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Nuclear Physics overview

Full chapter

Nuclear Physics

All 9 topics and the revision summary on one page.

01

Evidence for the nucleus

Rutherford scattering connects a pattern of observations to an atomic model. Most alpha particles pass almost straight through a thin foil; rare large deflections reveal a small concentration of positive charge and mass.

A narrow beam of positively charged alpha particles encounters a thin metal foil, and a detecting screen records their outgoing directions. Distinguish the observed deflection pattern from the explanation inferred from it.

A narrow beam, a thin foil and a detecting screen

A narrow beam, a thin foil and a detecting screenA shielded alpha source at the left sends a narrow beam through a collimating slit towards a thin foil. A curved detecting screen surrounds much of the foil, with an opening for the incident beam. Dashed radial guides identify forward, sideways and backward detector positions; they indicate observation directions, not a claim that alpha trajectories kink at the foil. Most detected alpha particles have small deflection; a small fraction have large deflection and very few return towards the source. Source, foil and screen are schematic apparatus sizes. The following enlarged nucleus view has a separate scale.αSourceSlitThin foilDetecting screenGuides mark observation directions.Apparatus is schematic.

The screen records where particles arrive. Most have little deflection; a small fraction turn through large angles. The selected direction guides do not represent the relative number of events or the detailed path near a nucleus.

Enlarged view: repulsion bends the paths continuously

Enlarged view: repulsion bends the paths continuouslyFour selected model alpha trajectories enter from the left with rightward initial motion and pass a positive nucleus at the centre. Their acceleration is everywhere radially outward. None touches the symbolic nuclear surface. The near-central path turns continuously and emerges towards the left; more distant paths turn less. The horizontal and vertical distances share one scale. This enlarged comparison does not represent atom-to-nucleus sizes or measured scattering frequencies, and adds no required scattering formula.Selected enlarged pathsIncident motion is initially rightward.+Positive nucleusNo hard-ball impact is drawn.Paths are not a frequency sample.

A positive alpha particle is repelled by concentrated positive charge. These smooth paths show the mechanism of deflection. The symbolic nucleus and the surrounding atom are not drawn on one size scale, and four selected paths cannot indicate how often each deflection occurs.

The first view identifies the source, narrow beam, foil and detecting screen. The second is an enlarged view of continuous repulsive paths near a positive nucleus. The selected paths illustrate directions, not the relative numbers of events; the nucleus and whole atom are not shown on one size scale.
Scattering observations and the nuclear model they support
ObservationInference
Most alpha particles have little deflection.Most of the atom's volume does not contain the concentrated positive charge and mass.
A small fraction have large deflections; very few scatter backwards.A close approach to a small, massive, positive nucleus can strongly repel an alpha particle.

Large deflections are not explained by alpha particles striking individual light electrons. Nor does this experiment by itself identify the neutron or establish all electronic energy levels. Its central result is the existence of a nucleus much smaller than the whole atom.

Read nuclide notation

A nucleus is represented by . The nucleon number A, also called mass number, counts protons plus neutrons. The proton number Z, also called atomic number, counts protons and identifies the element.

Protons = Z
Neutron number N = A - Z
Electrons in a neutral atom = Z

Here N denotes the neutron number of one nucleus. When N later denotes a sample population, the stated context changes its meaning.

Isotopes have the same Z but different neutron numbers. For example, chlorine-35 and chlorine-37 both have 17 protons, but contain 18 and 20 neutrons respectively.

Worked isotope and ion count

Changing electrons does not change the isotope

A neutral atom has 17 protons, 20 neutrons and 17 electrons. Losing one electron produces a singly positive ion with 16 electrons. Its nucleus still has A = 37 and Z = 17.

Do not subtract an electron from the nuclear proton number. A is a particle count, not an exact measured mass of A atomic mass units.

Optional check Most alpha particles pass through a thin foil with little deflection, while very few are scattered through large angles or backwards. Which inference is supported?
Most alpha particles pass through a thin foil with little deflection, while very few are scattered through large angles or backwards. Which inference is supported?

02

Count atoms, nuclei and nucleons

The mole counts specified entities. Calculate the number of atoms or molecules first, then use their composition to count nuclei, protons, neutrons or electrons.

One mole contains 6.02 × 1023 entities. With amount of substance n, sample mass m, molar mass M and Avogadro constant NA:

n = m/M
N = nNA
NA = 6.02 × 1023 mol-1

Use matching mass units: g with g/mol, or kg with kg/mol. N counts the entity named by the molar mass. It is not automatically the count of every nucleus or nucleon within a molecule.

Separate relative mass from mass and molar mass

Relative atomic mass Ar is the mean atomic mass relative to one twelfth of the mass of a neutral carbon-12 atom. Relative molecular mass Mr is the corresponding ratio for a molecule. Both are dimensionless.

An individual atomic mass ma has a unit such as kg or u; molar mass M has a unit such as kg/mol or g/mol. The familiar numerical correspondence between relative mass and molar mass in g/mol is sufficient for the rounded supplied examples, but it does not give Ar or Mr a g/mol unit.

Worked argon sample

One nucleus per atom, forty nucleons per nucleus

A few times 10-4 mol should contain order 1020 entities. Now take 0.0200 g of pure neutral argon-40, with supplied M = 40.0 g/mol:

n = 0.0200/40.0 = 5.00 × 10-4 mol
N = (5.00 × 10-4)(6.02 × 1023)
= 3.01 × 1020 atoms or nuclei

The same calculation uses 2.00 × 10-5 kg and 0.0400 kg/mol. Each argon-40 nucleus contains 18 protons and 22 neutrons:

Particle counts in the supplied argon-40 sample
EntityCalculationSample count
Protons18N5.418 × 1021
Neutrons22N6.622 × 1021
Electrons18N5.418 × 1021
Nucleons40N1.204 × 1022

The electron count uses the stated neutrality. At three significant figures, the proton and neutron totals are 5.42 × 1021 and 6.62 × 1021. Use the supplied molar mass rather than treating mass number 40 as an exact molar-mass measurement.

This stable-isotope counting example is separate from the radioactive models later in the chapter.

A molecule can contain several nuclei

For H2O, supplied Ar(H) = 1.0 and Ar(O) = 16.0 give Mr = 2(1.0) + 16.0 = 18.0, without a unit. Using supplied M = 18.0 g/mol, a 1.80 g sample contains:

n = 1.80/18.0 = 0.100 mol
Nmolecules = 6.02 × 1022
Nnuclei = 3Nmolecules = 1.806 × 1023

There are three atomic nuclei in each molecule. The molecule count and nucleus count therefore differ even though they describe the same sample.

Optional check A 0.0200 g sample of pure argon-40 has supplied molar mass 40.0 g/mol. With N_A = 6.02 x 10^23 mol^-1, how many argon nuclei does it contain?
A 0.0200 g sample of pure argon-40 has supplied molar mass 40.0 g/mol. With N_A = 6.02 x 10^23 mol^-1, how many argon nuclei does it contain?

03

Alpha, beta and gamma

Radiation type determines what is emitted and how it interacts. Penetration and ionisation comparisons also depend on its energy and the material through which it travels.

The nature and characteristic interactions of three nuclear emissions
EmissionNature and chargeTypical comparison in a stated absorber
Alpha, αA helium-4 nucleus: two protons and two neutrons, charge +2e, mass approximately 4 u.Dense ionisation over a short range; ordinary alpha radiation can be absorbed by paper or a thin barrier.
Beta-minus, β-An electron produced in a nuclear process, charge -e, electron rest mass.Intermediate penetration and ionisation in the comparison; a suitable thin metal or plastic absorber can greatly reduce transmission.
Gamma, γAn electromagnetic photon with no charge or rest mass.Generally much more penetrating, with less direct ionisation along the path. Thick dense shielding attenuates it.

A beta electron is not a pre-existing orbital electron simply leaving the atom. Gamma emission reduces nuclear energy without changing its nucleon or proton count. Zero photon rest mass does not mean zero photon energy or momentum.

Compare transmission with a stated absorber

Compare transmission with a stated absorberThree separate schematic rows share the same source and detector positions. The alpha row places a thin paper barrier between them and has no transmitted alpha arrow. The beta-minus row uses a suitable plastic or metal absorber and retains a transmitted arrow labelled reduced count. The gamma row uses thick dense shielding and retains a transmitted arrow labelled reduced count: shielding attenuates gamma rather than making transmission identically zero. Barrier width and radiation arrow width are qualitative, with no supplied material thickness, energy, count efficiency or universal stopping range.α: alphaThin paperNo alpha throughβ-: beta-minusSuitable plastic / metalReduced countγ: gammaThick dense shieldingReduced countSourceDetector

Absorption and attenuation depend on the emitted energy and the material. These ordinary school-source comparisons do not assign a universal range. A reduced transmitted count is not a statement that every surviving photon has lost the same fraction of its energy.

The sign of charge determines the initial bending direction

The sign of charge determines the initial bending directionRadiations initially travel right into a magnetic field directed into the page, shown by crosses. A positive alpha trajectory curves upwards; a beta-minus electron trajectory curves downwards; the gamma ray remains straight. The two charged curves use equal illustrative radii only to show opposite signs, not to compare real radii. The field does not exert an electric-charge magnetic force on gamma photons. Momentum, charge and field strength, not charge alone, set a charged particle path radius.B into the pageEntryα (+)β- (-)γNo radius comparison is implied.

For the same rightward entry and field into the page, positive charge is forced up and negative charge down. The curves are schematic: their radii do not compare the particles' momenta. A cross here denotes magnetic field into the page, not current.

The absorber comparison keeps the geometry fixed and shows attenuation, not an absolute universal stopping thickness. The separate field view has entry to the right and B into the page: alpha bends up, beta-minus down, and gamma is not magnetically deflected. Curvatures are qualitative; charge alone does not determine their relative radii.

Charged alpha and beta particles also experience electric forces in opposite directions in a given electric field. Gamma has no electric charge force. Charged-particle speeds depend on their kinetic energies and are below c; gamma travels at c in vacuum.

A penetration statement must name the radiation energy and absorber context. Gamma is not completely stopped by one universal thickness, and short alpha range does not imply weak ionisation. For living tissue, an internal alpha-emitting material can deposit energy densely close to its location.

Optional check Why does the short range of ordinary alpha radiation not make an alpha-emitting substance inside the body harmless?
Why does the short range of ordinary alpha radiation not make an alpha-emitting substance inside the body harmless?

04

Random decay and measured counts

An individual decay is unpredictable, while a large collection has a statistical trend. A detector records counts, which must be interpreted using the observation time, its response and the background.

Spontaneous means nuclear decay does not need an external triggering event. Random means we cannot predict the exact time at which a particular unstable nucleus will decay. Ordinary changes of temperature, pressure or chemical state do not provide a way to set that time in this model.

Activity is not automatically count rate

Activity is the rate of nuclear decays. Its unit is the becquerel:

1 Bq = 1 decay/s
Detector count rate = recorded counts / counting time

Some emissions miss the detector or are not registered, while background adds other counts. An activity and a count rate therefore need not be numerically equal. Bq is a decay-rate unit, not a radiation-dose unit. Here activity A is distinct from a nuclide's nucleon number A.

Identify the background contribution

Background includes cosmic radiation and natural radionuclides in the environment, materials and bodies. Artificial sources can also contribute. A comparable background measurement excludes the investigated source while retaining the relevant detector and surroundings.

Keep the detector conditions fixed for the background comparison

Keep the detector conditions fixed for the background comparisonTwo schematic arrangements use the same detector, counter connection and source position. In the upper view the source is present and points towards the detector window; the fixed spacing is marked d. In the lower view the investigated source is absent, shown by an empty dashed position, while the detector and counter remain in the same places. Background still contributes counts. The cable connects each detector to its corresponding counter; these are two conditions of the same arrangement, not two independent efficiencies or simultaneous sources.Source presentDetectord unchangedCounterSourceInvestigated source absentDetectorBackgroundd unchangedCounterRemoved

Measure source-plus-background and background as separate conditions with otherwise comparable settings. Use each actual interval to calculate a rate. The source is absent for background; subtracting an unmatched raw count would not make the rates comparable.

Equal intervals fluctuate around a statistical mean

Equal intervals fluctuate around a statistical meanFive separate supplied source-plus-background counts for successive equal ten-second intervals are 412, 389, 405, 394 and 400. Points are not joined by a smooth decay curve. The horizontal axis is interval number one to five. The count axis deliberately starts at 380, not zero, and ends at 420. A horizontal dashed line at 400 identifies the mean count, not a measured constant at every instant. These constructed records are separate from the varying-duration workbook series.Count in 10.0 s380390400410420Mean 40012345Interval numberVertical axis starts at 380 counts.

Each point is a count in 10.0 s. Their mean is 400 counts per interval, giving a total rate of 40.0 counts/s. Scatter about this mean does not by itself imply detector failure or a smoothly falling source activity.

The source-detector arrangement identifies the quantities to hold fixed; the background condition excludes the investigated source. The separate points are the supplied equal-10 s counts, not a smooth decay curve. Their count-axis lower limit is shown explicitly.

Worked background correction

Convert counts to rates before subtracting

A rate of order 40 counts/s over 10 s gives hundreds of counts; background of order 4 counts/s gives tens. These constructed teaching records describe a source whose expected rate changes negligibly during the short observations. Each interval lasts 10.0 s.

Separate source-plus-background and background records, each counted for 10.0 s
IntervalSource + backgroundBackground
141238
238942
340536
439444
540040
Total rate = 2000 counts / 50.0 s = 40.0 counts/s
Background rate = 200 counts / 50.0 s = 4.0 counts/s
Net source rate = 40.0 - 4.0 = 36.0 counts/s

The corresponding means are 400 and 40 counts per 10.0 s. If durations differ, subtract their rates, not the unmatched raw counts. Pool a stable-rate series using total counts divided by total duration rather than an unweighted mean of unequal-duration rates.

Fluctuations between equal intervals are evidence of the random nature of decay, not proof that the detector is broken. A smooth expected decay curve does not mean each nucleus decays on schedule.

Evaluate the counting method

Compare rates with source-detector positions, absorber, detector window and timing method fixed. A longer observation can reduce fractional random scatter when the expected rate is sufficiently stable, but it can blur a rapid decay change. Repetition does not correct an omitted background, incorrect time interval or changed geometry.

Changing detector efficiency and dead time at high rates can limit the result. Background can fluctuate or drift too. Retain the supplied records and explain a specific limitation rather than deleting points simply to obtain a smoother pattern.

Optional check Constructed source-plus-background records total 2000 counts in 50.0 s. Comparable background records total 200 counts in 50.0 s. What is the net source count rate?
Constructed source-plus-background records total 2000 counts in 50.0 s. Comparable background records total 200 counts in 50.0 s. What is the net source count rate?

05

Half-life and corrected curves

Half-life is the time for the expected number of undecayed parent nuclei, or the activity, to fall to half its initial value. For a detector reading, use the source contribution and check that the conditions make it proportional to the decaying population.

For a single unchanged radionuclide, the expected number of undecayed parent nuclei and the activity halve over the same interval. A proportional net source count rate does too when detection conditions remain fixed. Half-life is not the lifetime of every individual nucleus.

After one, two and three half-lives, the remaining fractions are 1/2, 1/4 and 1/8. A few half-lives leave a fraction, not zero. Match time units before comparing: 0.2 min is 12 s, while 2 min is 120 s.

Separate total rate from the source trend

In this supplied model, net source rate starts at 80.0 counts/s and halves every 120 s. Background remains 4.0 counts/s. These are instantaneous expected rates at the named times, not long-window measured averages.

Expected net and total count rates in the 120 s half-life model
t / sNet / counts s-1Total / counts s-1
08084
1204044
2402024
3601014
48059

Total expected rate approaches the background

Total expected rate approaches the backgroundBoth decay panels use identical time and rate scales, over zero to 480 seconds. The expected source rate begins at 80 counts per second and halves every 120 seconds. Adding a constant background of four counts per second gives total values 84, 44, 24, 14 and 9 at successive 120-second times. The curve tends towards the dashed background line; it never becomes that value at a finite time. The marked values are ideal instantaneous model rates, not raw counts or long-window averages.Total rate / counts s-10204060800120240360480t / sBackground: 4

The same fixed background is added at every time. Half of the initial raw total is 42 counts/s; that is not the correct halving target for the source. At 120 s the total is 44 counts/s.

Corrected expected rate halves in equal times

Corrected expected rate halves in equal timesBoth decay panels use identical time and rate scales, over zero to 480 seconds. The expected source rate begins at 80 counts per second and halves every 120 seconds. Subtracting the background gives 80, 40, 20, 10 and 5 counts per second at successive 120-second times. Guides identify 80 to 40 and 40 to 20 with two adjacent 120-second brackets. The smooth expected net rate tends towards zero, not to a negative activity and not to zero at a finite plotted time. The marked values are ideal instantaneous model rates, not raw counts or long-window averages.Net rate / counts s-10204060800120240360480t / s120 s120 s

The two marked intervals each last 120 s even though the absolute decrease is smaller in the second interval. A proportional corrected rate shares the half-life only while source and detection conditions remain fixed.

Both expected-rate plots use the same time and rate scales. Total approaches the 4 counts/s background; net approaches zero. Equal 120 s spans halve the corrected rate from different starting points.

Half of the initial raw total is 42 counts/s, but the total at one half-life is 44 counts/s: 40 from the source plus 4 background. Read half-life from the corrected quantity. The falling trend becomes less steep as fewer parents remain; it is not a constant rate of loss.

Worked transfer problem

Count the halvings

A corrected rate decreases from 72 to 9 counts/s in 15 min under fixed detection conditions:

72 → 36 → 18 → 9
Three half-lives = 15 min
Half-life = 5.0 min

One eighth of the initial rate represents three halvings, not eight. A small negative background-subtracted estimate near zero can arise from fluctuations; it is not negative physical activity.

Supplied fractional model

Use an exponential between whole half-lives

For the 120 s model, use the following supplied continuous relationship, with t in seconds and r0 = 80 counts/s:

r/r0 = (1/2)t/120
= exp[-ln(2)t/120]

The exponent and the rate ratio are dimensionless. At 60 s, calculate EXP(-LN(2)*60/120) = 0.7071068. The expected net rate is 56.5685 counts/s and total rate is 60.5685 counts/s. Half a half-life does not remove half of the amount lost during a whole half-life.

At 180 s the fraction is 0.3535534 and net rate is 28.2843 counts/s. To find when only 10.0% remains, take logarithms of the supplied ratio:

ln(0.10) = -ln(2)t/120
t = -120 ln(0.10)/ln(2)
= 398.631 s ≈ 399 s

The same model gives a straight transformed relationship:

ln(r/r0) = -[ln(2)/120]t

A logarithm turns the supplied fractional model into a straight line

A logarithm turns the supplied fractional model into a straight lineThe corrected rate ratio is r divided by r-zero, with r-zero eighty counts per second. The vertical quantity is its natural logarithm and is dimensionless. At times 0, 120, 240, 360 and 480 seconds its values are 0, minus ln 2, minus 2 ln 2, minus 3 ln 2 and minus 4 ln 2. The time scale matches the preceding curves. The straight line has decreasing numerical value as time increases; the arrowed time axis is at logarithm zero. No zero or negative rate is logged.ln(r/r0) (dimensionless)00-0.693120-1.386240-2.079360-2.773480t / sPositive ratios only.

This graph is the supplied model ln(r/r0) = -ln(2)t/120, with time in seconds. Its gradient is about -0.00578 per second. The logarithm acts on a dimensionless positive ratio; the zero ordinate means r = r0, not r = 0.

This is the logarithmic graph of the supplied fractional model. The ordinate is a dimensionless positive-rate ratio's logarithm; the gradient has unit s-1. No point represents taking the logarithm of zero.

The logarithm requires a positive corrected rate. Keep seconds in this expression; substituting a time in minutes without converting changes the result.

Optional check With fixed detection conditions, a background-corrected rate falls from 72 to 9 counts/s in 15 min. What is the half-life?
With fixed detection conditions, a background-corrected rate falls from 72 to 9 counts/s in 15 min. What is the half-life?

06

Choose radiation for a task

A useful source must interact in the required way and remain suitable for the time involved. Explain both its penetration or ionisation and its half-life, using the stated task.

Measure sheet thickness

Place a radiation source and detector on opposite sides of a sheet. For a suitable beta source and sheet material, some particles are absorbed and some are transmitted. Greater thickness then produces a lower corrected detector rate when the geometry and other conditions stay fixed.

A source absorbed almost completely by even the thinnest sheet gives little useful contrast. A source that passes through all the relevant thicknesses almost unchanged also gives poor sensitivity. Choose radiation that is partially transmitted over the required thickness range.

The source half-life should be long enough for useful output over the operating and calibration period, or the expected decay must be allowed for. A falling rate caused by source decay must not be interpreted automatically as a thicker sheet.

A partially transmitting sheet allows a thickness comparison

A partially transmitting sheet allows a thickness comparisonTwo schematic comparison rows use the same source and detector positions, identical incident direction and the same sheet centre. The upper sheet has drawing width eight and the lower has width twenty-six; these are qualitative thicknesses, not material data. A beta source sends radiation through the sheet towards a detector. Under otherwise fixed conditions the thicker sheet gives a lower transmitted count rate. Both rows show some transmission; neither complete absorption nor almost unchanged transmission is the intended sensitive regime. This is a conceptual industrial gauge, not a handling procedure.Same source and detector geometryβThinner sheetHigher rateSourceβThicker sheetLower rateSourceBoth cases allow partial transmission.

Compare thickness only while the source output, detector geometry, sheet material and other relevant conditions are controlled. A falling source activity could also reduce the count, so it must not be mistaken for a thicker sheet.

The source is inside; the detector receives photons outside

The source is inside; the detector receives photons outsideA schematic body outline contains a small internal tracer marker. Four gamma-photon direction arrows begin at that marker and leave in different directions; one reaches the external detector on the right. The detector is physically outside the body, separated from its outer edge. The figure does not place the source in the detector, make all emissions travel towards it, or specify a clinical dose or organ-specific uptake. Body proportions, source size and arrow counts are schematic.DetectorTracerOnly some emitted photonsreach the detector.

The emitted photons must escape the tissue to be detected externally. Only part of the emission reaches and is registered by the detector. A suitable half-life must also fit preparation and measurement, alongside the stated chemical and biological requirements.

The thickness gauge compares partial transmission with fixed source and detector positions. The tracer view places the emitting material inside the body and the detector outside: photons must escape the tissue to be detected. Arrows identify radiation paths, not movement of the radioactive material.

Detect an internal tracer

For external detection of a tracer within the body, suitable emitted photons must penetrate tissue and reach the detector. The distribution of the tracer provides information about the stated organ or process; its chemical suitability also matters.

The half-life must allow preparation and measurement while avoiding unnecessarily prolonged radioactive activity. Physical half-life describes nuclear decay; it does not by itself give the time for biological clearance from the body.

Compare supplied candidate properties

Meet both detection and timing conditions

In a simplified selection, all other suitability conditions are supplied as equal. A detector outside the body must receive photons, and at least half the initial activity must remain after 60 min. Consider fictional candidates emitting alpha with a 2 h half-life, gamma with a 1 min half-life, or gamma with a 2 h half-life.

The gamma emitter with a 2 h half-life meets both conditions. After 60 min it retains 2-1/2, approximately 0.707, of its initial activity. The alpha choice fails the specified external photon-detection requirement; the 1 min gamma choice decays far too quickly for the stated time.

This comparison uses the supplied conditions. The longest half-life is not automatically best for every application.

Deliver radiation to a treatment target

In external treatment, sufficiently penetrating radiation can reach a target within tissue. Ionisation deposits energy and can damage cells, which is the intended effect in the target but also a concern for surrounding tissue. Direction and exposure distribution matter alongside radiation type.

The source must maintain suitable output over its operating period, with decay allowed for. A long half-life reduces the rate of output change but also means the source remains radioactive for longer. Half-life alone does not determine a treatment's suitability.

Sterilise items through packaging

Penetrating gamma radiation can reach microorganisms inside suitable final packaging. Ionisation damages them without requiring the package to be opened. The source must provide useful output over the processing interval, and its changing activity must be allowed for as it ages.

Explain the hazard in context

Ionisation can damage living cells and genetic material. Relevant factors include activity, emitted energy and radiation type, exposure time, distance, shielding, and whether radioactive material enters the body. Short alpha range can limit an external exposure while an internal alpha emitter deposits energy densely near its location.

Irradiation means exposure to radiation. Contamination means radioactive material is present on or inside an object. Ordinary irradiation does not necessarily make the object radioactive; do not confuse the radiation reaching an object with the source material moving onto it.

Optional check In a fictional tracer comparison, an outside detector must receive emitted photons and at least half the initial activity must remain after 60 min. Under otherwise equal stated suitability, which candidate meets both criteria?
In a fictional tracer comparison, an outside detector must receive emitted photons and at least half the initial activity must remain after 60 min. Under otherwise equal stated suitability, which candidate meets both criteria?

07

Balance nuclear equations

A nuclear equation accounts for nucleon number and electric charge. Balance both before identifying an unknown product; energy and momentum must also be conserved in the complete process.

In , A counts nucleons and Z counts protons. For an emitted particle, the lower number records its charge in units of +e. An electron therefore has lower number -1 even though it contains no negative proton.

Alpha emission

+

An alpha particle carries four nucleons and charge +2e. The daughter therefore has A reduced by 4 and Z reduced by 2. Here 238 = 234 + 4 and 92 = 90 + 2.

Beta-minus emission

+

This displays the nucleon and charge account: 14 = 14 + 0 and 6 = 7 - 1. Within the nucleus, a neutron changes into a proton. The total nucleon number stays the same, while the daughter's proton number rises by one.

The emitted electron is produced by the nuclear process, rather than removed from an atomic shell. The displayed A/Z account is not a claim that only two products carry energy and momentum: an additional uncharged particle also participates.

Gamma emission

+ γ

The star indicates a higher-energy nuclear state. A gamma photon carries away energy and momentum; the nucleus has the same A and Z afterwards. The energy change does not require a change of element or isotope.

A reaction can have an incoming particle

++

Both sides contain 18 nucleons and charge +9e. A and Z balance across all incoming and outgoing particles, not just the two nuclei that look most prominent.

Conserved nucleon number does not mean conserved rest mass. A change of total rest mass can accompany a change in kinetic or radiation energy while total mass-energy remains conserved.

Optional check In beta-minus decay, a parent nucleus has nucleon number A and proton number Z. What are the daughter values in the nucleon/charge account?
In beta-minus decay, a parent nucleus has nucleon number A and proton number Z. What are the daughter values in the nucleon/charge account?

08

Mass defect and binding energy

A bound nucleus has less rest mass than its separated nucleons. The corresponding energy difference is the energy required to separate the nucleus completely.

Use mass units consistently

The unified atomic mass unit u is defined as one twelfth of the mass of a neutral carbon-12 atom. For these supplied calculations use 1 u = 1.66 × 10-27 kg. Nucleon number A is dimensionless; it is not an exact mass in u.

Worked conversions

Keep the type of particle in the mass label

A neutral chlorine-37 atom has supplied mass 36.966 u:

ma = 36.966(1.66 × 10-27)
= 6.136356 × 10-26 kg
≈ 6.14 × 10-26 kg

This is an atomic mass, including electrons, not a bare nuclear mass. Conversely, a supplied electron mass 9.11 × 10-31 kg is:

me/u = (9.11 × 10-31)/(1.66 × 10-27)
me ≈ 5.49 × 10-4 u

The rounded nuclear masses used below contain no orbital electrons. Proton and neutron masses are close to, but not exactly, 1 u.

Supplied nuclear masses for the binding and fusion calculations
Particle or nucleusMass / u
Proton, mp1.00727647
Neutron, mn1.00866492
Deuteron, hydrogen-2 nucleus2.01355321
Triton, hydrogen-3 nucleus3.01550072
Alpha particle, helium-4 nucleus4.00150618

Convert an energy difference from mass

Mass and energy are equivalent through E = mc2, where c is the speed of electromagnetic waves in vacuum; use the supplied value 3.00 × 108 m/s. A change of rest mass Δm corresponds to an energy change ΔE = Δmc2. For a complete nuclear process, a decrease in total rest mass can supply kinetic energy and radiation while total mass-energy is conserved.

One electronvolt is the energy transferred to a charge of magnitude e through a potential difference of 1 V. With supplied e = 1.60 × 10-19 C:

1 eV = 1.60 × 10-19 J
1 MeV = 1.60 × 10-13 J
Supplied conversion: 1 u c2 = 931.5 MeV

A mass difference of order 10-3 u gives an energy of order 1 MeV, about 10-13 J per nucleus. This is a useful scale check before a precise subtraction. An eV is a unit of energy, not electric potential.

Binding energy is a positive separation energy

For a nucleus with Z protons and A - Z neutrons, the positive mass defect is:

Δm = Zmp + (A - Z)mn - mnucleus
Binding energy = Δmc2

The separated nucleons have greater rest energy. Forming the bound nucleus releases this energy difference; separating it completely requires the same amount to be supplied. A nucleus does not keep radiating its binding energy merely because it remains bound.

The same nucleons have different total rest energies

The same nucleons have different total rest energiesA convenient energy reference puts a separated free proton and neutron at zero. The bound deuteron is lower by 2.22459 megaelectronvolts. A downward brown arrow represents energy released when those free nucleons bind; an equal-length upward blue arrow represents the energy input required to separate the deuteron completely into the same proton and neutron. Both arrows span the same levels. The diagram deletes no nucleon and draws no ongoing emission from a nucleus that simply remains bound. The numerical zero is a relative energy reference, not zero rest mass.Relative energy / MeV0-2.22459Free proton + neutronBound deuteron (p + n)ReleaseInputEqual energy difference:2.22459 MeVNo nucleon disappears.

Binding releases the positive energy difference; complete separation requires that same energy input. The deuteron still contains one proton and one neutron. Its total binding energy is about 2.22 MeV, whereas its binding energy per nucleon is about 1.11 MeV.

The separated proton and neutron define relative energy zero. The bound deuteron is lower by 2.22459 MeV. Equal-sized arrows show energy released on formation and energy required for separation; vertical position represents energy, not physical height.

Worked deuteron

Total and per-nucleon binding energy

A deuteron contains one proton and one neutron. Using the supplied nuclear masses:

Δm = 1.00727647 + 1.00866492 - 2.01355321
= 0.00238818 u
Binding energy = 0.00238818(931.5)
= 2.22458967 MeV ≈ 2.22 MeV

In joules this is approximately 3.56 × 10-13 J. There are two nucleons, so the binding energy per nucleon is 2.22458967/2 = 1.11229 MeV per nucleon, approximately 1.11 MeV per nucleon.

Keep the mass digits until after subtraction: the defect is much smaller than either original mass. Use one consistent conversion route rather than mixing differently rounded constants. If a problem supplies atomic masses instead, account for electron masses consistently; do not silently mix atomic and nuclear masses.

Optional check A deuteron contains two nucleons and has binding mass defect 0.00238818 u. Use 1 u c^2 = 931.5 MeV. Which statement distinguishes total from per-nucleon binding energy?
A deuteron contains two nucleons and has binding mass defect 0.00238818 u. Use 1 u c^2 = 931.5 MeV. Which statement distinguishes total from per-nucleon binding energy?

09

Energy from fission and fusion

A reaction releases energy when its products have less total rest mass than its reactants. The binding-energy-per-nucleon curve helps explain why both the fusion of light nuclei and the fission of heavy nuclei can do this.

Read the binding-energy curve

Binding energy per nucleon rises rapidly across the light nuclei, reaches a broad maximum near nucleon numbers 50-60 in the iron/nickel region, then decreases gradually for heavier nuclei. It measures an average binding energy, not the total binding energy of a nucleus.

Binding energy per nucleon on the actual nucleon-number scale

Binding energy per nucleon on the actual nucleon-number scaleThe vertical axis is binding energy per nucleon from zero to nine megaelectronvolts per nucleon. The horizontal axis is nucleon number 1 to 240 on a uniform numerical scale. The values at nucleon numbers 2 and 4 correctly remain close to the left boundary rather than being artificially spread across the graph. The approximate anchors pair nucleon numbers 1, 2, 4, 16, 56, 92, 141 and 235 with binding energies per nucleon 0, 1.11, 7.07, 8.0, 8.8, 8.6, 8.3 and 7.6 MeV respectively. The trend rises sharply for light nuclei, has a broad maximum near nucleon number 56 and then falls gradually for heavy nuclei. The smooth connecting curve is an approximate trend through supplied anchors, not a precision measured isotope dataset. Individual isotopes can have additional local structure.Binding energy per nucleon/ MeV per nucleon0369160120180240Peak regionNucleon number AApproximate trend; equal A spacing.

The steep light-nucleus rise is compressed by the true horizontal scale. Read the separate enlarged panel below for A = 2 and A = 4. Energy release for a reaction still requires the complete, balanced initial and final account.

Light nuclei: a separately enlarged nucleon-number range

Light nuclei: a separately enlarged nucleon-number rangeThe vertical axis is binding energy per nucleon from zero to nine megaelectronvolts per nucleon. This separately labelled horizontal enlargement covers nucleon number zero to twenty. The trend begins with a single nucleon; no nucleus with zero nucleons is plotted. The same vertical scale as the main plot is retained. Approximate anchors are 1.11 MeV per nucleon at nucleon number 2, 7.07 at nucleon number 4, and 8.0 at nucleon number 16. The smooth connecting curve is an approximate trend through supplied anchors, not a precision measured isotope dataset. Individual isotopes can have additional local structure.Binding energy per nucleon/ MeV per nucleon03690248121620Nucleon number AHorizontal enlargement: A = 0 to 20

Only the horizontal range is enlarged here; the vertical quantity and scale are unchanged. A larger total binding energy is not automatically a larger binding energy per nucleon. The curve is an approximate trend, not an exact isotope mass table.

The main plot preserves the actual spacing of nucleon number A. The separate A = 0 to 20 view enlarges the light-nucleus region horizontally. These are approximate trend illustrations; the ordinate is binding energy per nucleon, not total binding energy.

Products with greater total binding energy have lower total rest energy for the same complete nucleon account. Multiply each nucleus's binding energy per nucleon by its own A before comparing totals. Local isotope differences exist, and this curve alone does not determine a reaction rate or every kind of nuclear stability.

Fusion combines light nuclei

In the supplied deuterium-tritium reaction, a hydrogen-2 nucleus and a hydrogen-3 nucleus form an alpha particle and a neutron:

++

Both sides have five nucleons and charge +2e. Use all products when comparing rest masses. The supplied nuclear masses, in u, are 2.01355321, 3.01550072, 4.00150618 and 1.00866492 in the displayed order.

Worked fusion energy

Include the outgoing neutron

Δm = (2.01355321 + 3.01550072)
- (4.00150618 + 1.00866492)
= 0.01888283 u
Q = 0.01888283(931.5)
= 17.5894 MeV ≈ 17.6 MeV

Here Q is the total energy released per complete reaction, using supplied 1 u c2 = 931.5 MeV. Leaving out the neutron would give a huge false mass deficit. Positive released energy does not remove the initial electric repulsion between the positively charged reactants.

Fission splits a heavy nucleus

One possible neutron-induced fission channel is:

+++ 3

The totals are A = 236 and charge +92e on both sides. Other fragment combinations are possible; this is one supplied channel, not a unique outcome of every fission event.

Approximate supplied binding energies for a fission estimate
NucleusABinding energy / MeV per nucleon
Uranium-2352357.60
Barium-1411418.30
Krypton-92928.60
Initial total binding = 235(7.60) = 1786.0 MeV
Final total binding = 141(8.30) + 92(8.60)
= 1170.3 + 791.2 = 1961.5 MeV
Energy released ≈ 1961.5 - 1786.0
= 175.5 MeV

Free neutrons contribute no nuclear binding energy. This is an estimate from the supplied rounded averages, not a precision energy value for the channel. The products are more tightly bound overall, so the decrease in total rest energy is available as kinetic energy and radiation.

For both fusion and fission, compare the complete initial and final accounts. Subtracting two per-nucleon values without weighting by nucleon number does not give the reaction energy.

Optional check For deuteron + triton -> alpha particle + neutron, supplied nuclear masses are 2.01355321, 3.01550072, 4.00150618 and 1.00866492 u respectively. With 1 u c^2 = 931.5 MeV, what is the total released energy?
For deuteron + triton -> alpha particle + neutron, supplied nuclear masses are 2.01355321, 3.01550072, 4.00150618 and 1.00866492 u respectively. With 1 u c^2 = 931.5 MeV, what is the total released energy?

Revision

Nuclear Physics at a glance

Identify what is being counted, what is actually detected and which energy account is complete. Those decisions come before substituting numbers.

Structure and evidence

Most alpha particles passing a thin foil are little deflected; rare large deflections support a small concentration of positive charge and mass. A nucleus has Z protons and A - Z neutrons. A neutral atom has Z electrons. Isotopes have the same Z and different neutron counts; ionisation changes electrons without changing the nucleus.

Count the named entity

n = m/M
N = nNA
NA = 6.02 × 1023 mol-1

Match the mass and molar-mass units. An atom has one nucleus; a molecule can have several. Relative atomic mass Ar and relative molecular mass Mr are dimensionless, while individual mass and molar mass have units.

Radiation and detection

Alpha is a helium-4 nucleus, beta-minus an electron from a nuclear process, and gamma an electromagnetic photon. Alpha and beta have opposite electric-force directions; gamma is uncharged. Penetration and ionisation comparisons need an energy and absorber context. Dense internal ionisation can be hazardous even for short-range radiation.

Count rate = recorded count / duration
Net source rate = total rate - background rate

Activity in Bq counts decays per second. Detector rate counts recorded events per second. They are related only through the stated emission and detection conditions. For unequal durations in a stable-rate series, pool counts over the total time; do not give every interval equal weight.

Decay is spontaneous and random for an individual nucleus. A large population has a predictable statistical trend, with fluctuations in finite counts. Repeats do not remove an incorrect background or changing geometry.

Half-life and time dependence

After one, two and three half-lives, the expected parent population, activity or proportional corrected rate is 1/2, 1/4 and 1/8 of its initial value. Halve a background-subtracted rate. The raw total tends towards background rather than zero.

For a supplied fractional-decay model with half-life t1/2:

r/r0 = (1/2)t/t1/2
= exp[-ln(2)t/t1/2]
t = -t1/2ln(r/r0)/ln(2)

The ratio and exponent are dimensionless. Use matching time units; a positive corrected ratio is required before taking its logarithm.

Applications need two kinds of reasoning

Explain the penetration or ionisation that makes the task possible, then relate half-life to its preparation, measurement or operating interval. A thickness gauge needs partial transmission over the useful range. An externally detected internal tracer needs radiation that escapes tissue. Treatment and sterilisation use damage from ionisation, with suitable penetration and controlled exposure.

Longer half-life is not universally preferable. Physical half-life is distinct from biological clearance. Irradiation is exposure to radiation; contamination is the presence of radioactive material.

Conservation and binding

Balance total nucleon number and charge across all particles. Alpha emission changes the daughter by ΔA = -4 and ΔZ = -2. Beta-minus leaves A unchanged and raises daughter Z by one. Gamma leaves A and Z unchanged while reducing nuclear energy.

Mass defect = Zmp + (A - Z)mn - mnucleus
Binding energy = Δmc2
Binding energy per nucleon = binding energy / A
Released energy Q = (minitial - mfinal)c2

Keep atomic and nuclear mass conventions consistent and retain digits until after subtraction. Binding energy is the positive energy required for complete separation. A fusion or fission energy account must include every reactant and product, including free neutrons.

The binding-energy-per-nucleon curve rises across light nuclei, peaks broadly near A = 50-60 and falls gradually for heavy nuclei. Compare total binding energies for a reaction by multiplying each per-nucleon value by its own nucleon count.

Nuclear quantities, symbols and units
QuantitySymbolUnit or meaning
Nucleon, proton and neutron numbersA, Z, NDimensionless counts; neutron N = A - Z
Number of specified particles or parent nucleiNDimensionless count
Amount of substancenmol
Avogadro constantNAmol-1
Molar massMkg/mol; g/mol when paired with g
Relative atomic and molecular massesAr, MrDimensionless ratios
Sample or particle massmkg; u for a specified particle
Atomic, electron, proton and neutron massesma, me, mp, mnkg or u; state which entity
ActivityABq = decays per second
Detected count ratercounts/s or counts/min
Time and half-lifet, t1/2s; convert min consistently
Energy, including binding energyEJ, eV or MeV
Energy released per reactionQJ or MeV
Vacuum light speedcm/s

A denotes a nucleon count in nuclide notation and activity in a decay-rate equation; identify the context and unit. N can denote neutron number A - Z or the remaining parent population. Dimensionless counts can also be labelled n or m under a stated convention; in this chapter n denotes amount in mol and m denotes mass. Read the definition and unit before substituting.

Supplied rounded conversions:
1 u = 1.66 × 10-27 kg
1 u c2 = 931.5 MeV
1 eV = 1.60 × 10-19 J
1 MeV = 1.60 × 10-13 J

Before reporting an answer, check the named entity, time unit, background correction, mass convention and whether the requested energy is total or per nucleon.