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Spec mapping: OCR H556 Module 6.4 — Nuclear and Particle Physics (Einstein's relation E=mc2; rest energy; the atomic mass unit and the MeV/c2; mass-energy conservation in nuclear processes; the reaction Q-value Q=Δmc2; conversion factor 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=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/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).
Einstein's equation says that any object with mass m has an associated rest energy:
E0=m0c2
where m0 is the rest mass (the mass of the object in its own rest frame) and c=3.00×108m s−1 is the speed of light in vacuum. The prefactor c2 is enormous — 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)(3×108)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 ΔE (becomes hotter, compresses a spring, gains kinetic energy), its mass increases by 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 2×10−12g), but in nuclear processes the mass changes are a measurable fraction of the rest masses involved — easily resolved on a mass spectrometer.
Working with kilograms is inconvenient at the nuclear scale, where masses are of order 10−27kg. Physicists use instead the atomic mass unit (u), defined so that a carbon-12 atom has a mass of exactly 12 u:
1u=1.661×10−27kg.
On this scale:
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 β+ decay you have to be careful — we shall flag it when the case arises.
Multiplying the atomic mass unit by c2 gives its rest energy. Converting to MeV (since 1MeV=1.60×10−13J):
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}$$ So: $$\boxed{\; 1\,\text{u} = 931.5\,\text{MeV}/c^{2} \;}$$ This conversion factor is one of the most useful numbers in nuclear physics. Whenever you see a mass *in atomic mass units* and need the corresponding rest energy *in MeV*, just multiply by 931.5. Particle physicists commonly use "MeV/$c^{2}$" (or "GeV/$c^{2}$") as a unit of mass directly, since it saves the conversion step. The rest energies of common particles are: | Particle | Rest energy (MeV) | Mass (u) | |---|---|---| | Electron | $0.511$ | $5.49 \times 10^{-4}$ | | Proton | $938.3$ | $1.00728$ | | Neutron | $939.6$ | $1.00867$ | | Alpha | $3727$ | $4.00150$ | | Pion $\pi^{+}$ | $139.6$ | $0.1499$ | | Muon $\mu^{-}$ | $105.7$ | $0.1135$ | Note that the neutron is slightly heavier than the proton — by about $1.3\,\text{MeV}/c^{2}$ — which is why a *free* neutron decays via $\beta^{-}$ to a proton plus electron plus antineutrino. The energy difference appears partly as the rest energy of the emitted electron ($0.511\,\text{MeV}$) and partly as the kinetic energy and antineutrino energy of the decay products. Inside a nucleus, of course, neutrons can be perfectly stable (Pauli blocking) — see the previous lesson's "Going further". --- ## Energy Released in a Nuclear Process In any nuclear process — decay, fission, fusion — the **total relativistic energy** is conserved. If the rest masses of the initial particles differ from those of the final particles, the mass difference is converted to (or from) kinetic energy of the products: $$\boxed{\; Q = \Big(\!\sum m_{\text{initial}} - \sum m_{\text{final}}\Big)\, c^{2} \;}$$ where $Q$ is called the **Q-value** of the reaction. A *positive* $Q$ means energy is released (the process is **exoergic**); a *negative* $Q$ means energy must be supplied (**endoergic**) and the reaction will not run spontaneously. <svg viewBox="0 0 540 240" xmlns="http://www.w3.org/2000/svg" role="img" aria-label="Mass-energy flow in a nuclear reaction showing rest mass before and after with the difference released as kinetic energy"> <rect x="0" y="0" width="540" height="240" fill="#fafafa" stroke="#bbb" stroke-width="1"/> <text x="20" y="28" font-family="Helvetica, Arial, sans-serif" font-size="13" fill="#222">Mass–energy conservation: initial rest mass = final rest mass + kinetic energy/c²</text> <rect x="40" y="70" width="160" height="110" fill="#e7f0ff" stroke="#1f6feb" stroke-width="1.5"/> <text x="120" y="100" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="13" fill="#222">Initial state</text> <text x="120" y="125" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#1f6feb">Σ m_initial · c²</text> <text x="120" y="155" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="11" fill="#444">(rest energy)</text> <text x="245" y="130" font-family="Helvetica, Arial, sans-serif" font-size="24" fill="#222">→</text> <rect x="280" y="50" width="180" height="70" fill="#e5f7eb" stroke="#2a8c4a" stroke-width="1.5"/> <text x="370" y="75" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="13" fill="#222">Final rest energy</text> <text x="370" y="100" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#2a8c4a">Σ m_final · c²</text> <rect x="280" y="130" width="180" height="50" fill="#fde2e2" stroke="#c0392b" stroke-width="1.5"/> <text x="370" y="155" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="13" fill="#222">Q = Δm · c²</text> <text x="370" y="172" text-anchor="middle" font-family="Helvetica, Arial, sans-serif" font-size="11" fill="#c0392b">(KE + photon energy)</text> <text x="20" y="218" font-family="Helvetica, Arial, sans-serif" font-size="11" fill="#444">Total relativistic energy is conserved. "Missing" rest mass appears as KE of products and/or photons.</text> </svg> ### 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({}^{206}\text{Pb}) = 205.97445\,\text{u}$, and $m({}^{4}\text{He}) = 4.00260\,\text{u}$. Compute the Q-value in MeV. $$\Delta m = 209.98286 - (205.97445 + 4.00260) = 0.00581\,\text{u},$$ $$Q = \Delta m \times 931.5\,\text{MeV/u} \approx 5.41\,\text{MeV}.$$ This 5.41 MeV appears mostly as kinetic energy of the emitted alpha particle (because the heavy daughter nucleus barely moves) and a tiny amount as recoil energy of the Pb-206 nucleus. The observed alpha energy of polonium-210 is indeed about $5.3\,\text{MeV}$ — the small discrepancy is exactly the recoil energy of the lead daughter, which momentum conservation says must take a fraction $4/210 \approx 1.9\,\%$ of the kinetic energy. ### Worked Example 2 — Q-value of beta-minus decay Carbon-14 $\beta^{-}$-decays to nitrogen-14. Atomic masses $m({}^{14}\text{C}) = 14.00324\,\text{u}$, $m({}^{14}\text{N}) = 14.00307\,\text{u}$. $$\Delta m = 14.00324 - 14.00307 = 0.00017\,\text{u},$$ $$Q = 0.00017 \times 931.5 \approx 0.158\,\text{MeV} = 158\,\text{keV}.$$ A subtle point: when using **atomic** masses for a $\beta^{-}$ decay, you do *not* need to add the mass of the emitted electron separately — the atomic masses already include all the electrons, and the electron added to the daughter atom exactly cancels the $\beta^{-}$ emitted from the parent nucleus. This 158 keV is the maximum kinetic energy that the emitted beta particle can carry. In reality the spectrum is continuous from 0 to 158 keV, because the antineutrino shares the available energy in random proportions. ### Worked Example 3 — Energy released in D-T fusion The deuterium-tritium fusion reaction is $^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{4}_{2}\text{He} + {}^{1}_{0}n$. Atomic masses $m({}^{2}\text{H}) = 2.01410\,\text{u}$, $m({}^{3}\text{H}) = 3.01605\,\text{u}$, $m({}^{4}\text{He}) = 4.00260\,\text{u}$, $m(n) = 1.00867\,\text{u}$. $$\Delta m = (2.01410 + 3.01605) - (4.00260 + 1.00867) = 5.03015 - 5.01127 = 0.01888\,\text{u},$$ $$Q = 0.01888 \times 931.5 \approx 17.6\,\text{MeV}.$$ This 17.6 MeV is released as kinetic energy, shared (by momentum conservation, since the neutron is much lighter than the alpha) as $14.1\,\text{MeV}$ to the neutron and $3.5\,\text{MeV}$ to the helium nucleus. It is the target reaction of experimental fusion reactors such as JET and ITER. ### Worked Example 4 — Q-value in joules and the kg route Express the 5.41 MeV Q-value of the Po-210 alpha decay in joules, and verify by computing $\Delta m \, c^{2}$ in SI. $$Q = 5.41 \times 1.60 \times 10^{-13}\,\text{J} \approx 8.66 \times 10^{-13}\,\text{J}.$$ Cross-check: $\Delta m = 0.00581 \times 1.661 \times 10^{-27}\,\text{kg} = 9.65 \times 10^{-30}\,\text{kg}$, so $$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}.$$ The two routes agree to within the precision of the input data. The MeV-via-931.5 route is cleaner for nuclear-physics arithmetic; the SI route is the one to use if the question quotes masses in kg from the outset. ### Worked Example 5 — Pair-production threshold A high-energy gamma photon converts into an electron–positron pair near a heavy nucleus (the nucleus is required to soak up momentum). What minimum photon energy is needed? The reaction is $\gamma \to e^{-} + e^{+}$. At threshold, both produced particles are essentially at rest in the centre-of-momentum frame, and the photon must supply both of their rest energies: $$E_{\gamma,\,\text{min}} = 2 m_{e} c^{2} = 2 \times 0.511\,\text{MeV} = 1.022\,\text{MeV}.$$ This is a clean demonstration of $E = mc^{2}$ 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 $e^{-} + e^{+} \to 2\gamma$ produces two $0.511\,\text{MeV}$ back-to-back photons), in cosmic-ray cascades, and in any astrophysical environment where $\gamma$-ray photons interact with matter. --- ## Energy–Mass Conversions to Remember | Quantity | Value | |---|---| | $1\,\text{u}$ | $1.661 \times 10^{-27}\,\text{kg}$ | | $1\,\text{u} \times c^{2}$ | $1.495 \times 10^{-10}\,\text{J} = 931.5\,\text{MeV}$ | | Electron rest energy | $0.511\,\text{MeV}$ | | Proton rest energy | $938.3\,\text{MeV}$ | | Neutron rest energy | $939.6\,\text{MeV}$ | | $1\,\text{eV}$ | $1.60 \times 10^{-19}\,\text{J}$ | | $1\,\text{MeV}$ | $1.60 \times 10^{-13}\,\text{J}$ | | $1\,\text{GeV}$ | $1.60 \times 10^{-10}\,\text{J}$ | You should be able to convert between kg, u, J, MeV and GeV fluently. Most exam errors on this topic are unit-conversion errors. --- ## The Physical Interpretation Perhaps the most counter-intuitive part of $E = mc^{2}$ 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 $Q/c^{2}$, where $Q$ 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 $c^{2}$, is the **nuclear binding energy**, which we shall meet in the next lesson. --- ## Synoptic LinksSubscribe to continue reading
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