A Level Physics: Nuclear Physics — Revision Notes

Condensed for the final weeks. For the full explanation, use the Nuclear Physics study guide.

Mass defect and binding energy

mass defect       delta-m = (mass of separate nucleons) - (mass of nucleus)
binding energy    E = delta-m c^2

Binding energy is the energy required to separate a nucleus into its constituent nucleons — equivalently, the energy released when it forms.

Binding energy per nucleon is the measure of stability. The curve peaks near iron-56, which is why:

LIGHT nuclei  ->  FUSION   moves up the curve  ->  energy released
HEAVY nuclei  ->  FISSION  moves up the curve  ->  energy released

Both processes increase binding energy per nucleon. That single idea explains the whole shape of the curve.

Unit conversion: 1 u = 931.5 MeV; 1 eV = 1.60 × 10⁻¹⁹ J. This 931.5 MeV/u figure is a useful memorised shortcut, but it is not itself printed on the data sheet. The official/primary route is to convert via the constants the data sheet actually supplies — unified atomic mass constant u in kg, the speed of light c, and the elementary charge e — using E = Δm c² (in J) then dividing by e to convert to eV. Because 931.5 MeV/u is itself a rounded figure, working from the data-sheet constants directly can give a final answer that differs from the 931.5 MeV/u shortcut in the third significant figure; either method is acceptable, but know that a slight discrepancy can arise.

Worked example. A helium-4 nucleus has mass 4.00150 u; a proton is 1.00728 u, a neutron 1.00867 u.

mass of separate nucleons = 2(1.00728) + 2(1.00867) = 4.03190 u
mass defect  dm = 4.03190 - 4.00150 = 0.03040 u
binding energy  E = 0.03040 x 931.5 = 28.3 MeV
binding energy per nucleon = 28.3 / 4 = 7.08 MeV

Radioactive decay

Decay is random (you cannot predict which nucleus decays next) and spontaneous (unaffected by temperature, pressure or chemical state).

activity          A = lambda N          becquerels (Bq)
decay             N = N0 e^(-lambda t)
                  A = A0 e^(-lambda t)
half-life         t_half = ln 2 / lambda  =  0.693 / lambda

λ is the decay constant — the probability per unit time that a given nucleus decays.

To find λ graphically, plot ln A against t: the gradient is −λ.

Worked example. A source has half-life 8.0 days and initial activity 4.8 × 10⁵ Bq. Find the activity after 20 days, and the initial number of undecayed nuclei.

lambda = ln2 / t_half = 0.693 / 8.0 = 0.0866 day^-1
A = A0 e^(-lambda t) = 4.8x10^5 x e^(-0.0866x20) = 4.8x10^5 x 0.1769 = 8.49x10^4 Bq

Converting lambda to s^-1: 0.0866 / 86400 = 1.002x10^-6 s^-1
N = A / lambda = 4.8x10^5 / 1.002x10^-6 = 4.79x10^11 nuclei

Always convert λ to the same time unit as the answer requires — a day⁻¹ value must become s⁻¹ before it is combined with an activity in Bq (which is s⁻¹ by definition).

The three radiations

	Alpha	Beta	Gamma
Nature	Helium nucleus	Fast electron/positron	EM photon
Charge	+2e	∓e	0
Penetration	Paper	~3 mm aluminium	Thick lead
Range in air	Few cm	~1 m	No definite range — intensity falls as 1/d²
Ionising power	Strongest	Moderate	Weakest
Deflection in a field	Slight, one way	Large, opposite way	None

Ionising power and penetrating power are inversely related — alpha ionises strongly, so it loses energy fast and stops quickly.

Decay equations

alpha:   A -> (A-4) and Z -> (Z-2)
beta-:   A unchanged, Z -> (Z+1)     (n -> p + e- + antineutrino)
beta+:   A unchanged, Z -> (Z-1)
gamma:   no change to A or Z

Both nucleon number and proton number must balance on each side.

Exam traps

• Binding energy is the energy to separate nucleons, not the energy holding them “stored”.
• Mass defect: separate nucleons are heavier than the bound nucleus.
• Both fission and fusion release energy — by moving towards iron on the curve.
• λ and half-life are inversely related; a long half-life means a small λ.
• Activity requires the number of undecayed nuclei, not the original number.
• Background radiation must be subtracted before analysing experimental counts.
• Mixing time units for λ — a half-life in days gives λ in day⁻¹, which must be converted to s⁻¹ before combining with an activity in Bq.
• Forgetting to divide total binding energy by the number of nucleons, not just reporting the total.

Self-test

1. Define binding energy per nucleon and say why it matters.
2. Why do both fission and fusion release energy?
3. A sample has λ = 0.023 s⁻¹. Find its half-life.
4. Which radiation is most ionising, and why does that make it least penetrating?
5. Write the changes to A and Z for beta-minus decay.
6. A helium-4 nucleus has mass 4.00150 u (proton 1.00728 u, neutron 1.00867 u, 1 u = 931.5 MeV). Find its binding energy per nucleon.
7. A source has half-life 8.0 days and initial activity 4.8 × 10⁵ Bq. Find its activity after 20 days.

Answers: 1. The total binding energy of the nucleus divided by the number of nucleons; the higher it is, the more stable the nucleus. 2. Both move the products towards the peak of the binding-energy-per-nucleon curve near iron-56, so binding energy per nucleon increases and the surplus is released. 3. t½ = 0.693/0.023 = 30 s. 4. Alpha — its large charge and mass mean it interacts strongly with matter, losing energy rapidly over a short distance, so it is stopped by paper. 5. A is unchanged; Z increases by 1. 6. Mass of nucleons = 2(1.00728) + 2(1.00867) = 4.03190 u; Δm = 4.03190 − 4.00150 = 0.03040 u; E = 0.03040 × 931.5 = 28.3 MeV; per nucleon = 28.3 ÷ 4 = 7.08 MeV. 7. λ = 0.693 ÷ 8.0 = 0.0866 day⁻¹; A = 4.8 × 10⁵ × e^(−0.0866×20) = 8.49 × 10⁴ Bq.