Mass-Energy Equivalence

Understand E = mc², mass defect, atomic mass units, and how to calculate binding energies and Q-values of nuclear reactions.

AQA A-Level Physics · Section 8 · Nuclear Physics

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Apply E = mc² to nuclear processes

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Calculate mass defect for a given nucleus

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Convert between kg, u, and MeV/c²

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Calculate binding energy from mass defect

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Determine Q-values of nuclear reactions

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Explain why nuclear masses are less than the sum of constituent masses

Einstein's Mass-Energy Equivalence

Einstein's famous equation relates mass and energy:

E = mc²

Where:

E = energy (J)
m = mass (kg)
c = speed of light = 3.00 × 10⁸ m/s

Mass and energy are equivalent — any change in mass corresponds to a change in energy and vice versa. In nuclear reactions, the "missing" mass appears as kinetic energy of the products.

For nuclear physics, it is convenient to express energies in electronvolts (eV) or megaelectronvolts (MeV):

1 eV = 1.60 × 10⁻¹⁹ J 1 MeV = 1.60 × 10⁻¹³ J

Atomic Mass Units (u)

The unified atomic mass unit (u) is defined as 1/12 of the mass of a carbon-12 atom:

1 u = 1.661 × 10⁻²⁷ kg

Using E = mc², the energy equivalent of 1 u is:

1 u = 931.5 MeV/c² → 1 u ≡ 931.5 MeV

Key masses in atomic mass units:

Particle	Mass / u	Mass / kg
Proton (p)	1.007276 u	1.673 × 10⁻²⁷ kg
Neutron (n)	1.008665 u	1.675 × 10⁻²⁷ kg
Electron (e)	0.000549 u	9.109 × 10⁻³¹ kg
Alpha particle	4.001506 u	6.644 × 10⁻²⁷ kg

Mass Defect and Binding Energy

The mass defect Δm is the difference between the mass of the separate nucleons and the actual mass of the nucleus:

Δm = Z × m_p + (A−Z) × m_n − m_nucleus

The binding energy E_B is the energy equivalent of the mass defect:

E_B = Δm × c²

Binding energy represents the energy needed to completely separate a nucleus into its individual nucleons. It is also the energy released when nucleons combine to form the nucleus.

A larger binding energy means a more stable nucleus. Binding energy per nucleon (E_B/A) is the key measure of nuclear stability.

In calculations, use nuclear masses (not atomic masses) OR consistently use atomic masses (including electron masses in both sides). The AQA data sheet gives atomic masses — be careful about electron masses when high precision is needed.

Q-Values of Nuclear Reactions

The Q-value of a nuclear reaction is the energy released (positive Q) or absorbed (negative Q):

Q = (m_reactants − m_products) × c²

Equivalently in mass units:

Q = Δm × 931.5 MeV/u

If Q > 0: the reaction is exothermic (releases energy — spontaneous)
If Q < 0: the reaction is endothermic (requires energy input — threshold energy needed)

For radioactive decay, Q equals the total kinetic energy of the decay products (shared between them according to momentum conservation).

Worked Strategy for Mass-Energy Calculations

Write the nuclear equation for the reaction
Look up atomic/nuclear masses from data tables (in u)
Calculate total mass of reactants and products
Find Δm = m_reactants − m_products (in u)
Convert to energy: E = Δm × 931.5 MeV (if Δm in u) or E = Δm × c² (if Δm in kg)
Check sign: positive Δm means energy released; negative means energy absorbed

It is almost always easiest to work in atomic mass units (u) and convert using 1 u = 931.5 MeV. This avoids very small numbers in kg.

Example 1: Mass defect and binding energy of helium-4

Calculate the binding energy of ⁴₂He. Masses: m_p = 1.007276 u, m_n = 1.008665 u, m(⁴He) = 4.001506 u.

1 Sum of separate nucleon masses: 2 × 1.007276 + 2 × 1.008665 = 2.014552 + 2.017330 = 4.031882 u

2 Mass defect: Δm = 4.031882 − 4.001506 = 0.030376 u

3 Binding energy: E_B = 0.030376 × 931.5 = 28.30 MeV

E_B = 28.3 MeV (E_B per nucleon = 28.3/4 = 7.07 MeV/nucleon)

Example 2: Q-value of alpha decay

Radium-226 decays: ²²⁶Ra → ²²²Rn + ⁴He. Masses: m(Ra-226) = 226.0254 u, m(Rn-222) = 222.0176 u, m(He-4) = 4.0026 u.

1 Mass of reactants: 226.0254 u

2 Mass of products: 222.0176 + 4.0026 = 226.0202 u

3 Δm = 226.0254 − 226.0202 = 0.0052 u

4 Q = 0.0052 × 931.5 = 4.84 MeV

Q = 4.84 MeV (released as kinetic energy of Rn and alpha particle)

Example 3: Energy from mass conversion

A nuclear reaction converts 1.5 × 10⁻²⁸ kg of mass to energy. Calculate the energy released in MeV.

1 E = mc² = 1.5 × 10⁻²⁸ × (3.00 × 10⁸)² = 1.5 × 10⁻²⁸ × 9.00 × 10¹⁶ = 1.35 × 10⁻¹¹ J

2 Convert to MeV: 1.35 × 10⁻¹¹ / (1.60 × 10⁻¹³) = 84.4 MeV

E = 84 MeV

Example 4: Converting mass defect between units

A nucleus has a mass defect of 0.0500 u. Express this as (a) energy in MeV, (b) energy in joules, (c) mass in kg.

1 (a) E = 0.0500 × 931.5 = 46.6 MeV

2 (b) E = 46.6 × 1.60 × 10⁻¹³ = 7.45 × 10⁻¹² J

3 (c) m = 0.0500 × 1.661 × 10⁻²⁷ = 8.31 × 10⁻²⁹ kg (or use m = E/c² = 7.45×10⁻¹²/(9×10¹⁶) = 8.28×10⁻²⁹ kg ✓)

(a) 46.6 MeV (b) 7.45 × 10⁻¹² J (c) 8.30 × 10⁻²⁹ kg

Q1. The mass defect of a nucleus is 0.010 u. What is the binding energy?

Q2. What is the energy equivalent of 1 atomic mass unit (u) in MeV?

Q3. A nuclear reaction has a positive Q-value. What does this mean?

Q4. Which particle has the smallest mass in atomic mass units?

Q5. A nucleus of carbon-12 has mass exactly 12.000000 u (by definition). Its 6 protons (1.007276 u each) and 6 neutrons (1.008665 u each) have total mass 12.095646 u. What is the mass defect?

Challenge 1. Calculate the binding energy per nucleon for iron-56. Masses: m_p = 1.007276 u, m_n = 1.008665 u, m(Fe-56) = 55.934939 u. Iron has Z = 26.

Challenge 2. In a beta-minus decay: n → p + e⁻ + ν̄_e. Use masses m_n = 1.008665 u, m_p = 1.007276 u, m_e = 0.000549 u. Calculate the Q-value in MeV. Why can this decay happen spontaneously for a free neutron?

Challenge 3. A fusion reaction combines two deuterium nuclei: ²₁H + ²₁H → ³₂He + ¹₀n. Masses: m(²H) = 2.014102 u, m(³He) = 3.016029 u, m_n = 1.008665 u. Calculate (a) the Q-value in MeV, and (b) the energy released per kilogram of deuterium fuel.