define and use the terms mass defect and binding energy

Mass Defect and Nuclear Binding Energy

Learning Objectives

• Define mass defect (Δm) and binding energy (E_b).
• Convert a mass defect into binding energy using E = Δm c² and the factor 1 u c² = 931.5 MeV.
• Calculate the binding energy per nucleon and interpret its meaning for nuclear stability.
• Read and write nuclear equations in the Cambridge format A_ZX → ….
• Calculate Q‑values for α‑decay, fission and fusion reactions.
• Link the concepts of mass defect and binding energy to radioactive decay, half‑life and activity.

Key Definitions

Mass defect (Δm) – the “missing” mass when separate nucleons bind to form a nucleus.

• Z = number of protons, N = number of neutrons (A = Z + N)
• m_p = 1.007276 u (mass of a proton)
• m_n = 1.008665 u (mass of a neutron)
• m_nucleus = experimentally measured nuclear mass (in atomic mass units, u)

Binding energy (E_b) – the energy required to separate a nucleus into its constituent protons and neutrons (or, equivalently, the energy released when the nucleus is formed).

In practice we use the convenient conversion

Step‑by‑Step Procedure for Δm and E_b

1. Identify Z, N and A for the nucleus.
2. Write the masses of a free proton and neutron (see above).
3. Obtain the experimental nuclear mass m_nucleus from a table of atomic masses.
4. Calculate the total mass of the separated nucleons:
   

   Total nucleon mass = Z m_p + N m_n

5. Find the mass defect:
   

   Δm = (Z m_p + N m_n) – m_nucleus

6. Convert Δm to binding energy:
   

   E_b = Δm × 931.5 MeV

7. Binding energy per nucleon (useful for comparing stability):
   

   E_b/A

Worked Examples

Example 1 – Helium‑4 (⁴He)

Example 2 – Carbon‑12 (¹²C)

Given m_nucleus(¹²C) = 12.000000 u, Z = 6, N = 6.

• Total nucleon mass = 6 m_p + 6 m_n = 6(1.007276) + 6(1.008665) = 12.095862 u
• Δm = 12.095862 – 12.000000 = 0.095862 u
• E_b = 0.095862 × 931.5 ≈ 89.3 MeV
• E_b/A = 89.3 ÷ 12 ≈ 7.44 MeV per nucleon

Example 3 – Fusion of Deuterium and Tritium

Reaction: ²₁H + ³₁H → ⁴₂He + ¹₀n + Q

Mass of reactants = 2.014102 + 3.016049 = 5.030151 u

Mass of products = 4.002603 + 1.008665 = 5.011268 u

Mass defect Δm = 5.030151 – 5.011268 = 0.018883 u

Q‑value = 0.018883 × 931.5 ≈ 17.6 MeV (the textbook value)

Binding‑Energy‑per‑Nucleon Curve

Key points required by the syllabus:

• The curve rises sharply for the lightest nuclei, reaches a maximum near iron‑56, then falls slowly for heavier nuclei.
• Because the binding energy per nucleon is lower for nuclei with A < 56, they can release energy by fusion (moving up the curve).
• For nuclei with A > 56 the binding energy per nucleon is lower than that of iron, so they can release energy by fission (moving down the curve toward the peak).
• The shape of the curve explains why both nuclear fusion (e.g. D + T) and nuclear fission (e.g. ²³⁸U) are energy‑producing processes.

Standard Cambridge Nuclear‑Equation Notation (A_ZX)

Every reaction must be written with the mass number (A) as a superscript and the atomic number (Z) as a subscript, followed by the chemical symbol.

• α‑decay of uranium‑238: ²³⁸₉₂U → ²³₄₉₀Th + ⁴₂He
• β⁻‑decay of carbon‑14: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̅_e
• Fusion example (D + T): ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV

Q‑Value Calculations (Energy Released)

The Q‑value of a nuclear reaction is the difference in total mass of reactants and products, converted to energy:

Example – α‑decay of ²³⁸U

Atomic masses:

• m(²³⁸U) = 238.050788 u
• m(²³₄Th) = 234.043601 u
• m(⁴He) = 4.002603 u

Mass of reactants = 238.050788 u

Mass of products = 234.043601 + 4.002603 = 238.046204 u

Δm = 238.050788 – 238.046204 = 0.004584 u

Q = 0.004584 × 931.5 ≈ 4.27 MeV

Connections to Other Nuclear Topics

• Radioactive decay: The Q‑value determines the kinetic energy of emitted particles (α, β, γ).
• Half‑life and activity: The rate of decay (activity) is independent of the Q‑value but the energy released per decay is given by the Q‑value.
• Energy calculations in reactors: Total power = (Q‑value × decay rate) or (Q‑value × fission rate).

Practice Questions

1. Calculate the mass defect and binding energy of ¹²C using the data in Example 2. Show every step.
2. For the α‑decay reaction ²³⁸₉₂U → ²³₄₉₀Th + ⁴₂He, compute the Q‑value using the masses given above.
3. Using the binding‑energy‑per‑nucleon curve, explain why both the fusion of deuterium‑tritium and the fission of ²³⁸U release energy.
4. A nucleus has a measured mass defect of 0.0189 u. Determine:

   • its total binding energy in MeV, and
   • its binding energy per nucleon if A = 20.

5. Write the following reactions in Cambridge notation and state the Q‑value (use the masses given):

   • β⁻‑decay of ¹⁴C (m(¹⁴₆C)=14.003242 u, m(¹⁴₇N)=14.003074 u, m(e⁻)=0.000549 u).
   • Fusion of two deuterium nuclei: ²₁H + ²₁H → ⁴₂He + Q.

Summary

• Mass defect = (mass of free nucleons) – (actual nuclear mass).
• Binding energy = Δm c²; use 1 u c² = 931.5 MeV for conversion.
• Binding energy per nucleon indicates how tightly each nucleon is held; the curve peaks at iron‑56, explaining why light nuclei fuse and heavy nuclei fission to release energy.
• All nuclear reactions must be written in the form A_ZX → … for full marks.
• Q‑values are obtained by the same mass‑defect → energy conversion and are essential for quantifying the energy released in decay, fission and fusion.