Full chapter
Nuclear Physics
All 9 topics and the revision summary on one page.
01
Evidence for the nucleus
Make a guess. You are not marked.
In the alpha-scattering experiment, a very few alpha particles bounced almost straight back from thin gold foil. What did this show?
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The positive charge and most of the mass are in a tiny central nucleus.
Only a very dense, positively charged centre could turn a fast alpha particle right round. That so few did shows the nucleus is tiny compared with the atom.
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
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
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.
| Observation | Inference |
|---|---|
| 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.
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.
Check your understanding
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?
Show the answer
Positive charge and most atomic mass are concentrated in a small nucleus.
Most paths avoid a close approach to the small concentration, while a rare close approach to its positive charge can strongly repel an alpha particle. The observations support a nucleus much smaller than the whole atom.
02
Count atoms, nuclei and nucleons
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About how many atoms are in 1.0 g of carbon-12?
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About 5.0 x 1022, one twelfth of a mole.
12 g of carbon-12 is one mole, 6.02 x 1023 atoms. 1.0 g is one twelfth of that: 5.0 x 1022 atoms.
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 = 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 = (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:
| Entity | Calculation | Sample count |
|---|---|---|
| Protons | 18N | 5.418 × 1021 |
| Neutrons | 22N | 6.622 × 1021 |
| Electrons | 18N | 5.418 × 1021 |
| Nucleons | 40N | 1.204 × 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:
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.
Check your understanding
A 0.0200 g sample of pure argon-40 has supplied molar mass 40.0 g/mol. With N_A = 6.02 x 1023 mol-1, how many argon nuclei does it contain?
Show the answer
3.01 x 1020 nuclei.
n = m/M = 5.00 x 10-4 mol, so N = nN_A = 3.01 x 1020 atoms. Each atom contains one nucleus; its 40 nucleons are a different count.
03
Alpha, beta and gamma
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Alpha, beta and gamma radiation pass through the same magnetic field. Which is deflected most, and in which sense?
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Beta, much more than alpha, and in the opposite direction.
Beta particles have a tiny mass, so they curve far more tightly than alpha particles. Their charge is negative, so they curve the opposite way.
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.
| Emission | Nature and charge | Typical comparison in a stated absorber |
|---|---|---|
| Alpha, α | A 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, γ | An 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
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
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.
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.
Check your understanding
Why does the short range of ordinary alpha radiation not make an alpha-emitting substance inside the body harmless?
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It can produce dense ionisation close to the material inside tissue.
Range, ionisation and source location all matter. A thin external barrier can absorb ordinary alpha radiation, but an internal emitter can deposit energy in nearby living tissue.
04
Random decay and measured counts
Make a guess. You are not marked.
A detector records 100 counts in one minute from a long-lived source, then 89 counts in the next minute. Has the source changed?
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Not necessarily, because random decay makes equal-time counts vary.
Each decay is random, so counts in equal intervals scatter around a mean even when the source is steady. A change of 11 in 100 is within that scatter.
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:
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
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
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.
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.
| Interval | Source + background | Background |
|---|---|---|
| 1 | 412 | 38 |
| 2 | 389 | 42 |
| 3 | 405 | 36 |
| 4 | 394 | 44 |
| 5 | 400 | 40 |
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.
Check your understanding
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?
Show the answer
36.0 counts/s.
The total rate is 2000/50.0 = 40.0 counts/s and background rate is 200/50.0 = 4.0 counts/s. Their difference is 36.0 counts/s.
05
Half-life and corrected curves
Make a guess. You are not marked.
Two samples contain the same radioactive nuclide. Sample B has twice as many undecayed nuclei as sample A. How do their half-lives and activities compare?
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Same half-life. B has twice the activity.
Half-life is a property of the nuclide, not of the amount. Activity is lambda times the number of nuclei, so twice the nuclei means twice the decays per second.
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.
| t / s | Net / counts s-1 | Total / counts s-1 |
|---|---|---|
| 0 | 80 | 84 |
| 120 | 40 | 44 |
| 240 | 20 | 24 |
| 360 | 10 | 14 |
| 480 | 5 | 9 |
Total expected rate approaches the background
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
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.
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:
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.
Try it step by step
Work it out in steps: Number of half-lives, then Fraction of nuclei remaining, then Activity remaining in Bq.
Show the answer
Problem 1.
Number of half-lives: 24 / 8.0 = 3 half-lives
Fraction of nuclei remaining: Fraction = (1/2)3 = 0.125
Activity remaining in Bq: A = 1200 x 0.125 = 150 Bq
Problem 2.
Number of half-lives: 60 / 15 = 4 half-lives
Fraction of nuclei remaining: Fraction = (1/2)4 = 0.0625
Activity remaining in Bq: A = 5000 x 0.0625 = 313 Bq (3 sf)
Problem 3.
Number of half-lives: 10 / 2.0 = 5 half-lives
Fraction of nuclei remaining: Fraction = (1/2)5 = 0.03125
Activity remaining in Bq: A = 8000 x 0.03125 = 250 Bq
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:
= 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:
t = -120 ln(0.10)/ln(2)
= 398.631 s ≈ 399 s
The same model gives a straight transformed relationship:
A logarithm turns the supplied fractional model into a straight line
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.
The logarithm requires a positive corrected rate. Keep seconds in this expression; substituting a time in minutes without converting changes the result.
Try it step by step
Work it out in steps: Decay constant lambda = ln 2 / half-life in per hour, then Activity A = A0 e^(-lambda t) in Bq.
Show the answer
Problem 1.
Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 10 = 0.0693 per hour
Activity A = A0 e^(-lambda t) in Bq: A = 2000 x e^(-0.0693 x 5.0) = 2000 x 0.707 = 1410 Bq (3 sf)
Problem 2.
Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 6.0 = 0.116 per hour
Activity A = A0 e^(-lambda t) in Bq: A = 900 x e^(-0.116 x 4.0) = 900 x 0.630 = 567 Bq (3 sf)
Problem 3.
Decay constant lambda = ln 2 / half-life in per hour: lambda = 0.693 / 20 = 0.0347 per hour
Activity A = A0 e^(-lambda t) in Bq: A = 1500 x e^(-0.0347 x 30) = 1500 x 0.354 = 530 Bq (3 sf)
Check your understanding
With fixed detection conditions, a background-corrected rate falls from 72 to 9 counts/s in 15 min. What is the half-life?
Show the answer
5.0 min.
72 to 36 to 18 to 9 is three half-lives. Therefore each half-life is 15/3 = 5.0 min.
06
Choose radiation for a task
Make a guess. You are not marked.
Why is gamma radiation, not alpha, used to sterilise syringes in sealed packs?
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Gamma passes through the packaging to reach the syringes inside.
Gamma is very penetrating, so it reaches every part of the sealed pack and kills bacteria there. Alpha would be stopped by the packaging.
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
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 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.
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.
Check your understanding
A tracer inside a pipe must be detected from outside it, and at least half its initial activity must remain after 60 min. Which radiation and half-life meet both conditions?
Show the answer
A gamma emitter with a 2 h half-life.
It supplies photons and retains 2^(-60/120), about 0.707, of its activity after 60 min. Gamma photons pass through the pipe wall to the detector, so both conditions are met.
07
Balance nuclear equations
Make a guess. You are not marked.
Uranium-238, proton number 92, emits an alpha particle. What are the nucleon and proton numbers of the daughter nucleus?
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Nucleon number 234 and proton number 90.
An alpha particle is 4 nucleons and 2 protons. 238 - 4 = 234 and 92 - 2 = 90, which is thorium.
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
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.
Check your understanding
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?
Show the answer
A and Z + 1.
A neutron-to-proton change leaves the total nucleon count unchanged. The new proton raises nuclear Z by one, while the emitted electron carries charge number -1.
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:
= 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 ≈ 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.
| Particle or nucleus | Mass / u |
|---|---|
| Proton, mp | 1.00727647 |
| Neutron, mn | 1.00866492 |
| Deuteron, hydrogen-2 nucleus | 2.01355321 |
| Triton, hydrogen-3 nucleus | 3.01550072 |
| Alpha particle, helium-4 nucleus | 4.00150618 |
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 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:
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
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.
Worked deuteron
Total and per-nucleon binding energy
A deuteron contains one proton and one neutron. Using the supplied nuclear masses:
= 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.
Try it step by step
Work it out in steps: Mass defect in u, then Binding energy in MeV (1 u = 931.5 MeV), then Binding energy per nucleon in MeV.
Show the answer
Problem 1.
Mass defect in u: Mass defect = 2 x 1.007276 + 2 x 1.008665 - 4.001506 = 0.030376 u
Binding energy in MeV (1 u = 931.5 MeV): BE = 0.030376 x 931.5 = 28.3 MeV
Binding energy per nucleon in MeV: BE per nucleon = 28.3 / 4 = 7.07 MeV (3 sf)
Problem 2.
Mass defect in u: Mass defect = 6 x 1.007276 + 6 x 1.008665 - 11.996709 = 0.098937 u
Binding energy in MeV (1 u = 931.5 MeV): BE = 0.098937 x 931.5 = 92.2 MeV
Binding energy per nucleon in MeV: BE per nucleon = 92.2 / 12 = 7.68 MeV (3 sf)
Problem 3.
Mass defect in u: Mass defect = 8 x 1.007276 + 8 x 1.008665 - 15.990525 = 0.137003 u
Binding energy in MeV (1 u = 931.5 MeV): BE = 0.137003 x 931.5 = 127.6 MeV
Binding energy per nucleon in MeV: BE per nucleon = 127.6 / 16 = 7.98 MeV (3 sf)
Check your understanding
A deuteron contains two nucleons and has binding mass defect 0.00238818 u. Use 1 u c2 = 931.5 MeV. Which statement distinguishes total from per-nucleon binding energy?
Show the answer
About 2.22 MeV total and 1.11 MeV per nucleon.
Delta m c2 = 2.22459 MeV binds the whole deuteron. Dividing by its two nucleons gives 1.11229 MeV per nucleon.
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
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
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.
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.
For a sketch question
Label these features when you sketch the curve
Put nucleon number A on the x-axis and binding energy per nucleon (in MeV) on the y-axis. Then show these features:
- A steep rise for light nuclei. Hydrogen-2 is only about 1.1 MeV per nucleon.
- Helium-4 sits above its neighbours. At about 7.1 MeV per nucleon it is a spike above hydrogen-3, helium-3 and lithium-6. Helium-4 is unusually tightly bound.
- A broad peak of about 8.8 MeV per nucleon near A = 56. Label it iron or nickel. These are the most tightly bound nuclei.
- A slow fall for heavy nuclei. Uranium is down to about 7.6 MeV per nucleon at A = 235 to 238.
- Mark the two directions. Draw an arrow up the left side labelled "fusion" and an arrow up the right side, towards the peak, labelled "fission".
Explain fission and fusion in mark-scheme order
A common 3-mark question asks why both fission and fusion release energy. Write the links in this order:
- Fusion: light nuclei are on the rising part of the curve. When they join, the product has a higher binding energy per nucleon.
- Fission: a heavy nucleus is past the peak. When it splits into two medium nuclei, the products are nearer the peak and have a higher binding energy per nucleon.
- In both cases the total binding energy of the products is greater than that of the reactants. So the products have less total rest mass. This mass difference Δm is released as energy, E = Δmc2, mostly as kinetic energy of the products.
Iron and nickel sit at the peak. Neither fusing nor splitting them raises the binding energy per nucleon, so neither process releases energy from them.
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
- (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.
Put these in order
Explain why energy is released when a uranium-235 nucleus undergoes fission. Put the links in order.
Show the answer
1. A slow neutron is absorbed by a uranium-235 nucleus, making it unstable. The unstable nucleus cannot hold together and splits.
2. The nucleus splits into two smaller nuclei and two or three neutrons. The fragments sit nearer the peak of the binding energy per nucleon curve than uranium does.
3. The fragments have a higher binding energy per nucleon than uranium-235, so the total binding energy increases. A larger total binding energy means the products are lighter than the reactants.
4. The total mass of the products is less than the total mass of the reactants. The missing mass is released as energy according to E = (change in m) c2.
5. The mass difference appears as kinetic energy of the fragments and neutrons, E = (change in m) c2.
Fission splits a heavy nucleus
One possible neutron-induced fission channel is:
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.
| Nucleus | A | Binding energy / MeV per nucleon |
|---|---|---|
| Uranium-235 | 235 | 7.60 |
| Barium-141 | 141 | 8.30 |
| Krypton-92 | 92 | 8.60 |
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.
Check your understanding
For deuteron + triton -> alpha particle + neutron, supplied nuclear masses are 2.01355321, 3.01550072, 4.00150618 and 1.00866492 u respectively. With 1 u c2 = 931.5 MeV, what is the total released energy?
Show the answer
About 17.6 MeV.
Initial minus final rest mass is 2.01355321 + 3.01550072 - 4.00150618 - 1.00866492 = 0.01888283 u. Multiplying by 931.5 gives 17.5894 MeV.
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 = 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.
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:
= 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.
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 steeply across light nuclei, with helium-4 (about 7.1 MeV) above its neighbours. It peaks at about 8.8 MeV near A = 56 (iron and nickel) and falls slowly to about 7.6 MeV for uranium. Fusion of light nuclei and fission of heavy nuclei both move towards the peak. The products then have greater total binding energy and less rest mass, and the mass difference is released as energy. Compare total binding energies for a reaction by multiplying each per-nucleon value by its own nucleon count.
| Quantity | Symbol | Unit or meaning |
|---|---|---|
| Nucleon, proton and neutron numbers | A, Z, N | Dimensionless counts; neutron N = A - Z |
| Number of specified particles or parent nuclei | N | Dimensionless count |
| Amount of substance | n | mol |
| Avogadro constant | NA | mol-1 |
| Molar mass | M | kg/mol; g/mol when paired with g |
| Relative atomic and molecular masses | Ar, Mr | Dimensionless ratios |
| Sample or particle mass | m | kg; u for a specified particle |
| Atomic, electron, proton and neutron masses | ma, me, mp, mn | kg or u; state which entity |
| Activity | A | Bq = decays per second |
| Detected count rate | r | counts/s or counts/min |
| Time and half-life | t, t1/2 | s; convert min consistently |
| Energy, including binding energy | E | J, eV or MeV |
| Energy released per reaction | Q | J or MeV |
| Vacuum light speed | c | m/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.
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.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
Define the binding energy of a nucleus. [2]
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Exam-style question
A helium-4 nucleus has a mass of 4.00151 u. A proton has a mass of 1.00728 u and a neutron 1.00867 u. Taking 1 u as equivalent to 931.5 MeV, calculate the binding energy per nucleon of helium-4. [4]
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Exam-style question
Use the variation of binding energy per nucleon with nucleon number to explain why energy is released both when light nuclei fuse and when a heavy nucleus splits. [3]
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Exam-style question
Sketch a graph of binding energy per nucleon against nucleon number A for A from 1 to 240. Label the values and features an examiner would look for. [4]
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Exam-style question
Data-based question: The americium-241 smoke detector [15]
Many smoke detectors contain a tiny sample of americium-241 (proton number 95), which decays by alpha emission to neptunium-237 with a half-life of 432 years. The alpha particles ionise the air between two metal plates connected to a battery. The ions move to the plates, so a small current flows. When smoke enters, ions attach to the smoke particles and move much more slowly, the current falls, and the alarm sounds.
| mass of americium-241 | 0.30 µg |
| molar mass of americium-241 | 0.241 kg mol-1 |
| activity of source | 3.7 x 104 Bq |
| fraction of alpha particles that enter the air gap | 0.50 |
| kinetic energy of each alpha particle | 5.5 MeV |
| average energy needed to make one ion pair in air | 35 eV |
- (a) Write the nuclear equation for the decay of americium-241. [2]
- (b) Explain why an alpha emitter is used rather than a beta or gamma emitter. [2]
- (c) Calculate the number of americium-241 nuclei in the source. [2]
- (d) Calculate the number of ion pairs produced by one alpha particle that stops in the air gap. [1]
- (e) Calculate the maximum current in the air gap. [3]
- (f) State one assumption you made in (e). [1]
- (g) The maker says the source will not need replacing during the detector's 10-year life. Use the data to evaluate this claim. [3]
- (h) Even in clean air, the current fluctuates slightly about its mean value. Suggest why. [1]
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Quick recall
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?