calculate the energy released in nuclear reactions using E = c2∆m
23 Nuclear Physics
Learning Objectives
Calculate the energy released (or absorbed) in nuclear reactions using E = c^{2}\,\Delta m (the Q‑value).
Define and compute mass defect, binding energy, and binding‑energy per nucleon.
Interpret the binding‑energy curve and explain quantitatively why 56Fe is the most stable nucleus.
Describe and balance the four main types of radioactive decay (α, β⁻, β⁺, γ) and electron capture.
Identify the fundamental particles (quarks, leptons, gauge bosons) that make up nucleons and mediate nuclear processes.
Distinguish between fusion, fission and other nuclear reactions; calculate the Q‑value of any reaction.
Apply nuclear physics to real‑world contexts (medical imaging/therapy, power generation, dating techniques).
Plan and analyse a simple mass‑defect experiment (e.g. using a mass spectrometer or high‑precision balance).
23.1 Mass Defect and Nuclear Binding Energy
Key Definitions
Atomic mass unit (u): 1 u = 1.660 539 × 10⁻²⁷ kg.
Mass defect (Δm):
\[
\Delta m \;=\; \bigl(\sum m{\text{free nucleons}}\bigr) \;-\; m{\text{bound nucleus}}
\]
(the “missing” mass when nucleons bind together).
Binding energy (E_b):
\[
E_{b}= \Delta m\,c^{2}
\]
The energy equivalent of the mass defect.
Binding‑energy per nucleon:
\[
\frac{E_{b}}{A}\quad\text{(where }A\text{ is the mass number)}
\]
A direct measure of nuclear stability.
Why a Mass Defect Occurs
When protons and neutrons combine to form a nucleus, part of their total rest mass is converted into the energy that holds the nucleons together (the strong nuclear force). The “missing” mass is the mass defect, and its energy equivalent is the binding energy.
Step‑by‑Step Calculation of Δm and Energy Released
Write the balanced nuclear reaction and list all reactants and products.
Obtain atomic masses (in u) from a reliable table.
Calculate the total mass of reactants (Mreact) and of products (Mprod).
Find the mass defect:
\[
\Delta m = M{\text{react}} - M{\text{prod}}
\]
(Δm > 0 ⇒ energy released; Δm < 0 ⇒ energy must be supplied.)
Mprod = 140.914411 + 91.926156 + 3(1.008665) = 235.966562 u
Δm = 0.086033 u
Q = 0.086033 u × 931.5 MeV ≈ 80.1 MeV (≈ 200 MeV when kinetic energy of neutrons is included; the exact value depends on the fragment masses used).
Binding‑Energy Curve and the Stability of 56Fe
The curve of binding energy per nucleon versus mass number has a pronounced maximum at 56Fe (≈ 8.8 MeV per nucleon). This means:
Any nucleus with A < 56 can increase its binding energy per nucleon by fusing with another light nucleus – the excess energy is released (fusion).
Any nucleus with A > 56 can increase its binding energy per nucleon by splitting into lighter fragments – the excess energy is released (fission).
Quantitative illustration for 56Fe:
Atomic mass of 56Fe ≈ 55.934937 u.
Mass of 26 free protons + 30 free neutrons:
\[
26(1.007825)+30(1.008665)=55.934965\;\text{u}
\]
Δm = 55.934965 − 55.934937 = 2.8 × 10⁻⁵ u.
Binding energy = 2.8 × 10⁻⁵ u × 931.5 MeV ≈ 26.1 MeV.
Binding‑energy per nucleon = 26.1 MeV ÷ 56 ≈ 8.8 MeV, the highest value on the curve.
Because 56Fe has the largest E_b/A, it is the most tightly bound and therefore the most stable nucleus. No nuclear reaction can increase its binding energy per nucleon, so it does not release energy by either fusion or fission.
Typical binding‑energy curve. The peak at iron‑56 explains why fusion powers stars (A < 56) and fission powers reactors (A > 56).
23.2 Radioactive Decay
Types of Decay
Decay
Particle(s) emitted
Change in (Z, A)
Typical Q‑value (MeV)
Example
α‑decay
⁴He nucleus (α‑particle)
(Z − 2, A − 4)
4 – 9
²³⁸U → ²³⁴Th + α
β⁻‑decay
Electron + antineutrino (e⁻ + \(\bar\nu_e\))
(Z + 1, A) (neutron → proton)
0.1 – 3
⁶⁰Co → ⁶⁰Ni + e⁻ + \(\bar\nu_e\)
β⁺‑decay (positron emission)
Positron + neutrino (e⁺ + νₑ)
(Z − 1, A) (proton → neutron)
1.022 + 0.1 – 3
¹⁴O → ¹⁴N + e⁺ + νₑ
Electron capture
Neutrino (νₑ)
(Z − 1, A) (proton + e⁻ → neutron)
≈ 0.1 – 3
⁵⁶Fe + e⁻ → ⁵⁶Mn + νₑ
γ‑decay
High‑energy photon (γ)
No change in (Z, A)
0.01 – 10
⁶⁰Co* → ⁶⁰Co + γ
Balancing a Decay Equation
Write the parent nucleus.
Identify the decay mode and the emitted particle(s).
Adjust Z and A on the right‑hand side so that both charge and nucleon number are conserved.
If required, calculate the Q‑value using the mass‑defect method (see Section 23.4).
Example – α‑Decay of 238U
Reaction: ⁽²³⁸⁾U → ⁽²³⁴⁾Th + ⁴He
Δm = (238.050788 u + 1.008665 u) − (234.043594 u + 4.002603 u) = 0.013156 u
Q = 0.013156 u × 931.5 MeV ≈ 12.3 MeV
23.3 Fundamental Particles Relevant to Nuclear Physics
Quarks
Six flavours: up (u), down (d), charm (c), strange (s), top (t), bottom (b).
Proton = uud, Neutron = udd.
Charges: +2/3 e (up‑type) or –1/3 e (down‑type); also carry colour charge (strong interaction).
Leptons
Charged: electron (e⁻), muon (μ⁻), tau (τ⁻).
Neutral (neutrinos): νₑ, νμ, ντ – very small masses, interact only via the weak force.
Gauge (Force‑Carrier) Bosons
Photon (γ) – electromagnetic interaction.
W⁺, W⁻, Z⁰ – weak interaction (responsible for β‑decay and electron capture).
Graviton – hypothetical; not required for the Cambridge syllabus.
Relevance to Nuclear Processes
The strong force (mediated by gluons) provides the binding energy that holds nucleons together. The weak force (mediated by W and Z bosons) governs β‑decay and electron capture. Understanding which particles are involved explains why certain reactions occur and why others are forbidden.
23.4 Nuclear Reactions, Q‑value and Energy Release
General Form of a Nuclear Reaction
A + a → B + b (where capital letters denote nuclei and lower‑case letters denote emitted particles such as neutrons, protons, α‑particles, etc.).
Reaction: ²H + ³He → ⁴He + ¹H + γ (often written as D + ³He → ⁴He + p + γ)
Mreact = 2.014102 u + 3.016029 u = 5.030131 u
Mprod = 4.002603 u + 1.007825 u = 5.010428 u
Δm = 5.030131 − 5.010428 = 0.019703 u
Q = 0.019703 u × 931.5 MeV ≈ 18.4 MeV
Fission Example – Simplified 235U Reaction
Reaction: ²³⁵U + ¹n → ⁹⁴Kr + ¹⁴¹Ba + 3 ¹n
Mreact = 235.043930 u + 1.008665 u = 236.052595 u
Mprod = 93.934140 u + 140.914411 u + 3(1.008665 u) = 235.966562 u
Δm = 0.086033 u
Q = 0.086033 u × 931.5 MeV ≈ 80.2 MeV (≈ 200 MeV when the kinetic energy of the fast neutrons is added).
Key Points for the Exam
Always use atomic masses (including electron masses) – the electrons cancel automatically in the mass‑defect calculation.
Remember the 931.5 MeV u⁻¹ conversion factor; it saves time.
Check the sign of Q to decide whether the reaction is energetically allowed.
When a problem asks for “energy released per kilogram of fuel”, first find Q per nucleus, then multiply by Avogadro’s number and the appropriate molar mass.
23.5 Applications of Nuclear Physics
Medical imaging and therapy: PET scans (β⁺ decay), radiotherapy with γ‑emitters, neutron capture therapy.
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