Mass-Energy Equivalence

Spec mapping: OCR H556 Module 6.4 — Nuclear and Particle Physics (Einstein's relation $E = mc^{2}$E=mc2; rest energy; the atomic mass unit and the MeV/$c^{2}$c2; mass-energy conservation in nuclear processes; the reaction Q-value $Q = \Delta m \, c^{2}$Q=Δmc2; conversion factor $1\,\text{u} = 931.5\,\text{MeV}/c^{2}$1u=931.5MeV/c2). Refer to the official OCR H556 specification document for exact wording.

Up to now in this course, mass and energy have been separate, conserved quantities. In mechanics, you conserved kinetic energy (in elastic collisions) and momentum; in thermodynamics, you conserved internal energy. Mass was a fixed property of each object. The great insight of special relativity was that mass and energy are not independent at all — they are two different expressions of the same underlying quantity, related by what is probably the most famous equation in all of physics:

$E = mc^{2}.$E=mc2.

This equation underpins the whole of nuclear and particle physics, from the fission reactor to the PET scanner. In this lesson we unpack what it means, introduce the atomic mass unit and the MeV/$c^{2}$c2, and apply it to simple nuclear processes. The material is the core of Module 6.4 of the OCR A-Level Physics A specification (H556).

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The Meaning of E = mc²

Einstein's equation says that any object with mass $m$m has an associated rest energy:

$E_{0} = m_{0}\, c^{2}$E0​=m0​c2

where $m_{0}$m0​ is the rest mass (the mass of the object in its own rest frame) and $c = 3.00 \times 10^{8}\,\text{m s}^{-1}$c=3.00×108m s−1 is the speed of light in vacuum. The prefactor $c^{2}$c2 is enormous — $9 \times 10^{16}\,\text{m}^{2}\,\text{s}^{-2}$9×1016m2s−2 — so even a small amount of mass corresponds to a huge amount of energy. A single gram of matter has a rest energy of:

$E = (10^{-3})\,(3 \times 10^{8})^{2} = 9 \times 10^{13}\,\text{J},$E=(10−3)(3×108)2=9×1013J,

which is the energy released by burning about two thousand tonnes of petrol. This is why nuclear processes are so energetic: they directly convert small amounts of mass into large amounts of energy.

The equation has a subtler meaning too. If an object absorbs energy $\Delta E$ΔE (becomes hotter, compresses a spring, gains kinetic energy), its mass increases by $\Delta m = \Delta E/c^{2}$Δm=ΔE/c2. Conversely, if it loses energy, its mass decreases. This effect is undetectably small in everyday processes (heating a cup of coffee from 20 °C to 60 °C increases its mass by about $2 \times 10^{-12}\,\text{g}$2×10−12g), but in nuclear processes the mass changes are a measurable fraction of the rest masses involved — easily resolved on a mass spectrometer.

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The Atomic Mass Unit

Working with kilograms is inconvenient at the nuclear scale, where masses are of order $10^{-27}\,\text{kg}$10−27kg. Physicists use instead the atomic mass unit (u), defined so that a carbon-12 atom has a mass of exactly 12 u:

$1\,\text{u} = 1.661 \times 10^{-27}\,\text{kg}.$1u=1.661×10−27kg.

On this scale:

• Proton mass: $m_{p} \approx 1.00728\,\text{u}$mp​≈1.00728u
• Neutron mass: $m_{n} \approx 1.00867\,\text{u}$mn​≈1.00867u
• Electron mass: $m_{e} \approx 5.49 \times 10^{-4}\,\text{u}$me​≈5.49×10−4u
• Alpha particle: $m_{\alpha} \approx 4.00150\,\text{u}$mα​≈4.00150u

The atomic mass unit is also called the dalton (Da) in chemistry, but OCR uses "u". You will see both in the literature.

A subtle distinction: "atomic mass" and "nuclear mass" are different. The atomic mass of a neutral atom includes its orbiting electrons; the nuclear mass does not. For most A-Level calculations using atomic masses, the electron bookkeeping cancels (because the same number of electrons appears on both sides of the equation), but for $\beta^{+}$β+ decay you have to be careful — we shall flag it when the case arises.

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MeV and MeV/$c^{2}$c2

Multiplying the atomic mass unit by $c^{2}$c2 gives its rest energy. Converting to MeV (since $1\,\text{MeV} = 1.60 \times 10^{-13}\,\text{J}$1MeV=1.60×10−13J):

$$\begin{aligned} 1\,\text{u} \times c^{2} &= (1.661 \times 10^{-27})(3.00 \times 10^{8})^{2}\,\text{J} \\ &= 1.495 \times 10^{-10}\,\text{J} \\ &= 931.5\,\text{MeV}. \end{aligned}$$1u×c2​=(1.661×10−27)(3.00×108)2J=1.495×10−10J=931.5MeV.​

So:

$\boxed{\; 1\,\text{u} = 931.5\,\text{MeV}/c^{2} \;}$1u=931.5MeV/c2​

This conversion factor is one of the most useful numbers in nuclear physics. Whenever you see a mass in atomic mass units and need the corresponding rest energy in MeV, just multiply by 931.5.

Particle physicists commonly use "MeV/$c^{2}$c2" (or "GeV/$c^{2}$c2") as a unit of mass directly, since it saves the conversion step. The rest energies of common particles are:

Particle	Rest energy (MeV)	Mass (u)
Electron	$0.511$0.511	$5.49 \times 10^{-4}$5.49×10−4
Proton	$938.3$938.3	$1.00728$1.00728
Neutron	$939.6$939.6	$1.00867$1.00867
Alpha	$3727$3727	$4.00150$4.00150
Pion $\pi^{+}$π+	$139.6$139.6	$0.1499$0.1499
Muon $\mu^{-}$μ−	$105.7$105.7	$0.1135$0.1135

Note that the neutron is slightly heavier than the proton — by about $1.3\,\text{MeV}/c^{2}$1.3MeV/c2 — which is why a free neutron decays via $\beta^{-}$β− to a proton plus electron plus antineutrino. The energy difference appears partly as the rest energy of the emitted electron ($0.511\,\text{MeV}$0.511MeV) and partly as the kinetic energy and antineutrino energy of the decay products. Inside a nucleus, of course, neutrons can be perfectly stable (Pauli blocking) — see the previous lesson's "Going further".

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Energy Released in a Nuclear Process

In any nuclear process — decay, fission, fusion — the total relativistic energy is conserved. If the rest masses of the initial particles differ from those of the final particles, the mass difference is converted to (or from) kinetic energy of the products:

$\boxed{\; Q = \Big(\!\sum m_{\text{initial}} - \sum m_{\text{final}}\Big)\, c^{2} \;}$Q=(∑minitial​−∑mfinal​)c2​

where $Q$Q is called the Q-value of the reaction. A positive $Q$Q means energy is released (the process is exoergic); a negative $Q$Q means energy must be supplied (endoergic) and the reaction will not run spontaneously.

Worked Example 1 — Q-value of alpha decay

Polonium-210 alpha-decays to lead-206. The relevant atomic masses are $m({}^{210}\text{Po}) = 209.98286\,\text{u}$m(210Po)=209.98286u, $m({}^{206}\text{Pb}) = 205.97445\,\text{u}$m(206Pb)=205.97445u, and $m({}^{4}\text{He}) = 4.00260\,\text{u}$m(4He)=4.00260u. Compute the Q-value in MeV.

$\Delta m = 209.98286 - (205.97445 + 4.00260) = 0.00581\,\text{u},$Δm=209.98286−(205.97445+4.00260)=0.00581u,

$Q = \Delta m \times 931.5\,\text{MeV/u} \approx 5.41\,\text{MeV}.$Q=Δm×931.5MeV/u≈5.41MeV.

This 5.41 MeV appears mostly as kinetic energy of the emitted alpha particle (because the heavy daughter nucleus barely moves) and a tiny amount as recoil energy of the Pb-206 nucleus. The observed alpha energy of polonium-210 is indeed about $5.3\,\text{MeV}$5.3MeV — the small discrepancy is exactly the recoil energy of the lead daughter, which momentum conservation says must take a fraction $4/210 \approx 1.9\,\%$4/210≈1.9% of the kinetic energy.

Worked Example 2 — Q-value of beta-minus decay

Carbon-14 $\beta^{-}$β−-decays to nitrogen-14. Atomic masses $m({}^{14}\text{C}) = 14.00324\,\text{u}$m(14C)=14.00324u, $m({}^{14}\text{N}) = 14.00307\,\text{u}$m(14N)=14.00307u.

$\Delta m = 14.00324 - 14.00307 = 0.00017\,\text{u},$Δm=14.00324−14.00307=0.00017u,

$Q = 0.00017 \times 931.5 \approx 0.158\,\text{MeV} = 158\,\text{keV}.$Q=0.00017×931.5≈0.158MeV=158keV.

A subtle point: when using atomic masses for a $\beta^{-}$β− decay, you do not need to add the mass of the emitted electron separately — the atomic masses already include all the electrons, and the electron added to the daughter atom exactly cancels the $\beta^{-}$β− emitted from the parent nucleus.

This 158 keV is the maximum kinetic energy that the emitted beta particle can carry. In reality the spectrum is continuous from 0 to 158 keV, because the antineutrino shares the available energy in random proportions.

Worked Example 3 — Energy released in D-T fusion

The deuterium-tritium fusion reaction is $^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{4}_{2}\text{He} + {}^{1}_{0}n$12​H+13​H→24​He+01​n. Atomic masses $m({}^{2}\text{H}) = 2.01410\,\text{u}$m(2H)=2.01410u, $m({}^{3}\text{H}) = 3.01605\,\text{u}$m(3H)=3.01605u, $m({}^{4}\text{He}) = 4.00260\,\text{u}$m(4He)=4.00260u, $m(n) = 1.00867\,\text{u}$m(n)=1.00867u.

$\Delta m = (2.01410 + 3.01605) - (4.00260 + 1.00867) = 5.03015 - 5.01127 = 0.01888\,\text{u},$Δm=(2.01410+3.01605)−(4.00260+1.00867)=5.03015−5.01127=0.01888u,

$Q = 0.01888 \times 931.5 \approx 17.6\,\text{MeV}.$Q=0.01888×931.5≈17.6MeV.

This 17.6 MeV is released as kinetic energy, shared (by momentum conservation, since the neutron is much lighter than the alpha) as $14.1\,\text{MeV}$14.1MeV to the neutron and $3.5\,\text{MeV}$3.5MeV to the helium nucleus. It is the target reaction of experimental fusion reactors such as JET and ITER.

Worked Example 4 — Q-value in joules and the kg route

Express the 5.41 MeV Q-value of the Po-210 alpha decay in joules, and verify by computing $\Delta m \, c^{2}$Δmc2 in SI.

$Q = 5.41 \times 1.60 \times 10^{-13}\,\text{J} \approx 8.66 \times 10^{-13}\,\text{J}.$Q=5.41×1.60×10−13J≈8.66×10−13J.

Cross-check: $\Delta m = 0.00581 \times 1.661 \times 10^{-27}\,\text{kg} = 9.65 \times 10^{-30}\,\text{kg}$Δm=0.00581×1.661×10−27kg=9.65×10−30kg, so

$Q = (9.65 \times 10^{-30})(3.00 \times 10^{8})^{2} = 9.65 \times 10^{-30} \times 9.00 \times 10^{16} \approx 8.68 \times 10^{-13}\,\text{J}.$Q=(9.65×10−30)(3.00×108)2=9.65×10−30×9.00×1016≈8.68×10−13J.

The two routes agree to within the precision of the input data. The MeV-via-931.5 route is cleaner for nuclear-physics arithmetic; the SI route is the one to use if the question quotes masses in kg from the outset.

Worked Example 5 — Pair-production threshold

A high-energy gamma photon converts into an electron–positron pair near a heavy nucleus (the nucleus is required to soak up momentum). What minimum photon energy is needed?

The reaction is $\gamma \to e^{-} + e^{+}$γ→e−+e+. At threshold, both produced particles are essentially at rest in the centre-of-momentum frame, and the photon must supply both of their rest energies:

$E_{\gamma,\,\text{min}} = 2 m_{e} c^{2} = 2 \times 0.511\,\text{MeV} = 1.022\,\text{MeV}.$Eγ,min​=2me​c2=2×0.511MeV=1.022MeV.

This is a clean demonstration of $E = mc^{2}$E=mc2 running in reverse: electromagnetic energy disappears and rest mass appears in its place. Below this threshold, pair production is forbidden by energy conservation. Above it, the cross-section grows with photon energy. The same threshold is relevant in PET imaging (the inverse reaction $e^{-} + e^{+} \to 2\gamma$e−+e+→2γ produces two $0.511\,\text{MeV}$0.511MeV back-to-back photons), in cosmic-ray cascades, and in any astrophysical environment where $\gamma$γ-ray photons interact with matter.

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Energy–Mass Conversions to Remember

Quantity	Value
$1\,\text{u}$1u	$1.661 \times 10^{-27}\,\text{kg}$1.661×10−27kg
$1\,\text{u} \times c^{2}$1u×c2	$1.495 \times 10^{-10}\,\text{J} = 931.5\,\text{MeV}$1.495×10−10J=931.5MeV
Electron rest energy	$0.511\,\text{MeV}$0.511MeV
Proton rest energy	$938.3\,\text{MeV}$938.3MeV
Neutron rest energy	$939.6\,\text{MeV}$939.6MeV
$1\,\text{eV}$1eV	$1.60 \times 10^{-19}\,\text{J}$1.60×10−19J
$1\,\text{MeV}$1MeV	$1.60 \times 10^{-13}\,\text{J}$1.60×10−13J
$1\,\text{GeV}$1GeV	$1.60 \times 10^{-10}\,\text{J}$1.60×10−10J

You should be able to convert between kg, u, J, MeV and GeV fluently. Most exam errors on this topic are unit-conversion errors.

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The Physical Interpretation

Perhaps the most counter-intuitive part of $E = mc^{2}$E=mc2 is that it applies to every form of energy, not just kinetic. A hot object has more mass than a cold one (by an amount $Q/c^{2}$Q/c2, where $Q$Q is the thermal energy added). A compressed spring has more mass than a relaxed one. A proton and an electron, bound in a hydrogen atom, have less combined mass than a free proton and electron — because binding energy is released when the atom forms, and the bound state has less total energy (and hence less mass) than the separated constituents.

The same principle — with vastly larger effects — applies in nuclei. A bound nucleus has less mass than the sum of its constituent free nucleons. The mass difference, times $c^{2}$c2, is the nuclear binding energy, which we shall meet in the next lesson.

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