Binding energy, fission and fusion flashcards

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Nuclear binding energy
The energy needed to separate a nucleus completely into its individual protons and neutrons.
Mass deficit (mass defect)
Total mass of the separate protons and neutrons minus the mass of the nucleus they form.
Binding energy per nucleon
Total binding energy of a nucleus divided by its nucleon number A; a measure of stability.
Atomic mass unit (u)
One twelfth of the mass of a carbon-12 atom: 1 u = 1.66 × 10⁻²⁷ kg.
Nuclear fission
The splitting of a heavy nucleus into two middle-mass nuclei, usually with a few free neutrons, releasing energy.
Nuclear fusion
The joining of two light nuclei to form a heavier nucleus, releasing energy.
Coulomb barrier
The electrostatic potential energy two nuclei must overcome to get close enough for the strong force to act.
Mass–energy equation
ΔE = c²Δm (ΔE in J, Δm in kg). (Yes)
Mass deficit of a nucleus with Z protons and A nucleons
Δm = Zmₚ + (A − Z)mₙ − m(nucleus). (No — learn it)
Energy equivalent of 1 u
1 u × c² = 1.49 × 10⁻¹⁰ J ≈ 931.5 MeV. (No — derive it)
1 MeV in joules
1 MeV = 10⁶ × 1.60 × 10⁻¹⁹ J = 1.60 × 10⁻¹³ J. (No — derive it)
Energy released in a reaction, from binding energies
Released = total binding energy after − total binding energy before (each total = per-nucleon value × A). (No — learn it)
Mean kinetic energy of a nucleus at temperature T
½m⟨c²⟩ = 3/2 kT. (Yes)
Electrostatic potential energy of two nuclei at separation r
E = kQ₁Q₂/r (from V = kQ/r multiplied by the second charge). (No — derive it)
Shape of the binding energy per nucleon curve
Steep rise for light nuclei, broad peak (about 8.8 MeV) near iron-56, slow fall to about 7.6 MeV for uranium.
Where on the curve does fusion release energy?
On the steep left side: light nuclei combine and move up towards the peak.
Where on the curve does fission release energy?
On the right side: a heavy nucleus splits into two nuclei nearer the peak.
Which nucleus sits unusually high on the left of the curve?
Helium-4 (about 7.1 MeV per nucleon) — far more tightly bound than its neighbours.
Potential energy against separation for two nuclei
Rises as kQ₁Q₂/r as they approach (the Coulomb barrier), then drops into a well once the strong force acts at about 10⁻¹⁵ m.
Why does a nucleus have less mass than its separate nucleons?
Energy is released when the nucleons bind, so the nucleus has less energy, and by ΔE = c²Δm less mass.
Why compare binding energy per nucleon, not total binding energy?
Total binding energy grows with the number of nucleons; per nucleon shows how tightly each is bound.
Why do fission and fusion both release energy?
Both raise the binding energy per nucleon; with A conserved, total binding energy rises, mass falls and Δm c² is released.
Why can't iron-56 release energy by fission or fusion?
It is at the peak; any change would lower the binding energy per nucleon and need an energy input.
Why does fusion need a very high temperature?
Nuclei repel electrostatically and must get within about 10⁻¹⁵ m; they need a large kinetic energy, and mean kinetic energy ∝ T.
Why does fusion need a high density?
More nuclei per unit volume means more collisions per second, so a higher rate of fusion.
Why is fusion more energy-rich per kilogram than fission, though it releases less per reaction?
Its fuel nuclei are very light, so many more reactions happen per kilogram (about 3.4 × 10¹⁴ J kg⁻¹ for D–T against 7.6 × 10¹³ J kg⁻¹ for U-235).
Mass deficit 0.0300 u. Binding energy in J and MeV?
0.0300 × 1.66 × 10⁻²⁷ × (3.00 × 10⁸)² = 4.48 × 10⁻¹² J = 28.0 MeV.
Lithium-7 has a mass deficit of 0.0423 u (1 u ≡ 931.5 MeV). Binding energy per nucleon?
0.0423 × 931.5 = 39.4 MeV; ÷ 7 = 5.63 MeV.
D + T → ⁴He + n. B/A: ²H 1.11, ³H 2.83, ⁴He 7.07 MeV. Energy released?
4 × 7.07 − (2 × 1.11 + 3 × 2.83) = 28.28 − 10.71 = 17.6 MeV (the neutron has no binding energy).
Two protons must reach 4.0 × 10⁻¹⁵ m. Barrier energy, and T for mean KE = half of it?
E = ke²/r = 5.8 × 10⁻¹⁴ J; 3/2 kT = 2.9 × 10⁻¹⁴ J → T ≈ 1.4 × 10⁹ K.
A power station releases 1.0 × 10⁹ J each second. Mass converted each second?
Δm = ΔE/c² = 1.0 × 10⁹ ÷ 9.00 × 10¹⁶ = 1.1 × 10⁻⁸ kg.
²³⁵U + n → ¹⁴⁰Xe + ⁹⁴Sr + 2n loses 0.1983 u. Energy released (1 u ≡ 931.5 MeV)?
0.1983 × 931.5 = 185 MeV (2.96 × 10⁻¹¹ J).
Rounding masses before finding a mass deficit
Don't. The deficit is a tiny difference of large numbers; keep every figure until after the subtraction.
Using ΔE = c²Δm with Δm in u
Gives a meaningless number. Convert u to kg first (× 1.66 × 10⁻²⁷), or use 931.5 MeV per u — not both.
Binding energy per nucleon: divide by what?
By the nucleon number A, not the proton number Z.
Free neutrons in a reaction equation
They have zero binding energy, but they must be counted in the mass balance.
Checking a nuclear equation before calculating
Nucleon number and proton number must balance on both sides; count the free neutrons.
Head-on approach of two moving nuclei
At closest approach both have stopped: potential energy = total kinetic energy of the pair.

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