BetasourceSheetDetectorSheetmotion

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

Gamma radiation travels from an internal tracer to an external detectorA rounded outline labelled body contains four filled brown marks representing radioactive tracer material. Three selected teal radiation arrows start inside the body and point outwards to a detector on the right. The material remains inside; the arrows represent emitted gamma radiation rather than transport of the tracer. Some gamma radiation reaches the external detector and can provide information about the source distribution. The paths do not show all emissions or guarantee that every emitted photon is detected.RadioactivetracerDetectorBody

The source is inside and the detector is outside. Some emitted gamma radiation travels out to the detector.

Irradiation: radiation reaches the object

External irradiation without transferring radioactive materialAn external source on the left contains filled brown marks for radioactive material. Teal radiation arrows reach a separate object on the right. No radioactive material marks are placed on or inside that object. The depicted event is irradiation: radiation reaches the object while the source material remains outside it. Exposure by itself does not establish contamination.ExternalsourceObjectSource material remains outside

Contamination: radioactive material is present

Radioactive material on or inside an objectThe object now has filled brown marks both on its upper surface and inside it. Grey label leaders identify those locations; they are not movement arrows. Teal arrows start at the radioactive material and represent emitted radiation. This is contamination because source material itself is present on or in the object. No source-handling procedure or route of transfer is illustrated.ObjectOn the surfaceInsideBrown dots represent source material

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.

A transmission gauge compares radiation arriving through a sheet. An internal tracer emits towards an external detector. The exposure examples distinguish radiation reaching an object from radioactive material being present on or inside it.

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?

  1. Anything stopped by paper is harmless in every location, inside or outside the body.
  2. Gamma radiation is the one that damages cells, because it is the most penetrating.
  3. It emits strongly ionising particles straight into nearby living cells.
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?

  1. S, with a half-life of 2 h.
  2. T, with a half-life of 8 days.
  3. R, with a half-life of 20 min.
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?

  1. Heat from burning fuel, which splits the nucleus apart.
  2. A neutron absorbed by the nucleus, often one released by an earlier fission.
  3. A proton fired at the nucleus from an accelerator in the reactor.
Show the answer

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 uranium fuel nucleus splits into smaller nuclei and releases neutronsAn incoming neutron initiates one illustrative uranium-235 fuel-fission event. The heavy nucleus splits into two smaller nuclei and releases neutrons. The large outlined shapes represent entire nuclei; individual proton and neutron counts are not drawn or assigned to the fragments. The three emitted neutron symbols illustrate one possible outcome, not a fixed neutron number for every fission. Released energy is carried by moving products and radiation. No reaction energy or chain reaction is specified, and this initiating neutron does not define every possible fission process.Uranium-235 fuel exampleHeavynucleusnNeutronFissionSmallernucleusSmallernucleusNeutrons releasednnnMoving products and radiationcarry released energy.

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

Deuterium and tritium fuse, forming a helium nucleus and an emitted neutronTwo light hydrogen fuel nuclei, deuterium and tritium, join under suitable fusion conditions. The products are a heavier helium nucleus and a neutron, which carry released energy. The large outlined shapes represent whole nuclei, not individual nucleons or particle counts. Arrows indicate the process and products rather than measured trajectories or speeds. No numerical reaction energy is specified, and the example does not claim that every arbitrary joining of nuclei releases energy.Light hydrogen nucleiDeuteriumTritiumFusionHeliumnNucleusNeutronReleased energy is carriedby the products.

This suitable light-nucleus reaction releases energy. The change is in the nuclei; it is not ordinary chemical burning or simply mixing substances.

These are qualitative nuclear-fuel examples. The nucleus shapes represent whole nuclei rather than countable nucleons. In both suitable reactions, energy is carried away by the products and radiation.

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

Count rate = counts / counting time
Net source rate = total rate - mean background rate
Use 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 nuclides

Practise 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.

Table 1. Isotopes available
isotoperadiation emittedhalf-life
Aalpha6 hours
Bbeta15 hours
Cgamma6 hours
Dgamma30 years
Table 2. Count rate above the pipe
distance along pipe / mcount rate / counts per min
031
1029
2030
30186
4033
5028
  1. (a) Choose the most suitable isotope from Table 1. Explain your choice using both the radiation and the half-life. [3]
  2. (b) Calculate the corrected count rate at 30 m. [1]
  3. (c) State where the leak is and explain why the count rate is high there. [2]
  4. (d) State one assumption you made in (c). [1]
  5. (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

Card 1 of 7. Answer in your head, then check.

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?

Syllabus coverage

This chapter covers Radioactivity, topic 20(a-j), in the K323 syllabus (2027) and 6091 (2026).

Connect what changes in the nucleus to what a detector records and how the radiation is used.

Atoms, emissions, nuclear equations, background, half-life, source selection, hazards, fission and fusion.

View the learning outcomes
20(a)
Describe the positive nucleus, protons and neutrons, and negative electrons in an atom. Atoms, nuclei and isotopes
20(b)
Use proton number Z, nucleon number A and isotope. Atoms, nuclei and isotopes
20(c)
Interpret a nuclide and use its notation. Atoms, nuclei and isotopes
20(d)
Explain random, spontaneous decay in which an unstable nucleus loses energy by emitting radiation. Radioactive decay and radiation
20(e)
Compare the nature, ionising effect and penetration of alpha, beta-minus and gamma radiation. Radioactive decay and radiation
20(f)
Represent changes in nuclear composition with equations using nuclide notation. Nuclear decay equations
20(g)
Explain background radiation and account for it when interpreting detector readings. Background radiation and activity
20(h)
Use half-life in simple calculations, tables and decay curves. Half-life from numbers and graphs
20(i)
Discuss medical and industrial applications and hazards using both half-life and the penetration and ionisation of emissions. Uses and hazards
20(j)
Explain fission and fusion and relate suitable nuclear-fuel processes to energy release. Fission and fusion