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).

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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

1. Write the balanced nuclear reaction and list all reactants and products.
2. Obtain atomic masses (in u) from a reliable table.
3. Calculate the total mass of reactants (M_react) and of products (M_prod).
4. Find the mass defect: \[ \Delta m = M_{\text{react}} - M_{\text{prod}} \] (Δm > 0 ⇒ energy released; Δm < 0 ⇒ energy must be supplied.)
5. Convert Δm to kilograms (optional): \[ \Delta m\;(\text{kg}) = \Delta m\;(\text{u}) \times 1.660\,539 \times 10^{-27}\;\text{kg u}^{-1} \] Most Cambridge questions use the conversion factor 1 u c² = 931.5 MeV, so step 5 can be performed directly in MeV.
6. Calculate the released energy (the Q‑value): \[ Q = \Delta m\,c^{2} = \Delta m\;(\text{u}) \times 931.5\;\text{MeV} \] or, if Δm was converted to kg, use \(E = c^{2}\Delta m\) and then convert J → MeV (1 MeV = 1.602 18 × 10⁻¹³ J).
7. State the result in joules and/or mega‑electronvolts (MeV) as required.

Sample Atomic‑Mass Table (excerpt)

Nucleus	Atomic mass (u)	Z (protons)	N (neutrons)
¹H (proton)	1.007825	1	0
¹n (neutron)	1.008665	0	1
²H (deuterium)	2.014102	1	1
³He	3.016029	2	1
⁴He (α‑particle)	4.002603	2	2
²³⁵U	235.043930	92	143
¹⁴¹Ba	140.914411	56	85
⁹²Kr	91.926156	36	56
²³⁹Pu	239.052163	94	145

Worked Example 1 – Formation of ⁴He (Helium‑4)

Reaction: 2 ¹H + 2 ¹n → ⁴He

1. M_react = 2(1.007825) + 2(1.008665) = 4.03298 u
2. M_prod = 4.002603 u
3. Δm = 4.03298 − 4.002603 = 0.030377 u
4. Q = 0.030377 u × 931.5 MeV u⁻¹ ≈ 28.3 MeV
5. Binding energy per nucleon = 28.3 MeV ÷ 4 ≈ 7.1 MeV

Worked Example 2 – Fission of ²³⁵U

Typical fission reaction: ²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3 ¹n

1. M_react = 235.043930 + 1.008665 = 236.052595 u
2. M_prod = 140.914411 + 91.926156 + 3(1.008665) = 235.966562 u
3. Δm = 0.086033 u
4. 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:

1. Atomic mass of ⁵⁶Fe ≈ 55.934937 u.
2. Mass of 26 free protons + 30 free neutrons: \[ 26(1.007825)+30(1.008665)=55.934965\;\text{u} \]
3. Δm = 55.934965 − 55.934937 = 2.8 × 10⁻⁵ u.
4. Binding energy = 2.8 × 10⁻⁵ u × 931.5 MeV ≈ 26.1 MeV.
5. 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.

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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⁻ + \(\baru_e\))	(Z + 1, A) (neutron → proton)	0.1 – 3	⁶⁰Co → ⁶⁰Ni + e⁻ + \(\baru_e\)
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

1. Write the parent nucleus.
2. Identify the decay mode and the emitted particle(s).
3. Adjust Z and A on the right‑hand side so that both charge and nucleon number are conserved.
4. 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

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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, ν_μ, ν_τ – 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).
• Gluons (g) – strong interaction (binds quarks inside nucleons and nucleons inside nuclei).
• 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.

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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.).

Q‑value (Energy Released or Absorbed)

\[ Q = \bigl(\sum m_{\text{initial}} - \sum m_{\text{final}}\bigr)c^{2} = \Delta m\,c^{2} \]

• Q > 0 → exothermic (energy released, reaction can occur spontaneously).
• Q < 0 → endothermic (energy must be supplied, e.g. in many fusion reactions).

Conversion Shortcut

Because 1 u c² = 931.5 MeV, the Q‑value can be obtained directly from the mass defect in atomic mass units:

\[ Q\;(\text{MeV}) = \Delta m\;(\text{u}) \times 931.5\;\text{MeV u}^{-1} \]

Fusion Example – Deuterium–Helium‑3 Reaction

Reaction: ²H + ³He → ⁴He + ¹H + γ (often written as D + ³He → ⁴He + p + γ)

1. M_react = 2.014102 u + 3.016029 u = 5.030131 u
2. M_prod = 4.002603 u + 1.007825 u = 5.010428 u
3. Δm = 5.030131 − 5.010428 = 0.019703 u
4. Q = 0.019703 u × 931.5 MeV ≈ 18.4 MeV

Fission Example – Simplified ²³⁵U Reaction

Reaction: ²³⁵U + ¹n → ⁹⁴Kr + ¹⁴¹Ba + 3 ¹n

1. M_react = 235.043930 u + 1.008665 u = 236.052595 u
2. M_prod = 93.934140 u + 140.914411 u + 3(1.008665 u) = 235.966562 u
3. Δm = 0.086033 u
4. 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.

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23.5 Applications of Nuclear Physics

• Medical imaging and therapy: PET scans (β⁺ decay), radiotherapy with γ‑emitters, neutron capture therapy.
• Power generation: Fission reactors (U‑235, Pu‑239), prospective fusion reactors (D‑T, D‑He³).
• Dating techniques: Radiocarbon dating (¹⁴C β⁻ decay), uranium–lead dating (U‑238 → Pb‑206).
• Industrial uses: Neutron activation analysis, gamma radiography.

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Quick Reference Table

Quantity	Symbol	Value	Units
Atomic mass unit	1 u	1.660 539 × 10⁻²⁷	kg
Speed of light	c	2.998 × 10⁸	m s⁻¹
Energy–mass conversion	1 u c²	931.5	MeV
1 MeV in joules		1.602 18 × 10⁻¹³	J