9478 / 2027
Nuclear Physics overview

Lesson 9 of 10

Mass defect and binding energy

  • I can explain mass defect.Syllabus

    Syllabus 9478, 20(p). Explain mass defect.

  • I can use the mass-energy equivalence E = mc^2.Syllabus

    Syllabus 9478, 20(q). Use the mass-energy equivalence E = mc^2.

  • I can explain nuclear binding energy and its relation to mass defect.Syllabus

    Syllabus 9478, 20(r). Explain nuclear binding energy and its relation to mass defect.

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

Use mass units consistently

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

Worked conversions

Keep the type of particle in the mass label

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

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

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

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

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

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

Convert an energy difference from mass

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

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

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

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

Binding energy is a positive separation energy

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

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

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

The same nucleons have different total rest energies

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

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

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

Worked deuteron

Total and per-nucleon binding energy

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

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

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

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

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.

  1. Problem 1. A helium-4 nucleus has mass 4.001506 u. Proton mass = 1.007276 u, neutron mass = 1.008665 u. Find its binding energy per nucleon.
  2. Problem 2. A carbon-12 nucleus has mass 11.996709 u. Proton mass = 1.007276 u, neutron mass = 1.008665 u. Find its binding energy per nucleon.
  3. Problem 3. An oxygen-16 nucleus has mass 15.990525 u. Proton mass = 1.007276 u, neutron mass = 1.008665 u. Find its binding energy per nucleon.
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?

  1. About 1.11 MeV total and 2.22 MeV per nucleon.
  2. About 2.22 MeV for both quantities.
  3. About 2.22 MeV total and 1.11 MeV per nucleon.
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.

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]

Saved on this browser only. Anyone using this browser can see it.

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]

Saved on this browser only. Anyone using this browser can see it.