Mass-energy relation, mass defect and binding energy
A nucleus weighs less than its separate nucleons. This mass defect becomes the binding energy Eb = ΔM c², with 1 u equal to 931.5 MeV.
Mass is a form of energy
Einstein showed that mass and energy are equivalent: E = mc², where c is the speed of light. A small mass is worth a huge energy, because c² is so large. In any reaction, the total of mass and energy is conserved, not mass alone. In chemical reactions the change of mass is far too small to measure. In nuclear reactions it is large enough to measure.
Mass defect and binding energy
Careful measurements show that the mass of a nucleus is less than the total mass of its free protons and neutrons. The difference is the mass defect ΔM. When nucleons come together to form a nucleus, energy is released; that energy carries away the mass ΔM. To pull the nucleus apart into free nucleons, the same energy must be supplied. This energy is the binding energy Eb = ΔM c². A larger binding energy means a more tightly held nucleus.
How to find the binding energy, step by step
- 1Count Z (protons) and N = A − Z (neutrons).
- 2Find the total mass of the free nucleons: Zmₚ + Nmₙ.
- 3Subtract the mass of the nucleus: ΔM = Zmₚ + Nmₙ − M.
- 4Convert to energy: Eb (in MeV) = ΔM (in u) × 931.5.
- 5Divide by A to get the binding energy per nucleon. (If atomic masses are given, use the mass of a hydrogen atom, 1.00783 u, in place of mₚ, so the electron masses cancel.)
Worked example: binding energy of helium-4
The mass of a ⁴₂He nucleus is 4.00151 u. Find its binding energy and binding energy per nucleon. (mₚ = 1.00727 u, mₙ = 1.00866 u, 1 u = 931.5 MeV)
- Z = 2, N = 2.
- Mass of 2 protons = 2 × 1.00727 = 2.01454 u; mass of 2 neutrons = 2 × 1.00866 = 2.01732 u.
- Total = 4.03186 u.
- ΔM = 4.03186 − 4.00151 = 0.03035 u.
- Eb = 0.03035 × 931.5 MeV = 28.3 MeV.
- Per nucleon: 28.3 MeV ÷ 4 = 7.07 MeV.
Eb ≈ 28.3 MeV; about 7.07 MeV per nucleon
Worked example: energy of 1 g of mass, and 1 u in MeV
(a) Find the energy equivalent of 1 g of mass. (b) Show that 1 u is equivalent to about 931.5 MeV. (1 u = 1.66054 × 10⁻²⁷ kg, c = 2.9979 × 10⁸ m/s, 1 MeV = 1.6022 × 10⁻¹³ J)
- (a) E = mc² = 10⁻³ kg × (3 × 10⁸ m/s)² = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ J.
- (b) E = 1.66054 × 10⁻²⁷ × (2.9979 × 10⁸)² J = 1.66054 × 10⁻²⁷ × 8.9874 × 10¹⁶ J ≈ 1.4924 × 10⁻¹⁰ J.
- In MeV: 1.4924 × 10⁻¹⁰ ÷ 1.6022 × 10⁻¹³ ≈ 931.5 MeV. (Exact constants are needed here; rounded ones give about 934 MeV.)
(a) 9 × 10¹³ J; (b) 1 u ≈ 931.5 MeV
Fully turning just 1 g of mass into energy would give 9 × 10¹³ J, about 2.5 crore units (kWh) of electricity. Nuclear power stations release only a small part of this, yet it is still millions of times more than burning the same mass of coal.
2 marks: “Define mass defect and binding energy of a nucleus. How are they related?” Mass defect: free-nucleon mass minus nuclear mass, ΔM = Zmₚ + Nmₙ − M. Binding energy: energy to split the nucleus into free nucleons. Eb = ΔM c².
3 marks: “Calculate the binding energy per nucleon of ⁴₂He. (Nuclear mass 4.00151 u, mₚ = 1.00727 u, mₙ = 1.00866 u, 1 u = 931.5 MeV)” ΔM = 2(1.00727) + 2(1.00866) − 4.00151 = 0.03035 u. Eb = 0.03035 × 931.5 ≈ 28.3 MeV. Per nucleon ≈ 7.07 MeV.
1 marks: “Write the energy equivalent of 1 u in MeV.” 931.5 MeV.
Lost mass is locked energy: ΔM (in u) × 931.5 = Eb (in MeV).