Apply Einstein's mass-energy equivalence $E = mc^2$ (rest energy) and the relativistic energy $E = \gamma m c^2$ (total energy) to nuclear reactions, particle physics and astrophysics

What this dot point is asking

QCAA wants you to apply Einstein's mass-energy equivalence ($E = mc^2$) in nuclear and particle-physics contexts: nuclear reactions, fission, fusion, and the rest energies of particles. Cross-link: see the mass-energy calculator.

Rest energy

Einstein's special relativity predicts that even an object at rest has an energy associated with its mass:

This is the rest energy. For ordinary matter, $E_0$ is enormous. The rest energy of 1 kg of any substance is $9 \times 10^{16}$ J, equivalent to the energy of about 21 megatonnes of TNT.

The rest energy is not (in classical contexts) accessible. To liberate it would require converting all the matter to other forms of energy (e.g. radiation). This happens in matter-antimatter annihilation (100 percent conversion) but is otherwise rare.

Standard particle rest energies

These values are used throughout particle and nuclear physics. The proton and neutron rest energies are nearly equal but the neutron is slightly heavier (its instability to beta decay is consistent with this).

Total relativistic energy

The total relativistic energy of a particle in motion is:

This includes rest energy plus kinetic energy. For non-relativistic speeds, $E \approx m c^2 + \frac{1}{2} m v^2$ (rest plus classical kinetic energy). For relativistic speeds, the kinetic energy is $E - m c^2 = (\gamma - 1) m c^2$, which differs from $\frac{1}{2} m v^2$ at high speeds.

Behaviour at high speeds

As $v \to c$, $\gamma \to \infty$ and $E \to \infty$. Infinite energy would be required to accelerate a particle with rest mass to the speed of light; no such particle ever reaches $c$.

Photons (rest mass zero, $m = 0$) travel at $c$. Their energy is $E = hf$ (photon model). Their momentum is $p = E/c = hf/c = h/\lambda$.

The energy-momentum relation

The full relativistic energy-momentum relation:

This holds for any object, with or without rest mass:

• Rest mass = 0 (photons): $E = pc$.
• Particle at rest ($p = 0$): $E = m c^2$, the rest energy.
• Non-relativistic ($pc \ll mc^2$): $E \approx mc^2 + p^2 / (2m)$, recovering classical kinetic energy.

Mass-energy equivalence in nuclear reactions

When nuclei react (fission, fusion), the products are typically lighter than the reactants. The mass difference $\Delta m$ is converted to energy by $\Delta E = \Delta m \cdot c^2$.

Fission example. $^{235}\text{U} + \text{n} \to ^{141}\text{Ba} + ^{92}\text{Kr} + 3\text{n}$. Mass deficit approximately 0.2 atomic mass units (amu). Energy released approximately 200 MeV per fission event. Fission powers nuclear reactors and the atomic bomb.

Fusion example. $^2\text{H} + ^3\text{H} \to ^4\text{He} + \text{n}$. Mass deficit approximately 0.0188 amu. Energy released approximately 17.6 MeV per fusion event. Fusion powers the sun and the hydrogen bomb.

Atomic mass unit (amu) conversion

$1$ amu $= 1.661 \times 10^{-27}$ kg. The rest energy of 1 amu is:

$E = m c^2 = 1.661 \times 10^{-27} \times (3 \times 10^8)^2 = 1.49 \times 10^{-10}$ J $\approx 931.5$ MeV.

So 1 amu $\equiv 931.5$ MeV (a useful conversion for nuclear physics).

A mass defect of 0.0188 amu in the DT fusion reaction corresponds to $0.0188 \times 931.5 = 17.5$ MeV, agreeing with the direct calculation above.

Pair production and annihilation

Pair production. A high-energy photon (energy at least $2 m_e c^2 = 1.022$ MeV) can convert into an electron-positron pair in the presence of a nucleus (which carries away momentum to conserve momentum). Total mass appears from total energy.

Annihilation. An electron and a positron annihilate to two photons (back-to-back, each with energy $0.511$ MeV in the centre-of-mass frame). All rest mass is converted to electromagnetic energy. This is the basis of PET (positron emission tomography) medical imaging.

Both processes are direct demonstrations of mass-energy equivalence.

Astrophysics

Solar luminosity

The sun converts about $4 \times 10^9$ kg of mass to energy per second through hydrogen fusion. Total solar luminosity: $4 \times 10^{26}$ W. The sun has about $2 \times 10^{30}$ kg of mass and will continue fusing hydrogen for another 5 billion years.

Supernovae

Stellar core collapse converts substantial mass to energy, producing the most luminous events in the universe. Type Ia supernovae are used as "standard candles" because their peak luminosity is calibrated.

Black hole accretion

Matter falling into a black hole can convert up to about 40 percent of its rest energy to radiation, the most efficient natural energy source known.

Worked example. Electron-positron annihilation

An electron and positron at rest annihilate. Calculate the total energy of the resulting two photons and their wavelengths.

Total rest energy: $2 \times 0.511$ MeV $= 1.022$ MeV $= 1.022 \times 10^6 \times 1.6 \times 10^{-19}$ J $= 1.635 \times 10^{-13}$ J.

Two photons, each with energy $0.511$ MeV $= 8.17 \times 10^{-14}$ J.

Wavelength: $\lambda = h c / E = (6.626 \times 10^{-34} \times 3 \times 10^8) / 8.17 \times 10^{-14} = 2.43 \times 10^{-12}$ m $= 2.43$ pm.

This wavelength (2.43 pm) is the Compton wavelength of the electron, characteristic of annihilation gamma rays.

Examples in context

Example 1. Nuclear fusion in solar plasma converts $4.0 \times 10^9 \text{ kg s}^{-1}$ of mass into energy via $E = mc^2 = 4.0 \times 10^9 \times (3 \times 10^8)^2 = 3.6 \times 10^{26} \text{ W}$. A Sunshine Coast solar-panel array intercepting $\sim 1 \text{ kW m}^{-2}$ ultimately draws on this $E = mc^2$ accounting. QCAA EA Unit 4 thematic items often link astrophysical and terrestrial.

Example 2. A $^{235}\text{U}$ fission releases $\Delta m = 0.215 \text{ u} = 3.57 \times 10^{-28} \text{ kg}$, giving $E = \Delta m c^2 = 3.21 \times 10^{-11} \text{ J} = 200 \text{ MeV}$. ANSTO research at OPAL applies precisely this mass-energy relation to compute fission product yields, the same arithmetic QCAA students perform in Unit 4 EA Paper 2.

Try this

Q1. State Einstein's mass-energy equivalence. [2 marks]

• Cue. $E = mc^2$ (rest energy); $E = \gamma m c^2$ (total relativistic energy).

Q2. A mass defect of $0.030 \text{ u}$ accompanies a fusion reaction. Calculate the energy released in MeV. [3 marks]

• Cue. $0.030 \times 931.5 = 27.9 \text{ MeV}$.

Q3. A $^{235}\text{U}$ fission has $\Delta m = 0.215 \text{ u}$. (a) Calculate the energy released in J and MeV. (b) Calculate the energy per kilogram of $^{235}\text{U}$ ($m_{nucleus} = 235.04 \text{ u}$). (c) Compare with the energy per kilogram of solar fusion ($\Delta m / m \approx 0.0072$). [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $3.21 \times 10^{-11} \text{ J} = 200 \text{ MeV}$; (b) $8.2 \times 10^{13} \text{ J kg}^{-1}$; (c) fusion $\approx 6.5 \times 10^{14} \text{ J kg}^{-1}$, factor $\sim 8$ higher.