23.1 Mass defect and nuclear binding energy
- Syllabus
- 9702–2028–2029
- Topic
- 23.1
- Level
- A2
Mass is a form of stored energy. A system of mass m has rest energy E=mc²; when its mass changes by Δm, the corresponding energy change is ΔE=Δmc².
E=mc2,m=E/c2,c=3.00×108ms−1UseminkgforEinJ.
For 1.00 u=1.66×10⁻²⁷ kg, E=(1.66×10⁻²⁷)(3.00×10⁸)²=1.49×10⁻¹⁰ J=931 MeV.
E=mc² is rest-energy equivalence, not the classical kinetic-energy formula. Convert g or u to kg before using SI c, unless using the established 1 u≈931 MeV/c² bridge.
In nuclide notation ᴬ_ZX, A is nucleon number and Z is proton number. Across a nuclear reaction, the total A and total Z are each conserved.
Write two balances: ΣA(left)=ΣA(right) and ΣZ(left)=ΣZ(right). Solve both before identifying the missing nuclide or particle from its A and Z.
¹74N+24He→817O+11HA:14+4=17+1;Z:7+2=8+1
Useful entries: neutron ¹₀n, proton ¹₁p, beta-minus ⁰₋₁e, beta-plus ⁰₊₁e and gamma ⁰₀γ.
Do not conserve the number of written symbols or electrons as in chemistry. A beta-minus particle contributes A=0 and Z=−1 to the equation.
Mass defect Δm is the mass of the constituent nucleons when separated to infinity minus the mass of the bound nucleus.
Binding energy is the minimum energy required to separate all nucleons of the nucleus to infinity. The same energy is released when the nucleus forms from those separated nucleons.
Δm=Zmp+(A−Z)mn−mnucleusEbinding=Δmc2;bindingenergypernucleon=Ebinding/A
A positive mass defect means the bound state has lower mass-energy. More binding energy must be supplied to dismantle that nucleus into the same free nucleons.
Mass is not lost from conservation laws: the lower rest mass corresponds to energy released. State the infinitely-separated reference state in definitions.
Horizontal axis: nucleon number A, from about 1 to 250. Vertical axis: binding energy per nucleon, usually in MeV.
Starting near A=1, draw a steep rise to one broad maximum near A≈56 (iron/nickel region), then a much shallower continuous decrease toward heavy nuclei.
The right-hand branch stays well above zero at A≈250. Light fusion candidates lie left of the peak; heavy fission or alpha-decay candidates lie right of it.
Greater binding energy per nucleon means nucleons are, on average, more tightly bound and more energy per nucleon is required to separate the nucleus.
Do not draw a symmetric bell curve or return the heavy-nucleus end to zero. The graph is binding energy per nucleon, not total binding energy.
Nuclear fusion is the joining of two light nuclei to form one heavier nucleus, usually requiring very high temperature so nuclei can approach despite electrostatic repulsion.
Nuclear fission is the splitting of one heavy nucleus into two or more lighter nuclei, commonly after the heavy nucleus absorbs a neutron; further neutrons may be released.
Fusion: light + light → heavier. Fission: heavy → lighter + lighter (and often neutrons). Both conserve total nucleon number and proton number.
Energy release is a consequence, not the definition. Do not call alpha decay fission merely because a heavy nucleus emits a small particle.
A reaction releases energy when its products have greater total binding energy than its reactants. The products then have lower total rest mass; the difference appears as released energy.
Fusion starts with light nuclei left of the A≈56 peak. Joining them moves the product toward the peak, increasing binding energy per nucleon and total binding energy.
Fission starts with a very heavy nucleus right of the peak. Splitting it into medium-mass products also moves nucleons toward the peak and increases average binding.
releasedenergy=totalbindingenergy(products)−totalbindingenergy(reactants)wheretotalbindingenergy=A×(bindingenergypernucleon)foreachnucleus
Compare totals when nuclei have different A. Energy is released because binding energy increases and mass-energy decreases—not because binding energy is consumed.
Δm=Σmreactants−ΣmproductsEreleased=Δmc2
For ²₁H+²₁H→⁴₂He with masses 2.013553 u and 4.001505 u: Δm=2(2.013553)−4.001505=0.025601 u.
Convert Δm=(0.025601)(1.66×10⁻²⁷)=4.25×10⁻²⁹ kg, then E=(4.25×10⁻²⁹)(3.00×10⁸)²=3.82×10⁻¹² J per helium nucleus.
For 1.00 mol of helium nuclei, multiply by 6.02×10²³: E=2.30×10¹² J. For reaction rate R, power P=RE_reaction.
If binding energies are given, calculate Ereleased=ΣBEproducts−ΣBEreactants. If binding energy per nucleon is given, multiply each value by that nucleus's A before summing.
Include every coefficient in the mass sum. A positive reactant-minus-product Δm means release; keep per-reaction, per-nucleus, per-mole and per-second quantities distinct.