Full chapter
Nuclear Physics
All 10 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
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.
For a sample population, N = (m/M)NA, with sample mass m and supplied molar mass M in matching units. NA = 6.02 × 1023 mol-1; a few times 10-4 mol therefore contain order 1020 entities. Name those entities: one atom has one nucleus, but usually many nucleons. The particle-counting explanation gives the mole and relative-mass foundations.
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?
02
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.
| 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.
Optional check Why does the short range of ordinary alpha radiation not make an alpha-emitting substance inside the body harmless?
03
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:
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.
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?
04
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.
| 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.
The exponential model relates this shape to decay constant, activity and a logarithmic data fit.
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?
05
Activity and exponential decay
The decay constant describes the chance of decay per nucleus per unit time. A large population follows a predictable expected trend even though each individual decay remains random.
Let N be the number of undecayed parent nuclei and λ the decay constant. For a sufficiently short interval Δt, the probability that one parent nucleus decays is approximately λΔt. The unit of λ is s-1 when time is in seconds.
N = N0e-λt
A = A0e-λt
Activity is the expected decay rate, in Bq: one decay per second. A constant λ means a constant fractional decay rate, not a constant number of decays per second. As N decreases, A decreases. The cumulative number that have decayed is N0 - N, which increases.
These equations describe one parent radionuclide with constant decay constant. They do not make its individual nuclei decay at predetermined times, and a mixture need not have one exponential total activity.
Connect decay constant and half-life
At t = t1/2, half the original parent population remains:
ln(1/2) = -λt1/2
λ = ln(2)/t1/2
A half-life of order 100 s suggests λ of order 10-2 s-1. For the supplied 120 s half-life, λ = 0.00577623 s-1. Convert minutes to seconds before using that value with a time.
Worked activity and detector rate
Decay rate is a property of the source
A separate sample initially contains N0 = 6.00 × 105 parent nuclei and has half-life 120 s:
= 3465.74 Bq ≈ 3.47 × 103 Bq
After 240 s, two half-lives have elapsed. N = 1.50 × 105 and A = 866.434 Bq, approximately 866 Bq.
Suppose a supplied calibration for this unchanged setup counts 5.00% of decays, with one relevant emission per decay. Its initial net detector rate is then:
= 173.287 counts/s
This is a different source from the 80 counts/s model below. The fraction detected is supplied calibration information, not a universal conversion between counts/s and Bq. Detector geometry, response and attenuation matter.
Use ratios and logarithms
With constant detection conditions and correctly subtracted background, the source count rate is proportional to activity:
ln(r/r0) = -λt
The logarithm's argument is a dimensionless ratio. A plot of ln(r/r0) against t has gradient -λ. For the 120 s model, the time to reach 10.0% is:
= 398.631 s ≈ 399 s
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.
Zero and negative corrected rates have no real logarithm. A positive value very close to background can also have large relative uncertainty; being mathematically loggable does not make it a reliable fitting point.
Optional check A sample initially contains 6.00 x 10^5 parent nuclei and has half-life 120 s. With lambda = ln(2)/120 s^-1, what is its activity at 240 s?
Work with count records
Preserve the counts, then calculate rates
Open the counts and decay workbook, or import the raw count CSV. These are constructed teaching records. The working workbook contains formulas and charts; inspect and reproduce the calculations rather than treating the file as an experiment you performed.
For CSV import, choose comma separation and keep the start, duration and count columns numeric. Create sheets named Rates and Decay. Import the complete CSV into Rates starting at A10. If it opens in A1:F16 on a separate raw-import sheet, preserve that sheet and copy the entire block to Rates A10:F25. This places the headers and records at the cells used below.
You can also type the supplied records into those same cells manually. Retain the record IDs, series and data-status labels alongside the numbers. The Motion data method explains numeric import, formula copying and XY plots.
On Rates, headers are in A10:F10: record ID, series, start time in s, duration in s, total count and data status. Source-plus-background records S01-S05 occupy rows 11-15; background records B01-B03 occupy 16-18; decay records D01-D07 occupy 19-25. Each series uses its own time origin.
Keep these raw columns intact. A blank is not a measured zero. Resolve missing readings or non-positive durations before calculating or fitting; the stated ranges do not automatically expand when records are added.
Pool counts using the actual duration
The S records contain 4000 counts in 100 s; the B records contain 720 counts in 180 s. These separate stable-rate series give 40.0 total, 4.0 background and 36.0 net counts/s. The unweighted mean of the five S rates is 40.1533 counts/s, which gives equal weight to unequal counting durations.
- Rates B4: pooled background rate
=SUM(E16:E18)/SUM(D16:D18)- Rates B5: pooled total rate
=SUM(E11:E15)/SUM(D11:D15)- Rates B6: net stable-source rate
=B5-B4- G11: interval midpoint, in s
=C11+D11/2- H11: total rate, in counts/s
=E11/D11- I11: background-subtracted rate
=IF(B11="background","",H11-$B$4)
Copy G11:I11 down through row 25. The fixed reference $B$4 keeps the same background rate. Do not average the entire decaying series into one rate to estimate its time dependence.
Make the logarithmic selection explicit
On Decay, B4 contains reference rate 80 counts/s, B5 contains expected half-life 120 s, and B6 contains =LN(2)/B5. Label A10:F10 as record ID, midpoint, net rate, ratio, logarithm and status. Rows 11-17 link to the seven D records:
- A11: record ID; B11: midpoint
=Rates!A19=Rates!G19- C11: corrected rate; D11: dimensionless ratio
=Rates!I19=C11/$B$4- E11: logarithm only for positive rate
=IF(C11>0,LN(D11),"")- F11: retain the reason for an absent logarithm
=IF(C11>0,"positive","not logged")
Copy A11:F11 through row 17. The first five D records are rounded expected counts over finite windows; the final two deliberately illustrate fluctuations near background.
| Record | Midpoint / s | Net / counts s-1 |
|---|---|---|
| D01 | 2 | 79 |
| D02 | 120 | 40 |
| D03 | 240 | 20 |
| D04 | 360 | 10 |
| D05 | 480 | 5 |
| D06 | 960 | -0.1 |
| D07 | 1200 | 0 |
Keep all seven records. The last two are background-subtracted estimates, not negative or zero claims about the source's expected activity. Their logarithm cells remain blank, rather than being replaced with numerical zero.
- Decay E4: fitted decay constant, in s-1
=-SLOPE(E11:E15,B11:B15)- E5: free intercept of the log fit
=INTERCEPT(E11:E15,B11:B15)- E6: fitted half-life, in s
=LN(2)/E4
This specified fit uses D01-D05 only. It does not force the intercept to zero. For genuine records, preserve all readings and justify the range using the method and uncertainty; do not remove points merely to make a line straighter.
Compare the expected and fitted trends
A22:A82 contains model times 0 to 1200 s in 20 s steps. Use these formulas and copy B22:D22 through row 82:
- B22: expected net rate
=$B$4*EXP(-$B$6*A22)- C22: expected total including background
=B22+Rates!$B$4- D22: net rate from the free-intercept fit
=$B$4*EXP($E$5-$E$4*A22)
Use an XY chart with numeric midpoint time in s horizontally and corrected count rate in counts/s vertically. Show all seven records as individual markers, with the expected and fitted trends separately labelled. A second XY chart plots only E11:E15 against B11:B15, with dimensionless ln(r/r0) vertically and the fitted straight line. Do not join the record markers into an apparently observed continuous path.
Display the fitted numeric equation beside the log graph, using the intercept in E5 and negative gradient -E4. The working workbook does this in H43. For the supplied records it is approximately ln(r/r0) = -0.00061647 - 0.005774516t, with t in seconds and gradient unit s-1. Keep the underlying coefficient cells at full precision.
Compare the calculated results
The fitted λ is 0.005774516 s-1, the log intercept is -0.00061647, and the fitted half-life is 120.0355 s. These are close to the supplied generating values, but not identical.
A finite-window rate approximates the instantaneous midpoint rate. In this model, the 20 s early counting window changes that relation by about 0.05562%. Together with rounding to integer counts, this explains the small fitted difference; it is not evidence of a faulty real detector.
Keep full stored precision and format only the displayed values. Changing a supplied input should update its derived midpoint, rate and fit; preserve an original copy before exploring such changes. Long windows collect more counts but blur a rapidly changing rate.
With actual records, also consider random counting scatter, uncertainty or drift in background, timing, detector dead time, geometry and efficiency changes, and more than one radionuclide. Agreement with a constructed model is a calculation check. The measurement-record method explains how to preserve actual readings, instrument information and justified exclusions when analysing an investigation.
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
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.
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?
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.
The beta-decay evidence explains why the electron and daughter alone cannot account for the observations.
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.
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?
08
What beta decay reveals
Beta decay gives evidence for an additional particle through both the spread of electron energies and the momentum that is missing from an electron-and-daughter account.
Test the two-product prediction
Consider a parent nucleus initially at rest and a fixed decay branch to a specified daughter state. If only an electron and that daughter were produced, momentum conservation would require equal and opposite momenta. Together with the fixed energy available, this would fix the electron's energy.
Instead, beta electrons from that branch have a continuous range of energies up to an endpoint. The energy varies because it is shared with another emitted particle as well as the recoiling daughter. The extra particle is uncharged and difficult to detect; in beta-minus decay it is an antineutrino.
Electron energies form a continuous spread
A fixed two-body decay of a parent initially at rest would give a fixed electron energy for that branch. The observed spread motivates an additional energy carrier while preserving conservation. This schematic adds no independent event energy or isotope-specific endpoint.
Translate momenta head to tail to test the complete account
The electron and daughter momenta alone leave a downward resultant. An additional upward momentum closes the account. An unobserved neutral carrier can conserve energy and momentum; the diagram does not justify abandoning either conservation law.
Supplied momentum account
Find the missing vector
The parent is at rest. The supplied electron and daughter momenta in a planar example, in units of a common momentum p0, are:
pdaughter = (-3, -4)p0
Sum = (0, -4)p0
The total initial momentum is zero, so the missing particle must carry (0, +4)p0. Adding this vector closes the account. Adding only the two momentum magnitudes would discard their directions and give the wrong result.
The energy distribution and momentum imbalance support an additional carrier. They are not a reason to abandon energy or momentum conservation. Balancing nucleon number and charge alone would not have revealed the missing energy and momentum.
Optional check For one beta-decay branch of a parent initially at rest, electrons have a continuous energy spread and the measured electron-plus-daughter momentum does not close the account. What inference retains conservation?
09
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.
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?
10
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.
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.
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.
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?
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.
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.
N = N0e-λt
A = A0e-λt
λt1/2 = ln(2)
For fixed detection conditions, ln(r/r0) versus t has gradient -λ. Preserve zero and negative corrected records but do not logarithmically fit them as zero. Near-background positive points also need an uncertainty judgement. A fitted half-life from constructed values checks a model calculation; it does not establish experimental agreement.
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.
For a fixed beta-decay branch, a two-product model would fix the electron energy. Its continuous energy spectrum and a missing momentum vector support an additional emitted particle. Energy and momentum conservation apply to the complete account.
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.
| 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 |
| 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 |
| Decay constant | λ | s-1 for t in s |
| 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.