Mass defect and binding energy

A nucleus with atomic number $Z$ and mass number $A$ contains $Z$ protons and $N = A - Z$ neutrons. Add up the masses of these nucleons as free, separated particles, and the total always exceeds the actual measured mass of the bound nucleus. That shortfall is the mass defect:

where: $m_p$ is the proton mass, $m_n$ the neutron mass, and $M_{\text{nucleus}}$ the actual nuclear mass. In practice tables give atomic masses (nucleus plus $Z$ orbital electrons), so calculations normally use the hydrogen-atom mass in place of $m_p$ and the neutral atomic mass in place of $M_{\text{nucleus}}$ — the $Z$ electrons appear on both sides of the subtraction and cancel out. Einstein's mass–energy relation says any mass corresponds to an energy:

Applied to the missing mass, this gives the binding energy:

Physically, $BE$ is the energy that was released as the nucleons snapped together, and equally the energy an outside agent would have to supply to break the nucleus back into free, separated nucleons. It is not stored fuel waiting inside the nucleus; it is a measure of how much lower in energy the bound state sits compared to the scattered pieces. Working in SI units is clumsy because nuclear masses are so small, so nuclear physics uses the unified atomic mass unit, $1\,u = 1.66\times10^{-27}$ kg. Substituting this into $E = mc^2$ converts it into an energy:

so the working formula becomes $BE\,(\text{MeV}) = \Delta m\,(\text{in } u) \times 931.5$ — no separate $c^2$ calculation needed once $\Delta m$ is known in $u$. $E = mc^2$ is bigger than nuclear binding, though: it says mass and energy are two faces of the same currency, interconvertible in both directions. Two more syllabus examples make this concrete. In pair production, a high-energy photon passing near a nucleus vanishes and its energy reappears entirely as a particle–antiparticle pair, e.g. $\gamma \to e^- + e^+$; this needs at least the pair's combined rest-mass energy, $2 m_e c^2 = 1.02$ MeV, with anything extra becoming kinetic energy. In annihilation, the reverse happens: a particle meets its antiparticle, both vanish, and their entire rest mass reappears as photon energy.

Calculate the mass defect and binding energy of a deuteron ($^2_1\text{H}$), given: hydrogen atom mass $m_H = 1.007825\,u$, neutron mass $m_n = 1.008665\,u$, and deuterium atom mass $M_D = 2.014102\,u$. Then find how much energy is released if a deuteron additionally captures a free neutron to form tritium ($^3_1\text{H}$, mass $3.016049\,u$).

Step 1. Since $m_H$ and $M_D$ are both atomic masses, each carrying exactly one electron, the single electron cancels on subtraction, so the calculation can be done directly with the tabulated values:

Step 2. Convert to energy: $BE = \Delta m \times 931.5 = 0.002388 \times 931.5 = 2.224\ \text{MeV}$. This matches the well-known experimental deuteron binding energy of about 2.22 MeV — the energy needed to split a deuteron into a free proton and a free neutron.

Step 3. For the neutron-capture step $^2_1\text{H} + n \to {}^3_1\text{H}$, the energy released equals the mass lost in this reaction converted via $931.5$ MeV per $u$: $\Delta m' = (M_D + m_n) - M_T = (2.014102 + 1.008665) - 3.016049 = 0.006718\ u$, so the energy released is $0.006718 \times 931.5 = 6.258\ \text{MeV}$.

Answer: the deuteron has mass defect $0.002388\,u$ and binding energy $2.224$ MeV, and capturing one more neutron to form tritium releases a further $6.258$ MeV.

Competitive stretch

A 3.0 MeV photon undergoes pair production, creating an electron–positron pair (rest-mass energy $0.511$ MeV each). Rest-mass energy needed: $E_{\text{rest}} = 2 \times 0.511 = 1.022$ MeV. By energy conservation, $E_\gamma = E_{\text{rest}} + KE_{\text{total}}$, so $KE_{\text{total}} = 3.0 - 1.022 = 1.978\ \text{MeV}$ is shared as kinetic energy between the electron and positron. This is the same $E=mc^2$ bookkeeping as binding energy, run in reverse: energy converting into rest mass instead of mass converting into energy.

• Mixing atomic masses (which include electrons) with bare nuclear masses in the same subtraction — the electron masses then fail to cancel and introduce a small but graded-as-wrong error; always use atomic mass consistently on both sides.
• Confusing mass defect (a mass, in $u$ or kg) with binding energy (an energy, in MeV or J) — the two are related by a factor of $c^2$, they are not the same quantity or the same units.
• Using $c = 3\times10^8$ m/s but forgetting to square it, or ignoring the $1\,u = 931.5$ MeV shortcut and grinding through a much longer SI-unit calculation for no reason.
• Describing binding energy as fuel "stored inside" the nucleus that could be released like a battery — it is more accurate to say it is the energy required to pull the nucleus apart, i.e. how much lower in energy the bound state is compared to free nucleons.

Ask the tutor: want to work through the binding energy of a heavier nucleus like helium-4 or check your own pair-production calculation step by step?