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Notes/Physics/Paper 4/Nuclear Physics
CAIEA Level9702§23.1-23.2

Nuclear Physics

Mass-energy equivalence E = mc², mass defect and nuclear binding energy, the binding-energy-per-nucleon curve and what it says about nuclear fission and fusion, calculating the energy released in real fusion/fission reactions, electron-positron annihilation, and radioactive decay as a spontaneous, random process governed by the exponential law x = x₀e⁻λt.

440 min read 10 sub-topics
168
question parts
2021-2025 · 36 papers
9 marks
per paper
≈ 9% of the paper
1.9/3
avg difficulty
moderate
#4
most examined
of 16 topics by marks

A nucleus is billions of times denser than any ordinary matter, held together against the electrostatic repulsion of its own protons by a force strong enough to make that repulsion irrelevant at nuclear distances — and yet some nuclei are rock stable for the age of the universe, while others fall apart within microseconds. The key to both facts is a single, startling idea: mass and energy are the same physical quantity, related by Einstein's E=mc2E=mc^2, so that the "missing" mass of a bound nucleus — its mass defect — IS the energy that would be needed to tear it apart, its binding energy. Plotting that binding energy per nucleon against nucleon number produces one of the most information-dense graphs in the whole syllabus: a curve that peaks around iron and falls away on both sides, explaining in one picture why splitting a very heavy nucleus (fission) and joining two very light ones (fusion) can BOTH release energy, despite looking like opposite processes. The second half of this note turns to a completely different question: not how much energy a nucleus releases when it decays, but when — and the answer, backed by nothing more than a Geiger counter left running in a quiet room, is that individual nuclear decays are fundamentally random events, with only a statistical, exponential law, x=x0e−λtx=x_0e^{-\lambda t}, governing the population as a whole.

The bank places this among the paper's heaviest topics: 330 marks across 2021–2025 rank Nuclear Physics 4th of the 16 A2 topics by marks, carried by 168 question parts across 36 paper sittings — more heavily examined, in fact, than Quantum Physics itself (5th, 325 marks over the same window), even though it is taught immediately after it. The mean difficulty of 1.89 is close to Quantum Physics's 1.95: the challenge is rarely difficult algebra, and much more often keeping a chain of unit conversions (u ↔ kg, MeV ↔ J, and later years ↔ seconds) completely error-free across a long multi-part calculation, and reading a "show that" target correctly the first time.

The route through: §01 mass-energy equivalence itself, E=mc2E=mc^2 — §02 writing balanced nuclear reaction equations for these newly mass-changing processes — §03 mass defect and nuclear binding energy, the energy that would have to be supplied to separate a nucleus into its individual protons and neutrons — §04 the binding-energy-per-nucleon curve, and what its shape says about which nuclei are most stable — §05 nuclear fission and fusion, read directly off that curve — §06 turning all of this into actual energy-released calculations — §07 electron–positron pair annihilation, a synthesis of E=mc2E=mc^2 with antiparticles from AS and photon momentum from Quantum Physics — §08 the evidence that radioactive decay is spontaneous and random, from the erratic count rate of a real detector — §09 activity and the decay constant, A=λNA=\lambda N — and §10 half-life, λ=0.693/t1/2\lambda = 0.693/t_{1/2}, and the exponential decay law itself, x=x0e−λtx=x_0e^{-\lambda t}, together with the graphical and calculation techniques built on it.

Before you start you should be able to

  • Nuclide notation ZAX^A_Z\text{X}, the distinction between nucleon number and proton number, and balancing α\alpha- and β\beta-decay equations by conserving nucleon number and charge, from AS Particle Physics — this note's own reaction equations (§02) extend that notation without re-deriving it

  • Antiparticles — same mass, opposite charge — from AS Particle Physics, needed later in this note for electron–positron pair annihilation

  • The unified atomic mass unit, 1 u=1.66×10−27 kg1\ \text{u} = 1.66\times10^{-27}\ \text{kg}, and converting a mass between u and kg, from AS Particle Physics — every mass-defect and E=mc2E=mc^2 calculation in this note runs through this conversion, in both directions

  • Photon energy E=hfE=hf (equivalently E=hc/λE=hc/\lambda) and photon momentum p=E/cp=E/c from Quantum Physics — needed later in this note, both when a gamma-ray photon carries away part of a decay's energy and when electron–positron annihilation produces two gamma-ray photons

  • Basic AS mechanics and energy ideas — work done, kinetic energy Ek=12mv2E_k=\tfrac12mv^2, and conservation of momentum — needed whenever the energy released in a nuclear reaction has to be shared out as kinetic energy between the products

By the end of this page you can

  • understand the equivalence between energy and mass as represented by E=mc2E = mc^2 and recall and use this equation

  • represent simple nuclear reactions by nuclear equations of the form 714N+ 24He→ 817O+ 11H^{14}_{7}\mathrm{N} + \,^{4}_{2}\mathrm{He} \rightarrow \,^{17}_{8}\mathrm{O} + \,^{1}_{1}\mathrm{H}

  • define and use the terms mass defect and binding energy

  • sketch the variation of binding energy per nucleon with nucleon number

  • explain what is meant by nuclear fusion and nuclear fission

  • explain the relevance of binding energy per nucleon to nuclear reactions, including nuclear fusion and nuclear fission

  • calculate the energy released in nuclear reactions using E=c2ΔmE = c^2 \Delta m

  • understand that fluctuations in count rate provide evidence for the random nature of radioactive decay

  • understand that radioactive decay is both spontaneous and random

  • define activity and decay constant, and recall and use A=λNA = \lambda N

  • define half-life

  • use λ=0.693/t1/2\lambda = 0.693 / t_{1/2}

  • understand the exponential nature of radioactive decay, and sketch and use x=x0e−λtx = x_0 e^{-\lambda t}, where xx could represent activity, number of undecayed nuclei or received count rate

01

Mass-energy equivalence E = mc²

Syllabus requirement · §23.1

“

understand the equivalence between energy and mass as represented by E=mc2E = mc^2 and recall and use this equation; calculate the energy released in nuclear reactions using E=c2ΔmE = c^2 \Delta m

”

Mass as a form of stored energy

Every quantity met so far in this course has kept mass and energy firmly apart: mass is "how much stuff", measured in kilograms; energy is "how much stuff can do", measured in joules, and the two never converted into one another. Nuclear physics is where that separation breaks down. The central claim of this section is that mass is itself a form of stored energy — an object simply having mass, sitting perfectly still, already represents a fixed, enormous amount of energy locked up within it. When a physical process releases energy — a nuclear reaction, a radioactive decay, the annihilation of a particle and its antiparticle — that energy has to come from somewhere, and the somewhere is a measurable decrease in the total mass of the system. Weigh the products of a nuclear reaction precisely enough, and they weigh very slightly less than the reactants did; the "missing" mass has not vanished, it has been converted directly into the energy that was released.

This is not a small correction sitting on top of ordinary mechanics — it is the reason nuclear reactions can release such staggering quantities of energy from such tiny amounts of matter, and it is the idea this whole section exists to make precise.

Rest energy: every mass has an energy, every energy has a mass

Put the claim the other way round and it becomes even more useful. Not only does releasing energy cost a system some mass — every object that has mass, purely by virtue of having that mass, carries an associated amount of energy, called its rest energy, even while sitting perfectly still with no kinetic energy at all. And conversely, every amount of energy, of any kind, has an equivalent mass associated with it. Mass and energy are not two different things that occasionally get traded for one another; on this view they are the same physical quantity, just measured in two historically different units (kilograms and joules) that turn out to be connected by a single, fixed conversion factor. That factor is what Einstein's relation supplies.

Einstein's relation: E = mc²

E=mc2E = mc^2
  • EE — the energy, in joules (J), equivalent to a mass mm, or the energy released when a mass mm is converted away.
  • mm — the mass, in kilograms (kg).
  • cc — the speed of light in a vacuum, c=3.00×108 m s−1c = 3.00\times10^{8}\ \text{m s}^{-1}, a fixed constant of nature (given on the data sheet).

Nuclear physics almost never actually needs the total rest energy of an entire nucleus (an astronomically large number, of no practical use on its own). What is needed, again and again, is the energy released or absorbed when a system's mass changes by some amount Δm\Delta m — during a decay, a reaction, or an annihilation. Writing the same relation for a change in mass and the corresponding change in energy gives the form this note will use almost exclusively:

ΔE=c2Δm\Delta E = c^2\Delta m

Every symbol means the same thing as above, just applied to a difference rather than a total: Δm\Delta m is how much the system's mass has decreased (or increased), and ΔE\Delta E is the energy released (or absorbed) as a result.

Mass-energy equivalence — two forms of the same relation
E=mc2E = mc^2

total rest energy equivalent to a mass m

ΔE=c2Δm\Delta E = c^2\Delta m

energy released/absorbed for a mass CHANGE Δm — the form almost always needed

Why this matters: c² is enormous

The reason mass–energy equivalence produces such dramatic effects in nuclear physics, and essentially none in everyday life, comes down entirely to the size of c2c^2:

c2=(3.00×108)2=9.00×1016 m2s−2c^2 = (3.00\times10^{8})^2 = 9.00\times10^{16}\ \text{m}^2\text{s}^{-2}

Ninety thousand million million. Multiplying any mass change, however small, by a number this large turns even a mass difference too tiny to detect on the most sensitive laboratory balance into a genuinely large amount of energy. This is precisely why a chemical reaction — which also, in principle, involves a mass change, since chemical bond energies are real energies too — never shows a measurable mass change at all: chemical bond energies are of order electronvolts per reaction, corresponding to a mass change many orders of magnitude too small to weigh. A nuclear reaction, by contrast, releases energies of order millions of electronvolts (MeV) per reaction, corresponding to a mass change that, while still tiny in absolute terms, is now large enough to be measured directly — and it is exactly this measured mass change that the worked examples below use to calculate the energy released.

The one sentence to carry forward

Because c2≈9.00×1016 m2s−2c^2 \approx 9.00\times10^{16}\ \text{m}^2\text{s}^{-2} is so enormous, even a mass change far too small to register on a laboratory balance corresponds to an energy release large enough to matter on an industrial scale — this single fact is the entire reason nuclear reactions, not chemical ones, are the practical source of both nuclear power and nuclear weapons.

Invented demo — the energy equivalent of one microgram of mass

A mass of 1.0 μg1.0\ \mu\text{g} (1.0×10−9 kg1.0\times10^{-9}\ \text{kg}) — far too small to see, and about a thousandth of the mass of a grain of sand — is entirely converted into energy. Calculate the energy released, and compare it with a familiar quantity.

Show full working
  1. 1

    Identify the quantities:

    m=1.0×10−9 kg,c=3.00×108 m s−1m = 1.0\times10^{-9}\ \text{kg}, \qquad c = 3.00\times10^{8}\ \text{m s}^{-1}

    Writing down exactly what is given, in the right SI units, before touching any formula — the mass here is already in kg, so no conversion is needed yet.

  2. 2

    State the formula to use:

    E=mc2E = mc^2

    The question asks for the total energy equivalent to a given mass, not a change relative to some other mass — so the full form E = mc², not ΔE = c²Δm, is the right tool here.

  3. 3

    Calculate c2c^2 as its own step:

    c2=(3.00×108)2=9.00×1016 m2s−2c^2 = (3.00\times10^{8})^2 = 9.00\times10^{16}\ \text{m}^2\text{s}^{-2}

    Squaring a standard-form number is a common place to lose a power of ten — worth isolating as its own line rather than folding it into the next substitution.

  4. 4

    Substitute mm and c2c^2 into the formula:

    E=(1.0×10−9)×(9.00×1016)E = (1.0\times10^{-9})\times(9.00\times10^{16})

    Every value now plugged in directly, with nothing simplified yet — the multiplication itself is the next, separate step.

  5. 5

    Evaluate:

    E=9.0×107 J=90 MJE = 9.0\times10^{7}\ \text{J} = 90\ \text{MJ}

    Ninety megajoules from a mass smaller than a grain of sand — for comparison, a 3 kW electric heater run for a full 8-hour night uses about 8.6×10⁷ J, almost exactly this much energy. A genuinely microscopic mass, entirely converted, releases roughly a night's worth of home heating.

Answer

E = 9.0×10⁷ J (90 MJ) — comparable to running a 3 kW heater for about 8 hours.

Whenever a mass is described as being 'entirely converted' into energy (as happens, essentially exactly, in electron–positron annihilation later in this note), use the full E = mc² with the WHOLE mass. Whenever only a DIFFERENCE between two masses matters — the far more common situation in nuclear reactions — use ΔE = c²Δm with only that difference.

Real past-paper example: finding a missing nuclide mass

The most common way this idea is actually examined is the reverse of the demo above: instead of being given a mass and asked for the energy, a question gives the energy released in a real nuclear decay and asks for a mass — very often one entry missing from an otherwise-complete table of nuclide masses. The next example is exactly this, taken from a real Paper 4 question.

Real question — the missing mass of a uranium-238 nuclide

9702/41 O/N 2025 Q8(b)(i)3 marks

A stationary nucleus of uranium-238 (U-238, Z=92Z=92) undergoes alpha decay to produce a nucleus of thorium-234 (Th-234, Z=90Z=90). The kinetic energy of the emitted alpha particle is 4.200 MeV4.200\ \text{MeV}. A gamma-ray photon is also emitted during the decay. The rebound kinetic energy of the thorium nucleus is negligible.

Table 8.1 shows the masses of the nuclides involved. The mass of U-238 is missing.

nuclidemass / u
24α^{4}_{2}\alpha4.000407
90234Th^{234}_{90}\text{Th}233.915174
92238U^{238}_{92}\text{U}?

The total energy released in the decay of the U-238 nucleus is 4.274 MeV4.274\ \text{MeV}.

Calculate the mass, in u, of the uranium-238 nuclide. Give your answer to five decimal places.

Show full working
  1. 1

    Identify what is needed: the mass of U-238 equals the sum of the masses of everything it decays into — the thorium-234 nucleus, the alpha particle, AND the extra mass Δm\Delta m that was converted into the 4.274 MeV4.274\ \text{MeV} of energy released. That Δm\Delta m is found from

    ΔE=c2Δm⇒Δm=ΔEc2\Delta E = c^2\Delta m \quad\Rightarrow\quad \Delta m = \frac{\Delta E}{c^2}

    This is the step a rushed answer skips: naming exactly what Δm physically represents here — the mass that 'disappeared' as energy — and how it fits into reconstructing the missing total mass, before any arithmetic starts.

  2. 2

    Convert the given energy to joules:

    ΔE=4.274 MeV=4.274×(1.60×10−13)=6.838×10−13 J\Delta E = 4.274\ \text{MeV} = 4.274\times(1.60\times10^{-13}) = 6.838\times10^{-13}\ \text{J}

    1 MeV = 1.60×10⁻13 J (from the Quantum Physics note) — the formula ΔE = c²Δm needs SI joules, so this conversion is always the first arithmetic step, done on its own before anything else.

  3. 3

    Compute Δm\Delta m in kilograms:

    Δm=ΔEc2=6.838×10−139.00×1016=7.60×10−30 kg\Delta m = \frac{\Delta E}{c^2} = \frac{6.838\times10^{-13}}{9.00\times10^{16}} = 7.60\times10^{-30}\ \text{kg}

    Dividing the energy just found by c² — c² itself was already evaluated in the demo above, 9.00×10¹⁶ m² s⁻², and can be reused directly.

  4. 4

    Convert Δm\Delta m from kilograms to u, since every other mass in the table is given in u:

    Δm=7.60×10−301.66×10−27=4.58×10−3 u=0.00458 u\Delta m = \frac{7.60\times10^{-30}}{1.66\times10^{-27}} = 4.58\times10^{-3}\ \text{u} = 0.00458\ \text{u}

    The mark scheme's masses are all in u, so Δm must be converted into the SAME unit (dividing by 1.66×10⁻27, the mass of 1 u, from AS Particle Physics) before it can be added to anything else in the table.

  5. 5

    Add the three masses to reconstruct the mass of U-238:

    mU-238=mTh-234+mα+Δm=233.915174+4.000407+0.00458m_{\text{U-238}} = m_{\text{Th-234}} + m_\alpha + \Delta m = 233.915174 + 4.000407 + 0.00458 mU-238=237.92016 um_{\text{U-238}} = 237.92016\ \text{u}

    The parent nucleus (U-238) sits on the LEFT of the decay, before any energy is released, so it must be the HEAVIEST of the three masses — it is reconstructed by ADDING the daughter nucleus, the alpha particle, and the extra mass Δm that the released energy corresponds to, all onto one side.

Answer

m(U-238) = 237.92016 u.

In any 'missing mass' question like this one, work out FIRST which side of the decay equation the unknown nuclide sits on. The heavier, undecayed parent nucleus is always the SUM of everything it produces, including the mass-equivalent Δm of the energy released — get the direction of that addition backwards and every other step can still be perfectly correct while the final answer is wrong.

Real past-paper example: a "show that" energy release in fusion

The next example runs the same relation in the more direct direction — energy from a given mass change — but the mass change itself has to be built up from three separately-tabulated mass defects rather than read straight off the page. (Mass defect itself is defined properly in §03; for now, all that matters is that each value in the table below is a mass, in u, that can be added and subtracted exactly like any other.)

Real question — energy released in deuterium–tritium fusion

9702/42 M/J 2023 Q9(b)(ii)3 marks

Table 9.1 shows the mass defects of three nuclei.

nuclidemass defect / u
12H^{2}_{1}\text{H}0.002388
13H^{3}_{1}\text{H}0.009105
24He^{4}_{2}\text{He}0.030377

The nuclear fusion process in a particular star is described by

12H+ 13H→ 24He+X^{2}_{1}\text{H} + \,^{3}_{1}\text{H} \rightarrow \,^{4}_{2}\text{He} + X

where XX is a particle with no mass defect.

Show that the energy released when one nucleus of He-4 is formed in this fusion reaction is 2.8×10−12 J2.8\times10^{-12}\ \text{J}.

Show full working
  1. 1

    Identify Δm\Delta m from the three given mass defects: because the total number of protons and neutrons is exactly the same on both sides of the reaction (2 protons and 3 neutrons, either way), the mass "lost" to energy in the reaction is simply the AMOUNT BY WHICH the product's mass defect (He-4's) exceeds the combined mass defect of the two reactants:

    Δm=0.030377−0.002388−0.009105\Delta m = 0.030377 - 0.002388 - 0.009105

    This is the physical content of the whole question: a bigger mass defect means a MORE tightly bound, lower-mass nucleus, so if the products end up more tightly bound overall than the reactants were, that extra binding is exactly the mass that gets converted into the energy released.

  2. 2

    Evaluate this subtraction:

    Δm=0.018884 u\Delta m = 0.018884\ \text{u}

    Kept as its own step, separate from the physical reasoning above and from the unit conversion below, so each individual arithmetic move can be checked on its own.

  3. 3

    Convert Δm\Delta m from u to kilograms:

    Δm=0.018884×(1.66×10−27)=3.135×10−29 kg\Delta m = 0.018884\times(1.66\times10^{-27}) = 3.135\times10^{-29}\ \text{kg}

    E = Δmc² needs SI kilograms; the mass-defect table gave everything in u, so this conversion (×1.66×10⁻27, from AS Particle Physics) is the necessary bridge before the formula can be applied at all.

  4. 4

    Apply E=Δm c2E=\Delta m\,c^2:

    E=(3.135×10−29)×(3.00×108)2E = (3.135\times10^{-29})\times(3.00\times10^{8})^2

    Substituting the converted mass and the constant c straight into the formula — nothing evaluated yet, so the substitution itself can be checked before any arithmetic risk is introduced.

  5. 5

    Evaluate:

    E=(3.135×10−29)×(9.00×1016)=2.82×10−12 J≈2.8×10−12 J, as required.E = (3.135\times10^{-29})\times(9.00\times10^{16}) = 2.82\times10^{-12}\ \text{J} \approx 2.8\times10^{-12}\ \text{J, as required.}

    Matching the given 'show that' target to two significant figures — carrying the extra figure (2.82) before rounding is what makes a 'show that' answer trustworthy rather than a suspiciously exact coincidence.

Answer

E = Δmc² ≈ 2.8×10⁻12 J, as required.

A 'show that' question always rewards working to at least one more significant figure than the target before rounding at the very last step — stopping early and rounding twice is the most common way a technically-correct method still fails to convincingly 'show' the given value.

Common mistakes

  • Substituting the WHOLE mass of a nucleus into E=mc2E=mc^2 when a question only asks for the energy released in a reaction or decay.

    Use the CHANGE form, ΔE=c2Δm\Delta E = c^2\Delta m, with only the mass DIFFERENCE between the reactants and the products — nuclear physics questions almost never need the total rest energy of an entire nucleus.

    The total rest energy of a whole nucleus is an astronomically large, physically meaningless number for these purposes; only the mass CHANGE corresponds to energy actually released or absorbed.

  • Substituting an energy given in MeV (or eV) directly into ΔE=c2Δm\Delta E=c^2\Delta m, e.g. dividing 4.2744.274 by c2c^2 without converting to joules first.

    Convert to joules as an explicit first step — via 1 MeV=1.60×10−13 J1\ \text{MeV}=1.60\times10^{-13}\ \text{J} or 1 eV=1.60×10−19 J1\ \text{eV}=1.60\times10^{-19}\ \text{J} — before the energy goes anywhere near c2c^2.

    This is the single most common slip in this section: it produces a Δm that is out by many orders of magnitude, yet still 'looks like' a plausible standard-form answer without a sanity check.

  • Finding Δm\Delta m in kilograms and then adding or subtracting it directly against other masses that are tabulated in u, without converting.

    Convert Δm\Delta m into WHATEVER unit the other masses in the question are given in (usually u) before combining them — never mix kg and u in the same addition or subtraction.

    A mass difference of 0.00458 u looks nothing like 7.60×10⁻30 kg on the page — combining the two without converting produces an answer wrong by a factor of roughly 10²⁷.

Your turn

Two invented calculations running the formula in each direction (mass to energy, and energy to mass), plus a real continuation of the uranium-238 question above that carries the idea forward into a photon-energy calculation from the Quantum Physics note.

  1. 13 marks

    A particular nuclear reactor releases 9.00×1013 J9.00\times10^{13}\ \text{J} of energy from its fuel every day. Calculate the corresponding decrease in the mass of the fuel each day.

    Stuck? Show hint

    Rearrange ΔE = c²Δm for Δm, and remember c² = 9.00×10¹⁶ m² s⁻².

    Show solution
    1. 1

      Identify what is needed and rearrange the formula:

      ΔE=c2Δm⇒Δm=ΔEc2\Delta E = c^2\Delta m \quad\Rightarrow\quad \Delta m = \frac{\Delta E}{c^2}

      Naming the rearrangement before any numbers go in, exactly as in the worked examples above.

    2. 2

      Substitute the given energy and c2=9.00×1016 m2s−2c^2=9.00\times10^{16}\ \text{m}^2\text{s}^{-2}:

      Δm=9.00×10139.00×1016\Delta m = \frac{9.00\times10^{13}}{9.00\times10^{16}}

      Both values plugged in directly, with the division left to the next step.

    3. 3

      Evaluate:

      Δm=1.00×10−3 kg=1.00 g\Delta m = 1.00\times10^{-3}\ \text{kg} = 1.00\ \text{g}

      A single gram of mass lost, per day, is releasing tens of trillions of joules — a concrete sense of just how much energy is packed into an ordinary-sized mass, via c².

    Answer

    Δm = 1.00×10⁻3 kg (1.00 g) per day.

  2. 22 marks

    In a hypothetical experiment, a sample of matter loses 2.0×10−6 kg2.0\times10^{-6}\ \text{kg} of mass as its energy content is entirely released. Calculate the energy released, in joules.

    Stuck? Show hint

    This time the mass loss is given directly — apply ΔE = c²Δm straight away.

    Show solution
    1. 1

      State the formula and substitute directly:

      ΔE=c2Δm=(9.00×1016)×(2.0×10−6)\Delta E = c^2\Delta m = (9.00\times10^{16})\times(2.0\times10^{-6})

      The mass change is already given in kg, so no unit conversion is needed before substituting — the formula can be applied in a single, clean step here.

    2. 2

      Evaluate:

      ΔE=1.8×1011 J\Delta E = 1.8\times10^{11}\ \text{J}

      An enormous energy — of order a hundred thousand megajoules — from a mass loss of only two milligrams, reinforcing just how large the conversion factor c² really is.

    Answer

    ΔE = 1.8×10¹¹ J.

  3. 39702/41 O/N 2025 Q8(b)(ii)3 marks

    Continuing the uranium-238 decay above: the total energy released is 4.274 MeV4.274\ \text{MeV}, of which the alpha particle carries away 4.200 MeV4.200\ \text{MeV} as kinetic energy. The rest is carried away by a single gamma-ray photon.

    Determine a value for the wavelength of the gamma radiation emitted.

    Stuck? Show hint

    First find the gamma photon's energy as a difference of the two given energies, convert it to joules, then use E = hc/λ from the Quantum Physics note.

    Show solution
    1. 1

      Find the gamma photon's energy: the total energy released splits between the alpha particle's kinetic energy and the gamma photon (the thorium nucleus's own recoil kinetic energy was stated to be negligible), so

      Eγ=4.274−4.200=0.074 MeVE_\gamma = 4.274 - 4.200 = 0.074\ \text{MeV}

      Energy conservation applied to the decay as a whole — whatever energy the alpha particle does NOT carry away as kinetic energy must belong to the photon instead.

    2. 2

      Convert this energy to joules:

      Eγ=0.074×(1.60×10−13)=1.184×10−14 JE_\gamma = 0.074\times(1.60\times10^{-13}) = 1.184\times10^{-14}\ \text{J}

      The same MeV-to-J conversion factor used throughout this section, now applied to a much smaller energy.

    3. 3

      Apply E=hc/λE=hc/\lambda from the Quantum Physics note, rearranged for λ\lambda:

      λ=hcEγ\lambda = \frac{hc}{E_\gamma}

      A gamma-ray photon obeys exactly the same photon-energy relation as any other photon — the formula does not change just because the photon happens to originate from a nucleus rather than, say, a laser.

    4. 4

      Substitute and evaluate:

      λ=(6.63×10−34)×(3.00×108)1.184×10−14=1.7×10−11 m\lambda = \frac{(6.63\times10^{-34})\times(3.00\times10^{8})}{1.184\times10^{-14}} = 1.7\times10^{-11}\ \text{m}

      A wavelength far shorter than visible light or even X-rays, exactly as expected for a gamma-ray photon — a useful order-of-magnitude sanity check on the final answer.

    Answer

    λ ≈ 1.7×10⁻11 m.

Practise mass-energy equivalence questionsReal past-paper questions · Mass-energy equivalence E = mc^2

The rest of this note

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Can you do all of these?

  • E = mc² (and its change form ΔE = c²Δm) is NOT on the data sheet — only c is given as a constant; the equation itself must be recalled every time

  • c² ≈ 9.00×10¹⁶ m² s⁻² is enormous, so even a mass change far too small to register on a laboratory balance corresponds to a huge energy release — this is the entire reason nuclear reactions release so much more energy than chemical ones

  • Convert any energy to joules — from MeV via 1 MeV = 1.60×10⁻13 J, or from eV via 1 eV = 1.60×10⁻19 J — BEFORE substituting into E = mc² or rearranging for Δm = E/c²; mixing units here is the single most common slip

  • When Δm comes out of E = Δmc² in kilograms but the other masses in a question are tabulated in u, convert Δm to u (÷1.66×10⁻27) — or vice versa — before adding or subtracting; never combine a kg value with a u value directly

  • In a 'find the missing mass' question, check WHICH side of the decay the missing nuclide sits on before deciding whether Δm is added to or subtracted from the other given masses

  • Fusion moves light nuclei RIGHT along the curve's steep rising slope towards the peak; fission moves heavy-nucleus fragments LEFT along the shallow falling slope back towards the peak — BOTH directions increase binding energy per nucleon, which is why both release energy

  • Fusion needs extreme temperature/pressure to force nuclei close enough to overcome their mutual Coulomb repulsion before the short-range strong force can bind them; heavy nuclei can undergo fission far more readily because the long-range Coulomb repulsion between their many protons is already substantial

  • A nucleon (proton or neutron) on its own has ZERO mass defect — it is not bound to anything — so it contributes nothing when a mass-defect difference is used to find the energy released in a reaction that produces free protons/neutrons alongside a nucleus

  • Scaling a single reaction's energy release up to a bulk sample (a moles' worth of fuel, or a reactor's power output) is an extra multiplication by the Avogadro constant (or by a reaction rate) — always done as its own final step, after the single-reaction energy is already found

  • An antiparticle has the SAME mass and OPPOSITE charge to its particle — never a negative mass; a positron is the electron's antiparticle, from AS Particle Physics

  • Electron–positron annihilation converts essentially ALL of both particles' rest mass into photon energy (via E = mc², using the FULL mass, not a mass difference) — unlike most of this note's other E = mc² calculations, which use only a mass CHANGE

  • Annihilation must produce AT LEAST two photons, travelling in opposite directions with equal momentum, because the total momentum of an electron and positron approaching with equal and opposite momenta is zero, and a single photon can never have zero momentum unless it also has zero energy

  • 'Random' (cannot predict which nucleus, or when) and 'spontaneous' (unaffected by external/environmental factors) are SEPARATE recall points — give whichever the question actually names, and never blend the two definitions together

  • A nucleus's decay probability per unit time never depends on how long it has already survived undecayed — no nucleus 'wears out' or becomes more (or less) likely to decay with age

  • The standard evidence for random decay is FLUCTUATIONS in measured count rate around a smooth average — not a smooth, steadily falling reading — know this exact phrase, since it is what real mark schemes reward

  • Activity A is a RATE (disintegrations per unit time, in becquerel), never a running total of decays that have already happened — 'number of nuclear disintegrations per unit time' is the definition to give verbatim

  • The decay constant λ is a FIXED property of the isotope alone — it is N, not λ, that changes as a sample decays, and A = λN only falls because N falls, never because λ itself changes

  • A detector's measured count rate is always LESS than a sample's true activity: radiation is emitted in all directions but only a fraction reaches the detector, and some of what does arrive is absorbed or not registered

  • λ = ln2/t½, NOT 0.5/t½ — the constant is ln2 ≈ 0.693, from taking the natural log of one-half, not the number 0.5 itself

  • Convert a half-life (or any elapsed time) into the SAME time unit the rest of the calculation needs (often seconds) BEFORE substituting into λ = ln2/t½ or x = x₀e⁻λt — never mix minutes/hours/days/years with seconds

  • x = x₀e⁻λt applies identically whether x stands for N, activity A, or a detector's measured count rate — the same λ and the same half-life, whichever quantity a question happens to give

  • To solve x = x₀e⁻λt for λ (or for t) from two given values, isolate the exponential term COMPLETELY on one side first, THEN take ln of both sides — never take logs before isolating

  • The whole-half-lives shortcut (halving repeatedly) only works when an elapsed time is an EXACT whole-number multiple of the half-life — for any other elapsed time, go straight to x = x₀e⁻λt

  • A graph of ln(N/N₀) [or ln(A/A₀), or ln of count rate ratio] against t is the standard linearised form of exponential decay — its gradient is always −λ (a straight line through the origin), so the MAGNITUDE of the gradient is λ, never the gradient itself

Now do the questions
168 real Paper 4 parts from 2021-2025, sorted by difficulty, with mark schemes
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