Mass Defect & Nuclear Binding Energy

You’ll be able to

Compute the mass defect of a nucleus from its constituents
Convert mass defect to binding energy using E = mc²
Explain why both fusion of light nuclei and fission of heavy ones release energy

A nucleus weighs less than its parts

Add up the masses of the protons and neutrons in a nucleus and the total exceeds the measured mass of the nucleus itself. The difference is the mass defect, and by E = mc² it corresponds to the binding energy — the energy released when the nucleus formed, and equivalently the energy needed to pull it apart. The missing mass is not lost; it left as energy when the nucleus assembled.

Mass defect and binding energy

Δm = (Z·m_proton + N·m_neutron) − m_nucleus · E_binding = Δm·c² · in convenient units: 1 u = 931.5 MeV/c²

The conversion 1 u ↔ 931.5 MeV avoids ever handling c² explicitly, which is why nuclear physics is done in u and MeV.

Binding energy per nucleon is the meaningful number

Total binding energy grows simply because bigger nuclei have more nucleons, so it is a poor measure of stability. Binding energy per nucleon — total divided by mass number — is what indicates how tightly held each nucleon is. Plotted against mass number it rises steeply for light nuclei, peaks near iron-56 at about 8.8 MeV per nucleon, then declines slowly for heavy nuclei. Iron is therefore the most tightly bound nucleus, and that peak explains both of the energy-releasing nuclear processes.

Why fusion and fission both release energy

Any process that moves nucleons toward the iron peak increases binding energy per nucleon and therefore releases energy. Fusion of light nuclei — hydrogen into helium — climbs the steep left side, which is why it powers stars and releases far more energy per nucleon than fission. Fission of heavy nuclei — uranium splitting — climbs the gentle right side. Both move toward iron, from opposite directions. Fusing two iron nuclei, or splitting iron, would absorb energy, which is why stellar cores stop at iron and then collapse.

Conserved quantities in a nuclear equation

Balancing a nuclear reaction requires conserving mass number A (the superscript total) and charge Z (the subscript total). Note what is not conserved: total mass, which is exactly the point — the mass difference appears as released energy. An alpha decay reduces A by 4 and Z by 2; a beta-minus decay leaves A unchanged and raises Z by 1, since a neutron became a proton.

Worked example

A helium-4 nucleus has a mass of 4.0026 u. A proton is 1.00728 u and a neutron 1.00867 u. Find the mass defect and the binding energy per nucleon.

1.Helium-4 has 2 protons and 2 neutrons.
2.Sum of parts: 2(1.00728) + 2(1.00867) = 2.01456 + 2.01734 = 4.03190 u.
3.Mass defect: Δm = 4.03190 − 4.0026 = 0.02930 u.
4.Binding energy: (0.02930)(931.5) = 27.3 MeV.
5.Per nucleon: 27.3/4 = 6.8 MeV per nucleon.

Answer: Mass defect 0.0293 u, total binding energy 27.3 MeV, and 6.8 MeV per nucleon. That is high for such a light nucleus — helium-4 is unusually tightly bound, which is why it appears as the alpha particle in decay.

Watch out

Subtract the nuclear mass from the sum of the parts, in that order. Reversing it gives a negative defect and a negative binding energy, which would imply the nucleus falls apart spontaneously.

Checkpoint

The mass of a stable nucleus compared with the sum of its separated nucleons is:

Checkpoint

Both fusion of light nuclei and fission of heavy nuclei release energy because both:

Checkpoint

In beta-minus decay, the mass number A and atomic number Z change as follows:

On the exam

Work nuclear problems in atomic mass units and MeV using 1 u = 931.5 MeV. Converting to kilograms and joules is not wrong but it multiplies the opportunities for an exponent error.

Answer the 3 checkpoints as you read.