Free Pearson Edexcel International A Level Physics lesson

Nuclear binding, fusion and fission

Use mass differences to track nuclear energy changes.

What you’ll learn

  • Use consistent constituent and electron accounting before converting mass into binding energy.
  • Use the curve for explanation and a complete energy balance for a calculation.
  • Explain temperature, density and confinement separately before evaluating the energy output.

Bound nuclei have lower rest energy

A bound nucleus has less rest mass than the sum of its separated protons and neutrons. This difference is the mass deficit. Energy must be supplied to separate the nucleons completely, so binding energy is positive even though forming the nucleus lowers its rest energy. Compare exactly the same constituents before applying ΔE = Δmc².

The atomic mass unit is convenient for small masses: 1 u ≈ 1.66 × 10⁻²⁷ kg. With the rounded constants c = 3.00 × 10⁸ m s⁻¹ and 1 eV = 1.60 × 10⁻¹⁹ J, one u corresponds to approximately 934 MeV. If a question supplies the more precise 931.5 MeV per u, use that value consistently rather than mixing precision conventions.

Atomic masses include electrons; nuclear masses do not. Electron masses often cancel when neutral atomic masses are used on both sides of a balanced equation, but check their numbers. A comparison of free protons and neutrons with a nucleus requires nuclear mass or an explicit electron correction. Binding energy per nucleon is total binding divided by A, not by Z.

Δm = Zmp + (A − Z)mn − mnucleus Eb = Δmc² binding per nucleon = Eb/A

Symbols and units

  • A is nucleon number and Z is proton number, both dimensionless counts; A − Z is neutron number.
  • mp, mn and mnucleus are the rest masses of one proton, one neutron and the complete nucleus. Use all masses consistently in kg or in atomic mass units u.
  • Δm is separated-nucleon mass minus bound nuclear mass; Eb is the positive binding energy in J; c = 3.00 × 10⁸ m s⁻¹ is the speed of light.
  • Using this example’s rounded constants, 1 u = 1.66 × 10⁻²⁷ kg and 1 MeV = 1.60 × 10⁻¹³ J. Eb/A is binding energy per nucleon in J per nucleon or MeV per nucleon.

How the equation explains the idea

  1. Compare the same constituents: Δm = Zmp + (A − Z)mn − mnucleus. A bound nucleus has smaller rest mass, so Δm is positive; mixing atomic and nuclear masses without checking electrons changes the comparison.
  2. Convert the supplied deficit before using SI c: Δm = 0.0300 × 1.66 × 10⁻²⁷ = 4.98 × 10⁻²⁹ kg.
  3. Calculate Eb = Δmc² = (4.98 × 10⁻²⁹)(3.00 × 10⁸)² = 4.482 × 10⁻¹² J. Then Eb/(1.60 × 10⁻¹³ J per MeV) = 28.0 MeV.
  4. This energy must be supplied to separate the nucleus completely. If A were supplied, dividing 28.0 MeV by A would give binding per nucleon; division by Z would count only protons.
Binding energy is proportional to mass deficit

Conditions. Computed from Eb = Δmc², using c = 3.00 × 10⁸ m s⁻¹. Both axes are explicitly scaled SI quantities. Δm compares the same free nucleons with their bound nucleus, using consistent electron accounting.

Mass deficit Δm / (10⁻²⁹ kg); Binding energy / (10⁻¹² J).

Each 10⁻²⁹ kg of positive deficit corresponds to 0.900 × 10⁻¹² J of binding energy. The graph shows a mass–energy conversion, not measured nuclides or the binding-energy-per-nucleon curve; it does not establish reaction rates.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Convert a mass deficit

  1. Δm = 0.0300 u = 4.98 × 10⁻²⁹ kg using 1 u = 1.66 × 10⁻²⁷ kg.
  2. Eb = 4.98 × 10⁻²⁹ × (3.00 × 10⁸)² = 4.482 × 10⁻¹² J.
  3. Divide by 1.60 × 10⁻¹³ J per MeV to obtain 28.0 MeV.

A positive deficit corresponds to positive binding energy.

Keep this idea. Use consistent constituent and electron accounting before converting mass into binding energy.

Read the binding-energy curve

The broad binding-energy-per-nucleon curve rises sharply for light nuclei, reaches a broad maximum around the iron region and then falls slowly for heavier nuclei. Nuclear structure produces variations around that broad shape. Combining suitable light nuclei or splitting suitable heavy nuclei can therefore yield more tightly bound products.

Fusion combines nuclei; fission splits a heavy nucleus into smaller fragments, commonly releasing neutrons. To calculate released energy, compare total binding of every reactant and product. Multiply per-nucleon values by their respective nucleon counts before adding them. An upward movement on the per-nucleon curve suggests an energy-release possibility, but does not prove a fast reaction.

In a chain reaction, neutrons from one fission can induce further fissions. The balance between neutron production, absorption and escape decides whether the chain grows, remains steady or dies away. Energy is not released merely because something splits: the selected products must have lower total rest energy. Those products can also remain radioactive.

Released energy = total final binding − total initial binding

Symbols and units

  • A is nucleon number, a dimensionless count. Binding energy per nucleon is quoted in MeV per nucleon.
  • Total binding is A times binding per nucleon, in MeV. For several nuclei, add their individual total binding energies.
  • Q is released energy per reaction in MeV or J: Q = total final binding − total initial binding when the comparison preserves the same constituent rest-mass total.
  • Free emitted nucleons have zero nuclear binding energy; include them in the nucleon balance even though they add no binding term.

How the equation explains the idea

  1. The stated simplified reaction preserves all 200 nucleons and compares their mean binding. Initially, total binding is 200 × 7.5 = 1500 MeV.
  2. The final total is 200 × 8.3 = 1660 MeV. Therefore Q = 1660 − 1500 = 160 MeV, equivalently 200(8.3 − 7.5).
  3. Greater final binding means lower final rest energy for the same constituents; the 160 MeV difference becomes other energy, rather than appearing from destroyed nucleons.
  4. This balance tests an energy difference, not a reaction rate. Use actual product counts, including free neutrons, in a detailed reaction and consider the interaction conditions separately.

Worked example

A simplified fission balance

  1. A = 200 initially has binding 7.5 MeV per nucleon.
  2. Products preserve total A and have mean binding 8.3 MeV per nucleon.
  3. Released energy = 200(8.3 − 7.5) = 160 MeV.

Compare total energies for all conserved nucleons.

Keep this idea. Use the curve for explanation and a complete energy balance for a calculation.

Why fusion needs extreme conditions

Positive nuclei repel electrically. The strong interaction can bind suitable nuclei only when they come very close. High temperature raises the kinetic-energy distribution so that more collisions reach sufficiently close separations. Quantum tunnelling also contributes in stars, so a classical barrier height is not a strict energy threshold for every successful event.

High density increases the number of potential encounters per volume. Sustained fusion also requires confinement long enough for sufficient reactions to occur. Gravity confines stellar plasma, while its pressure responds to the balance between inward gravitational forces and outward pressure forces. Temperature, density and confinement do different jobs and cannot be substituted for one another.

When fusion products have smaller total rest mass, the difference appears as particle kinetic energy and radiation. Very hot plasma cannot simply touch an ordinary container without strong cooling and material damage. An explanation of the requirements is distinct from a claim that a particular laboratory system provides net useful electrical output; that would require a complete measured energy budget.

Q = (minitial − mfinal)c² released power = successful event rate × Q

Symbols and units

  • minitial and mfinal are summed rest masses of all initial and final particles in kg; Δm = minitial − mfinal is the rest-mass decrease.
  • Q is released energy per successful reaction in J; c = 3.00 × 10⁸ m s⁻¹ is the speed of light; 1 MeV = 1.60 × 10⁻¹³ J.
  • The successful event rate is the number of reactions per second in s⁻¹; multiplying it by Q gives released power in W.
  • Temperature (K), particle number density (m⁻³) and confinement time (s) describe different reaction conditions; Q alone specifies none of them.

How the equation explains the idea

  1. Use the balanced reaction’s decrease in rest mass, not the total reacting mass: Q = Δmc² = (3.0 × 10⁻²⁹)(3.00 × 10⁸)² = 2.7 × 10⁻¹² J.
  2. Convert the event energy: Q = (2.7 × 10⁻¹²)/(1.60 × 10⁻¹³) = 16.9 MeV. Rest energy falls while total energy, including product motion and radiation, is conserved.
  3. To find power, multiply Q by a separately known successful event rate. The example gives energy per event but no rate, so it cannot determine power or net useful electrical output.
  4. High temperature changes collision energies, density changes the number of possible encounters and confinement keeps reactants together. These conditions affect the rate; they are not derived from Q = Δmc².

Worked example

Energy per event

  1. A balanced reaction loses rest mass 3.0 × 10⁻²⁹ kg.
  2. Q = 3.0 × 10⁻²⁹ × 9.0 × 10¹⁶ = 2.7 × 10⁻¹² J.
  3. This is 16.9 MeV using 1.60 × 10⁻¹³ J per MeV.

Total energy is conserved while rest energy changes.

Keep this idea. Explain temperature, density and confinement separately before evaluating the energy output.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

Browse Physics Unit 5 lessons and resources