Summary to follow. 22 syllabus statements (9 HL) · 46 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
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Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 22 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Nuclide and nuclide notation
Nuclide and nuclide notation
A nuclide is a particular kind of nucleus, specified by its proton number Z (the number of protons) and its nucleon number A (the total number of protons and neutrons). It is written ᴬ_ZX, where X is the chemical symbol, and its neutron number is N = A − Z. For example, ¹⁴₆C has 6 protons and 8 neutrons.
Isotopes
Nuclides of the same element: they have the same proton number Z but different neutron numbers N, and so different nucleon numbers A. Isotopes of an element have the same number of electrons in the neutral atom and therefore the same chemical properties, but they have different masses and can differ in nuclear stability: ¹²₆C and ¹³₆C are stable, whereas ¹⁴₆C is radioactive.
Students often think The upper number in ᴬ_ZX is the number of protons and the lower number is the number of neutrons, so ¹⁴₆C has 14 protons and 6 neutrons. In fact 14 is the nucleon number A (protons plus neutrons) and 6 is the proton number Z. The neutron number is A − Z = 8.
Students often think Isotopes have different nuclei, so they have different chemical properties and could be separated by chemical reactions. In fact No. Chemical behaviour is set by the electrons, and a neutral atom has as many electrons as protons. All isotopes of an element have the same proton number, so they have the same chemical properties; they differ in mass and in nuclear properties such as stability.
Unified atomic mass unit (u) and the MeV c⁻²
Unified atomic mass unit (u) and the MeV c⁻²
The unified atomic mass unit is one twelfth of the mass of an atom of carbon-12: 1 u = 1.661 × 10⁻²⁷ kg. Nuclear masses are expressed in kg, in u, or in MeV c⁻², the mass whose rest energy is 1 MeV; 1 u = 931.5 MeV c⁻². For example, the mass of a proton is 1.673 × 10⁻²⁷ kg = 1.007276 u = 938.27 MeV c⁻².
Mass defect
The difference between the total mass of the separated nucleons of a nucleus and the mass of the nucleus itself: Δm = Z m_p + (A − Z) m_n − m_nucleus. It is positive for every bound nucleus. Unit kg (also expressed in u or MeV c⁻²).
Nuclear binding energy
The minimum energy needed to separate a nucleus completely into its individual protons and neutrons; equivalently, the energy released when the nucleus is formed from them. It equals the mass defect multiplied by c²: E_b = Δmc². Unit J (commonly given in MeV); a mass defect of 1 u corresponds to 931.5 MeV.
Students often think Binding energy is released when a nucleus is pulled apart into its nucleons, in the same way that breaking a chemical bond is thought to release energy. In fact No. The binding energy must be SUPPLIED to separate a nucleus completely into its individual nucleons. Equivalently, it is the energy released when the nucleus forms from separate nucleons.
Students often think A nucleus has more mass than its separate nucleons, because binding them together adds something, the 'glue' that holds them, to the mass. In fact No. The mass of a nucleus is LESS than the total mass of its separated nucleons. The difference is the mass defect, and the mass defect multiplied by c² is the binding energy.
Binding energy per nucleon
Binding energy per nucleon
The binding energy of a nucleus divided by its nucleon number, E_b/A. It measures how tightly, on average, each nucleon is bound: the higher the binding energy per nucleon, the more stable the nucleus. Unit J (commonly given in MeV).
Binding energy curve
The graph of binding energy per nucleon against nucleon number A. It rises steeply for light nuclei (about 1.1 MeV for ²₁H, with ⁴₂He unusually high at about 7.1 MeV), reaches a broad maximum of about 8.8 MeV near A ≈ 56–62 (iron and nickel) and then falls slowly to about 7.6 MeV for uranium. Energy is released when nuclei change into nuclei that lie higher on the curve: by fusion of light nuclei or by fission of heavy nuclei. The energy released is the increase in TOTAL binding energy, found by multiplying each binding energy per nucleon by the nucleon number.
Students often think The total binding energy of a nucleus shows how stable it is, and 'binding energy' and 'binding energy per nucleon' can be used interchangeably. In fact No. Stability is indicated by the binding energy PER NUCLEON, which peaks at about 8.8 MeV near A ≈ 56–62. Uranium-238 has a much greater total binding energy than iron-56 (about 1800 MeV against about 490 MeV) only because it has more nucleons; each of its nucleons is less tightly bound.
Students often think All fission reactions and all fusion reactions release energy, whatever nuclei are involved. In fact No. Energy is released only if the products have a greater total binding energy than the reactants. On the binding energy curve this means fusion of light nuclei and fission of heavy nuclei. Splitting an iron-56 nucleus, or fusing two medium nuclei into a heavy one, needs an energy input.
Mass–energy equivalence in nuclear reactions
Mass–energy equivalence in nuclear reactions
E = mc² relates the rest energy of a body to its mass. In a nuclear reaction or decay, total nucleon number and total charge are conserved, but the total rest mass of the products differs from that of the reactants. When the products have less rest mass, the energy released (mainly as kinetic energy of the products and as γ photons) is E = Δmc², where Δm is the decrease in total rest mass. With masses in u, the energy in MeV is Δm × 931.5.
Students often think In α decay the parent loses the mass of the α particle, so the energy released is the energy equivalent of the α particle's mass. In fact No. The α particle is emitted, not destroyed: its nucleons still exist. The mass converted is the small difference between the mass of the parent and the total mass of the products (daughter plus α particle), which for radium-226 is about 0.0052 u, not about 4 u.
Students often think The energy equivalent of a mass is found by multiplying the mass by the speed of light, c. In fact No. The energy equivalent is the mass multiplied by the SQUARE of the speed of light, c² = 9.00 × 10¹⁶ m² s⁻². Multiplying by c gives a value 3.00 × 10⁸ times too small, in kg m s⁻¹, which is not a unit of energy.
Strong nuclear force
Strong nuclear force
The attractive force between nucleons (proton–proton, proton–neutron and neutron–neutron) that holds nuclei together. At nucleon separations of about 10⁻¹⁵ m it is much stronger than the electric repulsion between protons, but it is short-range: it becomes negligible at separations beyond a few times 10⁻¹⁵ m, so it does not act between neighbouring nuclei.
Students often think The strong nuclear force is a kind of electric force: it depends on charge, so it repels like-charged nuclei and protons. In fact No. The strong nuclear force acts between nucleons whether or not they are charged: proton–proton, neutron–neutron and proton–neutron. It is attractive at nuclear separations and it is not an electric force.
Students often think The electrons around each nucleus shield neighbouring nuclei from its strong nuclear force. In fact No. Nothing blocks the strong force; it simply becomes negligible at separations much greater than about 10⁻¹⁵ m. Neighbouring nuclei in a solid are about 10⁻¹⁰ m apart, far beyond its range.
Random nature of radioactive decay
Random nature of radioactive decay
It is impossible to predict when a particular nucleus will decay. Every undecayed nucleus of a given nuclide has the same chance of decaying in a given time, however long it has already existed. For a large number of nuclei the fraction that decays in a given time is predictable, which is why a half-life can be defined; the randomness shows as fluctuations in the count rate about a smooth trend.
Spontaneous nature of radioactive decay
Radioactive decay happens without any external cause, and its rate is not affected by external conditions such as temperature, pressure or chemical combination, which change the electrons and motion of atoms but not their nuclei.
Students often think Radioactive decay is caused or speeded up by external conditions: a nucleus decays when it is heated, knocked by a collision, compressed or changed chemically. In fact No. Radioactive decay is spontaneous: it happens without any external cause, and its rate is not affected by temperature, pressure or chemical combination. These change the atoms' electrons or motion, not their nuclei.
Students often think Nuclei age like living things: a nucleus that has not decayed for a long time is 'due' to decay, and nuclei decay roughly in order of age. In fact No. Nuclei do not age. Every undecayed nucleus of a given nuclide has the same chance of decaying in the next second, however long it has already existed.
Alpha (α) particle and α decay
Alpha (α) particle and α decay
An α particle is a helium-4 nucleus, ⁴₂He or ⁴₂α (2 protons and 2 neutrons), with charge +2e. In α decay the parent nucleus loses 2 protons and 2 neutrons: Z decreases by 2 and A decreases by 4, so a different element is formed.
Beta-minus (β⁻) particle and β⁻ decay
A β⁻ particle is an electron, ⁰₋₁e, created and emitted by the nucleus. In β⁻ decay a neutron in the nucleus changes into a proton, and an electron antineutrino is also emitted: Z increases by 1 and A is unchanged.
Beta-plus (β⁺) particle and β⁺ decay
A β⁺ particle is a positron (the antiparticle of the electron), ⁰₊₁e, created and emitted by the nucleus. In β⁺ decay a proton in the nucleus changes into a neutron, and an electron neutrino is also emitted: Z decreases by 1 and A is unchanged.
Gamma (γ) radiation
A high-energy photon emitted when a nucleus in an excited state moves to a lower energy state. Z and A are unchanged. γ emission often follows α or β decay that has left the daughter nucleus in an excited state; an excited nuclide is marked with *, or with m if it is long-lived (metastable), as in ⁹⁹ᵐ₄₃Tc.
Students often think In β⁻ decay the atom ejects one of its orbiting electrons, so the nucleus itself does not change. In fact No. A β⁻ particle is an electron created and emitted by the NUCLEUS when a neutron changes into a proton. The orbital electrons are not involved, and the nucleus changes: Z increases by 1 and A is unchanged.
Students often think The changes are the other way round: in β⁻ decay a proton turns into a neutron, lowering Z, and in β⁺ decay a neutron turns into a proton, raising Z. In fact No. In β⁻ decay a neutron changes into a proton, so Z increases by 1. In β⁺ decay a proton changes into a neutron, so Z decreases by 1. In both, A is unchanged.
Radioactive decay equations
Radioactive decay equations
Nuclear equations in which total nucleon number and total charge are conserved. Examples: α decay ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂α; β⁻ decay ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ⁰₀ν̄; β⁺ decay ¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ⁰₀ν; γ emission ⁹⁹ᵐ₄₃Tc → ⁹⁹₄₃Tc + γ.
Students often think A neutrino (or antineutrino) is emitted in every type of radioactive decay, including α decay. In fact No. A neutrino accompanies β⁺ decay and an antineutrino accompanies β⁻ decay. α decay and γ emission do not produce neutrinos.
Neutrino (ν) and antineutrino (ν̄)
Neutrino (ν) and antineutrino (ν̄)
Uncharged particles with a very small mass (much smaller than that of the electron) that very rarely interact with matter; almost all pass straight through the Earth. An electron antineutrino is emitted with the electron in β⁻ decay, and an electron neutrino with the positron in β⁺ decay. Their nucleon number and charge are both zero: ⁰₀ν and ⁰₀ν̄.
Students often think A neutrino (or antineutrino) is a neutron, or a small kind of neutron, released from the nucleus in β decay. In fact No. A neutron is a nucleon with a mass of about 1 u. A neutrino is a different particle with no charge and a very small mass (far less than an electron's) that very rarely interacts with matter.
Students often think The antineutrino emitted in β⁻ decay is positively charged, because an antiparticle has the opposite charge and it balances the charge of the β⁻ particle. In fact No. Both the neutrino and the antineutrino are uncharged. Charge in β⁻ decay balances without them: 0 → (+1) + (−1).
Ionizing ability
Ionizing ability
The extent to which radiation removes electrons from the atoms along its path. α particles are the most strongly ionizing, β particles less so, and γ rays the least. Each ionization takes energy from the radiation, so strongly ionizing radiation gives up its energy within a short distance; this is why ionizing ability and penetration are inversely related.
Penetration
α particles are stopped by a sheet of paper or a few centimetres of air; β particles by a few millimetres of aluminium (about a metre of air); γ rays are not stopped completely, but their intensity is reduced significantly by several centimetres of lead or by thick concrete.
Irradiation and contamination
Irradiation is exposure of an object to radiation from a source outside it; exposure to α, β or γ radiation of the energies used in medicine and industry does not make the object radioactive. Contamination is the presence of radioactive material on or in an object, which then emits radiation itself until the material is removed or decays.
Choosing an isotope for an application
The radiation must have the right penetration and the half-life must suit the time scale of the use. Medical imaging tracers such as technetium-99m (γ, half-life 6.0 h) emit γ rays that leave the body and decay within hours; tracers for leaks in buried pipes such as sodium-24 (β⁻ and γ, half-life 15 h) emit γ rays that pass through soil; thickness gauges for paper and foil use β⁻ sources with long half-lives such as strontium-90 (29 years), so that the count rate changes only when the thickness does; radioactive dating uses a nuclide whose half-life is comparable with the age measured, such as carbon-14 (5730 years).
Students often think The 'strongest' radiation is both the most ionizing and the most penetrating, so the radiation that is best at one is best at the other. In fact No. The two go in opposite directions. α particles are the most strongly ionizing and the least penetrating; γ rays are the least ionizing and the most penetrating. Radiation that ionizes strongly loses its energy quickly and so does not travel far.
Students often think Anything exposed to radiation becomes radioactive itself, so irradiated food, instruments or people give off radiation afterwards. In fact No. Exposure to α, β or γ radiation of the energies used in medicine and industry (irradiation) does not make an object radioactive. An object is radioactive only if it contains radioactive material, for example because it has been contaminated.
Activity
Activity
The number of nuclei in a sample that decay per unit time. SI unit: becquerel, Bq (1 Bq = 1 decay per second).
Count rate
The number of particles or photons recorded by a detector per unit time (e.g. counts s⁻¹, counts min⁻¹). It is normally much smaller than the activity, because radiation leaves the source in all directions, some is absorbed before it reaches the detector, and the detector does not record everything that enters it. For a fixed arrangement, the corrected count rate is proportional to the activity.
Half-life
The time taken for the number of undecayed nuclei of a nuclide in a sample, or the activity (or corrected count rate) of the sample, to fall to half of its initial value. It is constant for a given nuclide, whatever the size of the sample and whenever the timing starts. It is determined from corrected count rates measured over time, as the time for any value to halve.
Students often think The half-life is half the time the sample takes to decay completely, so after two half-lives all of it has decayed. In fact No. In each half-life the number of undecayed nuclei halves, so after two half-lives a quarter remains, after three an eighth, and so on. For a large sample the decay is far from finished after two half-lives.
Students often think In one half-life the sample loses half of its mass, because half of it has 'gone' by decaying. In fact No. Half of the radioactive NUCLEI decay, but they become daughter nuclei that stay in the sample if the daughter is a solid. The sample's mass hardly changes; only the emitted particles and a tiny amount of mass-energy leave.
Decay over whole numbers of half-lives
Decay over whole numbers of half-lives
After n half-lives the number of undecayed nuclei, the activity and the corrected count rate are each multiplied by (½)ⁿ: ½ after one half-life, ¼ after two, ⅛ after three. For example, a source of activity 800 MBq has an activity of 50 MBq after 4 half-lives.
Students often think The number of half-lives and the factor by which the activity falls are the same number: a fall by a factor of 8 means 8 half-lives, and n half-lives divide the activity by n. In fact No to both. Each half-life divides the activity by 2, so n half-lives divide it by 2ⁿ. A fall by a factor of 8 = 2³ takes 3 half-lives, and after 4 half-lives the activity is divided by 2⁴ = 16, not by 4.
Background radiation and corrected count rate
Background radiation and corrected count rate
Background radiation is radiation from sources other than the one being studied, such as radon gas from rocks and building materials, cosmic rays, food and drink, and medical uses. The background count rate is measured with the source removed and subtracted from every reading; the result, the corrected count rate, is the count rate due to the source alone, and it is this that halves every half-life.
Students often think The background count is part of the source's count rate, so it halves every half-life along with the rest of the reading. In fact No. Background radiation comes from sources other than the sample, such as radon in the air, rocks and building materials, and cosmic rays, and it stays roughly constant. Only the count rate due to the source halves every half-life.
Students often think Once background has been measured and subtracted, it no longer affects the readings, so the counter will show only the corrected count rate. In fact No. The counter always records background as well as the source. Subtracting background is a step in a calculation; a prediction of what the counter will read must add the background back.
Evidence for the strong nuclear force HL
Evidence for the strong nuclear force
Nuclei containing many protons are stable, although neighbouring protons are only about 2 × 10⁻¹⁵ m apart, where their electric repulsion is about 60 N; the gravitational attraction between them (about 5 × 10⁻³⁵ N, some 10³⁶ times smaller) is far too weak to hold them together. A much stronger attractive force must therefore act between nucleons. That nuclei do not attract neighbouring nuclei, and that the binding energy per nucleon is roughly constant for large nuclei, show that this force is short-range.
Students often think Nucleons are packed so closely that the gravitational attraction between them is large enough to hold the nucleus together. In fact No. The gravitational attraction between two protons 2 × 10⁻¹⁵ m apart is about 5 × 10⁻³⁵ N, about 10³⁶ times smaller than their electric repulsion of about 58 N. Gravity is negligible in nuclei; the attraction is due to the strong nuclear force.
Neutron–proton ratio and nuclear stability HL
Neutron–proton ratio and nuclear stability
Stable light nuclides have N ≈ Z; stable heavier nuclides have progressively more neutrons than protons, up to N/Z ≈ 1.5 for the heaviest, because extra neutrons add strong-force attraction without adding electric repulsion. Nuclides with too many neutrons for stability tend to decay by β⁻ emission (a neutron becomes a proton, lowering N/Z); nuclides with too few neutrons tend to decay by β⁺ emission (raising N/Z); very heavy nuclides (A above about 200) commonly decay by α emission.
Students often think α decay is how a nucleus gets rid of surplus neutrons, since the α particle carries neutrons away. In fact No. α emission removes two protons and two neutrons, which does not reduce the neutron excess of a light nucleus: for carbon-14 it would raise N/Z from 1.33 to 1.5. A neutron-rich nuclide decays by β⁻ emission, in which a neutron changes into a proton, lowering N/Z.
Students often think Neutrons sit between the protons as spacers or insulators that shield the protons from one another's electric repulsion. In fact No. Neutrons do not shield or 'insulate' the protons: the electric repulsion between two protons is the same whatever lies between them. Neutrons help because they add strong-force attraction to their neighbours without adding any electric repulsion.
Approximate constancy of binding energy per nucleon above A ≈ 60 HL
Approximate constancy of binding energy per nucleon above A ≈ 60
For nucleon numbers above about 60 the binding energy per nucleon varies only between about 7.6 MeV and 8.8 MeV, so the total binding energy is roughly proportional to A. This shows that each nucleon is attracted only by its nearest neighbours, because the strong force is short-range: if every nucleon attracted every other, the binding energy per nucleon would keep rising with A. The slow decrease for heavy nuclei is due to the electric repulsion, which acts between every pair of protons.
Students often think The strong nuclear force is so much stronger than the electric force that the repulsion between protons plays no part in the binding of large nuclei. In fact No. The strong force acts only between near neighbours, but the electric repulsion acts between every pair of protons. In large nuclei the repulsion on each proton keeps growing, which is why the binding energy per nucleon falls slowly from about 8.8 MeV (A ≈ 56–62) to about 7.6 MeV (uranium) and why heavy nuclei need extra neutrons.
Discrete nuclear energy levels HL
Discrete nuclear energy levels
A nucleus can exist only in certain discrete energy states. The α particles emitted by a given nuclide have one or a few discrete energies, and the γ photons have discrete energies equal to differences between nuclear energy levels. For example, radium-226 emits α particles of 4.78 MeV and 4.60 MeV; the lower-energy decays leave radon-222 in an excited state 0.19 MeV above its ground state, which then emits a 0.19 MeV γ photon. (The two α energies differ by slightly less than 0.19 MeV because the recoiling radon nucleus takes a small share of the energy.)
Students often think Each nuclide emits α particles of a single energy, so two α energies from one sample mean that two different isotopes are present. In fact No. One nuclide can emit α particles of several discrete energies, because α decay can leave the daughter nucleus in its ground state or in one of its excited states. The difference between the α energies matches the energy of the γ photon emitted when the excited daughter drops to its ground state.
Students often think γ rays are produced in the same way as light: by electrons dropping between discrete energy levels of the atom. In fact No. γ photons are emitted by the NUCLEUS when it moves from an excited nuclear energy level to a lower one. Their energies match differences between the α energies from the parent, which depend on the daughter nucleus, not on its electrons.
Continuous beta spectrum HL
Continuous beta spectrum
β particles from a given nuclide are emitted with a continuous range of kinetic energies, from almost zero up to a maximum approximately equal to the energy released in the decay. If only two bodies (the β particle and the daughter nucleus) were produced, conservation of energy and momentum would give the β particle a single energy, as for α particles. The continuous spectrum therefore shows that a third particle, the antineutrino in β⁻ decay or the neutrino in β⁺ decay, shares the energy released.
Students often think β particles all leave the nucleus with the same energy, and the spread arises because they lose different amounts of energy colliding with atoms before they escape from the source. In fact No. The β particles leave the nucleus with a continuous range of energies, from almost zero up to a maximum set by the energy released in the decay. The energy released in each decay is shared in varying proportions between the β particle and an antineutrino (or neutrino).
Students often think The daughter nucleus can be left with any amount of energy, since nuclear energy levels are continuous, so β particles can have any energy. In fact No. Nuclei have discrete energy levels, as the discrete energies of α particles and γ photons show. The β spectrum is continuous because the fixed decay energy is shared with an antineutrino, not because the daughter nucleus can have any energy.
Decay constant HL
Decay constant
The constant λ in the radioactive decay law. For a short time interval Δt, λΔt approximates the probability that a given nucleus decays in that interval. Unit s⁻¹ (or h⁻¹, y⁻¹, etc.).
Radioactive decay law
The number of undecayed nuclei after time t is N = N₀e^(−λt), where N₀ is the number at t = 0. It applies to any time interval, not only to whole numbers of half-lives; rearranged, t = (1/λ) ln(N₀/N). A graph of ln N, or of ln(corrected count rate), against t is a straight line of gradient −λ.
Students often think Any logarithm undoes e^(−λt), so pressing the 'log' (base 10) key on a calculator gives the same result as 'ln'. In fact No. The decay law uses e, so t = (1/λ) ln(N₀/N). Using log₁₀ in place of ln gives a time that is too small by a factor of ln 10 ≈ 2.30.
Students often think Decay can only be worked out for whole numbers of half-lives, so the time must be rounded to the nearest whole number of half-lives. In fact No. 20 days is 2.5 half-lives, and the fraction remaining is (½)^2.5 = e^(−λt) with λ = ln 2/8.0 days, which gives 0.18, not 0.25. The decay law N = N₀e^(−λt) applies to any time t, not just to whole numbers of half-lives.
Decay constant and probability of decay HL
Decay constant and probability of decay
The probability that a given nucleus decays within a time t is 1 − e^(−λt). Only when λt is small (λt ≪ 1) is this approximately equal to λt, so λ approximates the probability of decay per unit time only in the limit of sufficiently small λt.
Students often think λ is exactly the probability of decay per unit time, so the probability that a nucleus decays within any time t is λt, and half the nuclei have decayed when λt = 0.5. In fact Only approximately, when λt is small. The probability of decay within time t is 1 − e^(−λt), which is close to λt when λt ≪ 1 but smaller than λt otherwise: for λt = 0.50 it is 0.39, not 0.50.
Students often think The expression e^(−λt) from the decay law gives the probability that a nucleus decays within time t. In fact No. e^(−λt) = N/N₀ is the fraction of nuclei that have NOT decayed after time t, which is the probability that a given nucleus survives. The probability that it decays is 1 − e^(−λt).
Activity and the decay constant HL
Activity and the decay constant
Activity is the rate of decay: A = λN = λN₀e^(−λt). The activity decreases exponentially with the same decay constant as N, and for equal numbers of undecayed nuclei it is proportional to λ.
Students often think Nuclides with long half-lives are more radioactive: for the same number of nuclei, a longer half-life gives a greater activity. In fact No. A = λN and λ = ln 2/T½, so for equal N a longer half-life means a smaller decay constant and a LOWER activity. A long-lived sample stays radioactive for longer, but at any instant it emits less.
Students often think The activity of a sample depends only on how many radioactive nuclei it contains, so equal numbers of nuclei give equal activities whatever the nuclide. In fact No. Activity depends on the decay constant as well as on the number of nuclei: A = λN. Samples of different nuclides with equal N have activities in the ratio of their decay constants.
Half-life and decay constant HL
Half-life and decay constant
T½ = ln 2/λ ≈ 0.693/λ, found by putting N = N₀/2 in N = N₀e^(−λt). The half-life of a nuclide can be determined from the gradient, −λ, of a graph of ln(corrected count rate) against time.
Students often think The half-life and the decay constant are reciprocals: T½ = 1/λ, or λ = 1/T½. In fact No. T½ = ln 2/λ ≈ 0.693/λ. The factor ln 2 comes from putting N = N₀/2 into N = N₀e^(−λt).
Students often think The decay constant is a number that does not depend on the unit of time used, so a value found in h⁻¹ or day⁻¹ can be reported in s⁻¹ unchanged. In fact No. λ = ln 2/T½ = 0.693/6.0 h = 0.12 h⁻¹. To give λ in s⁻¹ the half-life must first be converted to seconds: 6.0 h = 21 600 s, so λ = 0.693/21 600 s = 3.2 × 10⁻⁵ s⁻¹.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Carbon-12 (¹²₆C) and carbon-14 (¹⁴₆C) are isotopes of carbon. Which statement about these two nuclides is correct?
Answer and reasoning
They contain the same number of neutrons but different numbers of protons. — A student who reads the upper number as the number of protons and the lower number as the number of neutrons sees 12 and 14 on top and 6 underneath both. The upper number is the nucleon number A and the lower is the proton number Z: both nuclides have 6 protons.
Both contain the same number of protons, but different numbers of neutrons. — Both have proton number Z = 6, which is what makes them carbon. Carbon-12 has 12 − 6 = 6 neutrons and carbon-14 has 14 − 6 = 8, so they differ only in neutron number.
Their atoms have different chemical properties, as their nuclei differ. — A student who thinks different nuclei mean different chemistry picks this. Chemical properties depend on the electrons; both atoms have 6 electrons when neutral, so carbon-12 and carbon-14 react in the same way.
Their nuclei are identical, but their atoms have different numbers of electrons. — A student who confuses isotopes with ions picks this. Atoms of one element that differ in electron number are ions. These two nuclei are not identical: they contain different numbers of neutrons.
2 In the fusion reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, the binding energies per nucleon are 1.11 MeV for ²₁H, 2.83 MeV for ³₁H and 7.07 MeV for ⁴₂He. A free neutron has no binding energy. How much energy is released in one reaction?
Answer and reasoning
10.7 MeV — A student who treats binding energy as energy stored in the reacting nuclei adds the binding energies of ²H and ³H, 2(1.11) + 3(2.83) = 10.7 MeV. That energy would have to be SUPPLIED to separate them; the energy released is the gain in binding energy, 28.28 − 10.71 = 17.6 MeV.
17.6 MeV — The energy released is the increase in total binding energy: 4(7.07) − [2(1.11) + 3(2.83)] = 28.28 − 10.71 = 17.6 MeV.
28.3 MeV — A student who counts only the binding energy of the nucleus formed gives 4 × 7.07 = 28.3 MeV. The reactants were already bound by 10.71 MeV, so only the increase, 17.6 MeV, is released.
3.13 MeV — A student who subtracts binding energies per nucleon directly gets 7.07 − 1.11 − 2.83 = 3.13 MeV. Each value must first be multiplied by its nucleon number to give a total binding energy.
Working Total binding energy = A × (binding energy per nucleon). Reactants: ²H: 2 × 1.11 = 2.22 MeV; ³H: 3 × 2.83 = 8.49 MeV; total 10.71 MeV. Products: ⁴He: 4 × 7.07 = 28.28 MeV; neutron 0. Energy released = 28.28 MeV − 10.71 MeV = 17.57 MeV ≈ 17.6 MeV.
3 Nucleons in a nucleus are about 10⁻¹⁵ m apart, while the nuclei of neighbouring atoms in a solid are about 10⁻¹⁰ m apart. Why does the strong nuclear force bind the nucleons within each nucleus but have no effect between neighbouring nuclei?
Answer and reasoning
Its range is so short that it is negligible at separations much above 10⁻¹⁵ m. — The strong force is short-range: very strong between nucleons about 10⁻¹⁵ m apart, but negligible a few times further out. Nuclei in a solid are about 10⁵ times further apart than that, so they do not feel it.
It is a repulsive force between nuclei, as every nucleus carries positive charge. — A student who thinks the strong force depends on charge picks this. The strong force attracts nucleons whatever their charge; the repulsion between nuclei is the ELECTRIC force. The strong force is absent between nuclei because of its short range.
The electrons around each nucleus block the force from reaching other nuclei. — A student who thinks material in the way can block a force picks this. Nothing shields the strong force; it simply does not extend beyond a few times 10⁻¹⁵ m.
It falls as 1/r², so at 10⁻¹⁰ m it is 10¹⁰ times weaker than it is inside a nucleus. — A student who assumes every force follows an inverse-square law picks this. If the strong force fell as 1/r², it would fall in step with the electric force and stay stronger than the repulsion between nuclei, so nuclei in a solid would be pulled together. The strong force falls off far faster than 1/r².
Working The strong nuclear force acts only over distances of the order of 10⁻¹⁵ m. Neighbouring nuclei in a solid are about 10⁻¹⁰ m apart, 10⁵ times further, where the strong force is negligible.
4 A nucleus X undergoes α decay. The daughter nucleus then undergoes two β⁻ decays in succession. How does the final nucleus compare with X?
Answer and reasoning
It is an isotope of X whose nucleus has six fewer neutrons than X. — A student who thinks each β particle carries away a nucleon subtracts two more from A, giving N − 6. A β particle is an electron with nucleon number 0: in β⁻ decay A is unchanged while one neutron becomes a proton.
It is another element, with two fewer protons and two fewer neutrons. — A student who thinks β⁻ particles come from the electron shells leaves the nucleus unchanged after the α decay. β⁻ particles come from the nucleus: each one raises Z by 1, returning the nucleus to the element X.
It is another element, with four fewer protons and as many neutrons as X. — A student who thinks β⁻ decay turns a proton into a neutron lowers Z by a further 2 and restores the 2 neutrons. It is the other way round: β⁻ decay turns a neutron into a proton, making the nucleus more positive, as charge conservation requires.
It is an isotope of X whose nucleus has four fewer neutrons than X. — α decay removes 2 protons and 2 neutrons; each β⁻ decay turns a neutron into a proton. Overall Z is unchanged (−2 + 1 + 1) and N falls by 2 + 1 + 1 = 4, so the result is an isotope of X with a nucleon number 4 less.
Working α decay: Z − 2, N − 2 (A − 4). Each β⁻ decay: a neutron becomes a proton, so Z + 1 and N − 1 (A unchanged). Overall: Z − 2 + 1 + 1 = Z (same element); N − 2 − 1 − 1 = N − 4; A − 4. The final nucleus is an isotope of X with 4 fewer neutrons.
5 An electron antineutrino, ν̄, is emitted in every β⁻ decay. Which statement about the antineutrino is correct?
Answer and reasoning
It is a neutron that leaves the nucleus together with the β⁻ particle. — A student who confuses the similar names picks this. In β⁻ decay a neutron is used up, not emitted: it becomes a proton. The antineutrino has nucleon number 0 (⁰₀ν̄).
It is uncharged, has a very small mass and very rarely interacts with matter. — Antineutrinos, like neutrinos, have zero charge and a mass far smaller than the electron's, and they interact so rarely that almost all of them pass straight through the Earth.
It is positively charged, balancing the negative charge of the β⁻ particle. — A student who takes 'anti' to mean 'opposite charge' picks this. Charge balances without it: a neutron (0) becomes a proton (+1) and an electron (−1). The antineutrino is uncharged.
It is a high-energy photon, emitted as the new nucleus loses energy. — A student who merges the antineutrino with the γ photon picks this. A γ photon comes from an excited nucleus dropping to a lower energy state; the antineutrino is a separate particle created in β⁻ decay.
6 A hospital prepares a dose of technetium-99m (half-life 6.0 h) that has an activity of 800 MBq at 08:00. What will its activity be at 14:00 the next day?
Answer and reasoning
0 MBq — A student who thinks half-life is half the time to decay completely expects all the technetium-99m to have gone after 12 h. In fact the activity halves every 6.0 h and is still 25 MBq after 30 h.
160 MBq — A student who divides by the number of half-lives gets 800/5 = 160 MBq. Five half-lives divide the activity by 2⁵ = 32.
25 MBq — From 08:00 to 14:00 the next day is 30 h, which is five half-lives: 800 → 400 → 200 → 100 → 50 → 25 MBq.
12.5 MBq — A student who counts the six clock times 08:00, 14:00, 20:00, 02:00, 08:00 and 14:00 as six half-lives halves six times, giving 12.5 MBq. Those six times enclose only five 6-hour intervals.
Working From 08:00 to 14:00 the next day is 30 h = 30 h/6.0 h = 5 half-lives. A = 800 MBq × (½)⁵ = 800 MBq/32 = 25 MBq.
7 In a helium-4 nucleus the two protons are about 2 × 10⁻¹⁵ m apart. At this separation the electric repulsion between them is about 58 N, an enormous force on particles of mass 1.67 × 10⁻²⁷ kg. Yet helium-4 nuclei are stable. What does this evidence show? HL
Answer and reasoning
Gravitational attraction between protons this close together balances the repulsion. — A student who thinks nucleons are close enough for gravity to matter picks this. For two protons 2 × 10⁻¹⁵ m apart, Gm²/r² ≈ 5 × 10⁻³⁵ N, about 10³⁶ times smaller than the electric repulsion.
The neutrons attract the protons, but the two protons exert no attraction on each other. — A student who sees neutrons as the only 'glue' picks this. The strong force acts between all pairs of nucleons, including two protons; neutrons are not the only source of attraction.
An attraction stronger than the electric repulsion must act between nucleons. — The protons stay together despite a repulsion of about 58 N, so a still larger attractive force must act between nucleons about 10⁻¹⁵ m apart. This is the strong nuclear force.
A long-range attraction acts, which also binds the nuclei of neighbouring atoms together. — A student who assumes the strong force behaves like gravity or the electric force picks this. Neighbouring nuclei in a solid are not pulled together; the strong force acts only over distances of about 10⁻¹⁵ m.
8 For nuclides with nucleon number A above about 60, the binding energy per nucleon changes very little, staying between about 7.6 MeV and 8.8 MeV. What does this suggest about the strong nuclear force? HL
Answer and reasoning
Each nucleon attracts every other nucleon in the nucleus, as the force has a long range. — A student who assumes the strong force has a long range picks this. If every nucleon attracted every other, the number of attracting pairs, and so the binding per nucleon, would grow steadily with A; the near-constancy rules this out.
It gives every nucleus with A above 60 about the same total binding energy. — A student who confuses binding energy with binding energy per nucleon picks this. A roughly constant E_b/A means that the TOTAL binding energy grows roughly in proportion to A: about 490 MeV for iron-56 but about 1800 MeV for uranium-238.
It is so strong that the repulsion between protons has no effect in large nuclei. — A student who thinks the strong force completely overwhelms the repulsion picks this. The slow fall from 8.8 MeV to 7.6 MeV for heavy nuclei is caused by the electric repulsion, which acts between every pair of protons.
Each nucleon is attracted only by its nearest neighbours, as the force has a very short range. — If each nucleon is bound only to its near neighbours, adding nucleons adds about the same binding per nucleon, so E_b/A stays roughly constant and the total E_b is roughly proportional to A. This is evidence for the short range of the strong force.
9 In the β⁻ decay of a particular nuclide, the emitted electrons have a continuous range of kinetic energies, from almost zero up to a maximum value. Which statement correctly explains this? HL
Answer and reasoning
The energy released is shared in varying proportions between the electron and an antineutrino. — If only the electron and the daughter nucleus were produced, conservation of energy and momentum would fix the electron's energy. The continuous spread shows that a third particle, the antineutrino, carries off the rest of the energy released.
Electrons lose different amounts of energy in collisions with atoms before they leave the source. — A student who looks for an ordinary cause of the spread picks this. The spread is present at emission: each decay releases the same energy, and the electron receives a varying share of it, the rest going to the antineutrino.
The electrons come from different shells of the atom, which have different binding energies. — A student who thinks β⁻ particles are orbital electrons picks this. The electrons are created in the nucleus when a neutron changes into a proton; the atom's electron shells are not involved.
The daughter nucleus can be left with any energy, as nuclear energy levels are continuous. — A student who reads any continuous spectrum as evidence of continuous energy levels picks this. The discrete energies of α particles and γ photons show that nuclear energy levels are discrete; the energy is shared with an antineutrino instead.
10 The decay constant λ of a nuclide has the unit s⁻¹. Which statement about λ is correct? HL
Answer and reasoning
The probability that a nucleus decays in a short time Δt is approximately λΔt, provided λΔt is much less than one. — λ is the probability of decay per unit time, so for a short interval the probability of decay is about λΔt. The exact probability is 1 − e^(−λΔt), which equals λΔt only in the limit of small λΔt; for larger λΔt the approximation overestimates the probability.
The probability that a nucleus decays in a short time interval is approximately equal to λ, provided the interval is short. — A student who takes λ itself to be a probability picks this. A probability has no unit, but λ has the unit s⁻¹: it is a probability PER UNIT TIME, and must be multiplied by a (short) time interval to give a probability.
The probability that a nucleus decays within a time t is exactly λt, however long the interval t may be. — A student who takes λ to be exactly the probability per unit time for all t picks this. For large t, λt exceeds 1, which no probability can do. The exact probability is 1 − e^(−λt); λt is a good approximation only when λt is much less than one.
The value of λ for a sample falls as the sample decays, because each surviving nucleus becomes less and less likely to decay. — A student who thinks a falling activity means each nucleus is less likely to decay picks this. λ is a constant of the nuclide: every undecayed nucleus has the same probability of decay per unit time at all times. The activity falls because there are fewer undecayed nuclei, not because λ changes.
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36 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Which statement correctly describes the binding energy of a nucleus?
Answer and reasoning
The energy released when the nucleus is separated completely into its nucleons. — A student who thinks that breaking bonds releases energy picks this. The direction is reversed: energy is released when a nucleus forms from separate nucleons, and the same amount must be supplied to separate them.
The energy stored inside the nucleus, which it gives out when it later decays. — A student who pictures binding energy as a store held by the nucleus picks this. Binding energy is the energy needed to take the nucleus apart; a more tightly bound nucleus has less mass-energy than its separated nucleons, not more.
The minimum energy needed to pull the nucleus apart into separate nucleons. — Nucleons attract one another through the strong nuclear force, so energy must be supplied to pull them apart. The binding energy is this energy; it equals the mass defect multiplied by c².
The energy equivalent of the extra mass a nucleus has over its separate nucleons. — A student who thinks binding adds mass picks this. A nucleus has LESS mass than its separated nucleons; the binding energy is the energy equivalent of this mass defect.
2 The masses of a proton, a neutron and a helium-4 nucleus (⁴₂He) are 938.27 MeV c⁻², 939.57 MeV c⁻² and 3727.38 MeV c⁻² respectively. What is the binding energy per nucleon of helium-4?
Answer and reasoning
7.075 MeV — With masses in MeV c⁻², the mass defect in MeV c⁻² equals the binding energy in MeV: 2(938.27) + 2(939.57) − 3727.38 = 28.30 MeV. Dividing by the 4 nucleons gives 7.075 MeV per nucleon.
28.30 MeV — A student who treats binding energy and binding energy per nucleon as the same quantity stops at 28.30 MeV. That is the binding energy of the whole nucleus; per nucleon it must be divided by A = 4.
931.8 MeV — A student who takes the binding energy to be the energy contained in the nucleus divides its rest energy by 4: 3727.38/4 = 931.8 MeV. Binding energy comes from the mass DEFECT, the small difference between the separated nucleons and the nucleus.
317.5 MeV — A student who reads ⁴₂He as 4 protons and 2 neutrons calculates 4(938.27) + 2(939.57) − 3727.38 = 1904.84 MeV and divides by 6 nucleons. Helium-4 has 2 protons and 2 neutrons; A = 4 is the total number of nucleons.
Working Mass defect Δm = 2m_p + 2m_n − m_He = 2(938.27) + 2(939.57) − 3727.38 = 3755.68 − 3727.38 = 28.30 MeV c⁻². Binding energy E_b = Δmc² = 28.30 MeV. Helium-4 has A = 4 nucleons, so E_b/A = 28.30 MeV/4 = 7.075 MeV.
3 Approximate binding energies per nucleon are: ²₁H 1.1 MeV; ⁴₂He 7.1 MeV; ²⁸₁₄Si 8.4 MeV; ⁵⁶₂₆Fe 8.8 MeV; ¹¹⁸₅₀Sn 8.5 MeV; ²³⁶₉₂U 7.6 MeV. Which of these processes would release energy?
Answer and reasoning
Splitting one ⁵⁶₂₆Fe nucleus into two ²⁸₁₄Si nuclei — A student who thinks any fission releases energy picks this. Iron-56 is at the peak of the curve; the silicon-28 products have a lower binding energy per nucleon (8.4 MeV against 8.8 MeV), so about 56 × 0.4 ≈ 22 MeV must be supplied.
Joining two ¹¹⁸₅₀Sn nuclei into one ²³⁶₉₂U nucleus — A student who judges stability by total binding energy notes that one ²³⁶U nucleus (about 236 × 7.6 ≈ 1790 MeV) has more than one ¹¹⁸Sn nucleus (about 1000 MeV). But the two tin nuclei together have about 2010 MeV: forming uranium lowers the binding energy per nucleon from 8.5 MeV to 7.6 MeV and needs about 210 MeV.
Separating one ⁴₂He nucleus into its four free nucleons — A student who thinks binding energy is released when a nucleus is pulled apart picks this. Separating helium-4 into its nucleons needs its whole binding energy, 4 × 7.1 ≈ 28 MeV, to be supplied.
Fusing two ²₁H nuclei to form one ⁴₂He nucleus — Both deuterium nuclei move far up the binding energy curve: the total binding energy rises from 2 × 2 × 1.1 = 4.4 MeV to 4 × 7.1 = 28.4 MeV, so about 24 MeV is released.
Working Energy is released if the total binding energy (A × binding energy per nucleon) increases. 2 ²H → ⁴He: 4(7.1) − 2 × 2(1.1) = 28.4 − 4.4 = +24 MeV (released). ⁵⁶Fe → 2 ²⁸Si: 56(8.4) − 56(8.8) = −22 MeV (must be supplied). 2 ¹¹⁸Sn → ²³⁶U: 236(7.6) − 236(8.5) = −212 MeV (must be supplied). ⁴He → 4 free nucleons: 0 − 4(7.1) = −28 MeV (must be supplied).
4 Radium-226 decays by α emission: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α. The atomic masses are 226.025410 u for radium-226, 222.017578 u for radon-222 and 4.002603 u for helium-4 (the electron masses cancel). Take 1 u = 1.661 × 10⁻²⁷ kg = 931.5 MeV c⁻², 1 eV = 1.60 × 10⁻¹⁹ J and c = 3.00 × 10⁸ m s⁻¹. How much energy is released in the decay of one radium-226 nucleus?
Answer and reasoning
2.6 × 10⁻²¹ J — A student who multiplies the mass by c instead of c² gets 8.685 × 10⁻³⁰ × 3.00 × 10⁸ = 2.6 × 10⁻²¹. Its unit would be kg m s⁻¹, which is not an energy; E = mc² needs c² = 9.00 × 10¹⁶ m² s⁻².
6.0 × 10⁻¹⁰ J — A student who takes the mass converted to be the α particle's mass gets 4.002603 × 1.661 × 10⁻²⁷ kg × c² = 6.0 × 10⁻¹⁰ J. The α particle is emitted, not destroyed; only the 0.005229 u decrease in total mass becomes energy.
7.8 × 10⁻¹³ J — The decrease in rest mass is 226.025410 − (222.017578 + 4.002603) = 0.005229 u = 8.685 × 10⁻³⁰ kg, so E = Δmc² = 8.685 × 10⁻³⁰ kg × 9.00 × 10¹⁶ m² s⁻² = 7.8 × 10⁻¹³ J (4.87 MeV).
7.8 × 10⁻¹⁹ J — A student who finds 0.005229 × 931.5 = 4.87 MeV and converts it with 1.60 × 10⁻¹⁹ J gets 7.8 × 10⁻¹⁹ J. That factor converts eV; 1 MeV = 1.60 × 10⁻¹³ J, so 4.87 MeV = 7.8 × 10⁻¹³ J.
Working Δm = m(Ra) − [m(Rn) + m(He)] = 226.025410 u − (222.017578 u + 4.002603 u) = 0.005229 u = 0.005229 × 1.661 × 10⁻²⁷ kg = 8.685 × 10⁻³⁰ kg. E = Δmc² = 8.685 × 10⁻³⁰ kg × (3.00 × 10⁸ m s⁻¹)² = 7.82 × 10⁻¹³ J ≈ 7.8 × 10⁻¹³ J. Check in MeV: 0.005229 u × 931.5 MeV u⁻¹ = 4.87 MeV = 4.87 × 1.60 × 10⁻¹³ J = 7.79 × 10⁻¹³ J ≈ 7.8 × 10⁻¹³ J.
5 Radioactive decay is described as random and spontaneous. Which statement correctly describes what this means?
Answer and reasoning
A nucleus decays when a collision or a rise in temperature gives it the energy it needs. — A student who expects decay to need a trigger, like most chemical changes, picks this. Decay is spontaneous: heating and collisions involve energies of a few eV at most, far too little to affect a nucleus, and they do not change the decay rate.
Nuclei decay roughly in the order in which they formed, but their ages cannot be known. — A student who thinks nuclei age picks this. An undecayed nucleus has the same chance of decaying in the next second however long it has existed, so there is no order of decay.
The count rate varies so unpredictably that the half-life of a sample cannot be measured. — A student who takes 'random' to mean 'without any pattern' picks this. Individual decays are unpredictable, but with very many nuclei the fraction decaying in a given time is predictable, so the half-life can be measured.
When a given nucleus decays cannot be predicted, and outside conditions do not alter its chance. — Random: the moment at which a particular nucleus decays cannot be predicted. Spontaneous: decay needs no external trigger, and its rate is not changed by temperature, pressure or chemical state.
6 Radon-222 has a half-life of 3.8 days. A particular radon-222 nucleus in a sample has not decayed after 3.8 days. What is the probability that this nucleus decays during the next 3.8 days?
Answer and reasoning
Greater than ½ but less than 1, as it has survived one half-life — A student who thinks a nucleus becomes 'due' to decay picks this. A surviving nucleus is identical to a newly formed one; its past has no effect on its chance of decaying, which stays ½ per half-life.
½, the same as for any other undecayed radon-222 nucleus — Nuclei do not age. Every undecayed radon-222 nucleus, however long it has existed, has a probability of ½ of decaying within the next half-life of 3.8 days.
1, since radon-222 decays completely in two half-lives — A student who thinks half-life is half the time to decay completely expects every nucleus to have gone by 7.6 days, so this one must decay in the next 3.8 days. In fact a quarter of the nuclei survive two half-lives; each survivor has a probability of ½ of decaying in each 3.8 days.
¼, since the chance of decay halves with each half-life — A student who transfers the falling activity of the sample to each nucleus reasons that the chance of ½ in the first half-life halves to ¼ in the second. The activity falls because fewer nuclei remain; each remaining nucleus still has a probability of ½ of decaying per half-life.
Working 3.8 days is one half-life. The nucleus has survived one half-life, but nuclei do not age: the probability that any undecayed radon-222 nucleus decays within the next half-life is ½, so P = ½.
7 Which of these radioactive decay equations is correct?
Answer and reasoning
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ⁰₀ν — A student who pairs the positron with the antineutrino, and so the electron with the neutrino, picks this. β⁻ decay emits an electron and an electron ANTINEUTRINO: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ⁰₀ν̄.
¹⁸₉F → ¹⁸₁₀Ne + ⁰₊₁e + ⁰₀ν — A student who thinks β⁺ decay turns a neutron into a proton raises Z to 10. Charge is then not conserved (9 ≠ 10 + 1). In β⁺ decay a proton becomes a neutron: ¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ⁰₀ν.
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α — Nucleon number: 226 = 222 + 4. Charge: 88 = 86 + 2. An α particle is a helium-4 nucleus, so radium (Z = 88) becomes radon (Z = 86).
¹³⁷ᵐ₅₆Ba → ¹³⁷₅₅Cs + γ — A student who thinks γ emission changes the element lowers Z. A γ photon has no charge and no nucleon number, so charge would not balance (56 ≠ 55). The excited barium-137m nucleus becomes ¹³⁷₅₆Ba in its ground state.
Working Check each equation for conservation of nucleon number and charge, and for the right particles. α: 226 = 222 + 4 and 88 = 86 + 2, correct. β⁻ (carbon-14): the numbers balance, but β⁻ decay emits an antineutrino, ⁰₀ν̄, not a neutrino. β⁺ (fluorine-18): charge 9 ≠ 10 + 1; the daughter is ¹⁸₈O. γ: charge 56 ≠ 55 + 0; γ emission leaves the nuclide as ¹³⁷₅₆Ba.
8 The four decays below involve nuclides used in medicine and industry. Which decay equation is correct?
Answer and reasoning
¹³¹₅₃I → ¹³¹₅₄Xe + ⁰₋₁e + ⁰₀ν̄ — Nucleon number: 131 = 131 + 0 + 0. Charge: 53 = 54 + (−1) + 0. In β⁻ decay a neutron becomes a proton, and an electron and an electron antineutrino are emitted.
²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α + ⁰₀ν — A student who adds a neutrino to every decay picks this. The numbers balance, but α decay produces only the daughter and the α particle: ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α.
¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ⁰₀ν̄ — A student who pairs the positron with the antineutrino picks this. β⁺ decay emits a positron and an electron NEUTRINO: ¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ⁰₀ν.
⁹⁰₃₈Sr → ⁸⁹₃₉Y + ⁰₋₁e + ⁰₀ν̄ — A student who thinks a β particle carries away a nucleon lowers A to 89. Nucleon number is then not conserved (90 ≠ 89 + 0). In β⁻ decay A is unchanged: ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e + ⁰₀ν̄.
Working Iodine-131 (β⁻): 131 = 131 + 0 + 0 and 53 = 54 − 1 + 0, with an antineutrino as required: correct. Americium-241 (α): the numbers balance, but α decay emits no neutrino. Fluorine-18 (β⁺): the numbers balance, but β⁺ decay emits a neutrino, not an antineutrino. Strontium-90 (β⁻): charge balances (38 = 39 − 1) but nucleon number does not (90 ≠ 89 + 0); the daughter is ⁹⁰₃₉Y.
9 Technetium-99m emits γ rays only and has a half-life of 6.0 h. It is injected into patients so that the organs that take it up can be imaged by a γ camera outside the body. Which statement is a correct reason for choosing technetium-99m?
Answer and reasoning
Its γ rays are strongly ionizing, so they give a strong signal at the camera. — A student who thinks the most penetrating radiation is also the most ionizing picks this. γ rays are the least ionizing radiation, which is why they escape the body; strongly ionizing α particles would be absorbed within it.
Its half-life is short because so little is injected, and a small sample halves quickly. — A student who thinks half-life depends on the amount of material picks this. Half-life is a property of the nuclide: technetium-99m has a half-life of 6.0 h whether the sample is a trace or a large quantity.
Its γ rays make the organ's own tissue radioactive, so that the organ shows up. — A student who thinks irradiation makes things radioactive picks this. γ rays do not make tissue radioactive; the organ shows up because the technetium-99m itself collects there and emits γ rays.
Its γ rays are weakly ionizing and most pass out of the body to the camera. — γ rays are the most penetrating and least ionizing radiation, so most pass out of the body to the camera and relatively little energy is deposited in the patient. The short half-life then limits the dose after the scan.
10 A small amount of sodium-24 (half-life 15 h), which emits β⁻ particles and γ rays, is added to water flowing in a pipe buried 1 m underground. A detector moved along the surface above the pipe records a much higher count rate above one point. Which statement explains this reading?
Answer and reasoning
β⁻ particles from the leaked tracer have the energy to cross a metre of soil. — A student who overestimates the penetration of β particles picks this. β⁻ particles are stopped by a few millimetres of aluminium; a metre of soil absorbs them all. Only the γ rays reach the surface.
The tracer decays faster once it is out of the pipe, so it emits more there. — A student who thinks surroundings affect decay picks this. Sodium-24 decays at the same rate inside or outside the pipe; the count rate rises because more tracer has collected close to the detector.
γ rays from the tracer that has leaked into the soil pass up through the ground. — Tracer escaping at the leak collects in the soil there. Its γ rays are penetrating enough to pass through a metre of soil, so the count rate rises above the leak. The short half-life means the tracer does not remain active in the ground for long.
The water's γ rays make the soil around the leak radioactive, and this is detected. — A student who thinks irradiated material becomes radioactive picks this. γ rays do not make soil radioactive; the soil at the leak is contaminated by the tracer itself, and it is the tracer's γ rays that are detected.
11 A paper mill monitors the thickness of its paper, about 0.1 mm thick, with a radioactive source above the moving paper and a detector below it. The count rate is used to adjust the rollers. Which type of source is most suitable, and why?
Answer and reasoning
β⁻, as the paper absorbs part of the β⁻ radiation, and more when it is thicker. — A thickness gauge needs radiation that the paper absorbs partly. β⁻ particles are partly absorbed by paper, and more are absorbed when it is thicker, so the count rate changes measurably with thickness.
α, as α particles are the most ionizing and also the most penetrating radiation. — A student who credits α particles with being both the most ionizing and the most penetrating picks this. α particles are the least penetrating: paper stops them all, so the count rate would be background whatever the thickness.
γ, as γ rays pass through the paper easily and so give a high count rate. — A student who wants the highest count rate picks this. γ rays pass through paper almost unaffected, so a small change in thickness makes almost no change in the count rate; the gauge could not detect it.
γ, as γ rays are the least ionizing radiation and so the safest to use. — A student who ranks hazard by ionizing ability alone picks this. γ rays are also the most penetrating, so they reach workers at a distance, and they are unsuitable anyway because paper hardly absorbs them.
12 Sealed packs of surgical instruments are sterilized by exposing them to intense γ radiation from a cobalt-60 source, which kills bacteria. Which statement about this process is correct?
Answer and reasoning
The instruments become radioactive and must be stored until their activity has fallen. — A student who does not distinguish radiation from radioactive material picks this. Exposure to γ rays does not make the instruments radioactive; they emit nothing once they are removed from the source.
The instruments do not become radioactive: they are irradiated, not contaminated. — γ rays pass through the packs and kill bacteria by ionizing their molecules, but they do not make the instruments radioactive. No cobalt-60 touches the instruments, so they are irradiated, not contaminated.
Some of the γ radiation stays trapped inside each sealed pack until the pack is opened. — A student who pictures radiation as a substance picks this. γ photons are absorbed or pass through within a tiny fraction of a second; nothing is stored in the pack.
The cobalt-60 source would decay faster, and need replacing sooner, if it were kept hot. — A student who thinks outside conditions affect decay picks this. Thermal energies of a few eV per atom cannot affect a nucleus, where the energies are MeV; the source's activity halves every 5.3 years whatever its temperature.
13 Cobalt-60 has a half-life of 5.3 years and decays into nickel-60, a stable solid. A sample contains a large number of cobalt-60 nuclei. Which statement about the sample is correct?
Answer and reasoning
After 5.3 years, about half of the cobalt-60 nuclei present now will remain. — Half-life is the time for the number of undecayed nuclei to halve. With a large number of nuclei, the fraction that decays in 5.3 years is very close to one half.
After 10.6 years, all of the cobalt-60 nuclei will have decayed. — A student who thinks half-life is half the time to decay completely picks this. After two half-lives a quarter of the cobalt-60 nuclei remain, and the number halves again in every further half-life.
After 5.3 years, the sample's mass will be about half of its mass now. — A student who thinks half of the material disappears picks this. Each decaying cobalt-60 nucleus becomes a nickel-60 nucleus of almost the same mass, which stays in the sample, so the mass hardly changes.
A sample twice as large would take 10.6 years to fall to half. — A student who thinks a larger sample takes longer to decay picks this. Half-life is a property of the nuclide: a sample twice as large has twice the activity but still halves in 5.3 years.
Working Half-life is the time for the number of undecayed nuclei to halve: after 5.3 years N = N₀/2; after 10.6 years (two half-lives) N = N₀/4, not zero.
14 To determine the half-life of a nuclide, a student measures the count rate from a sample and the background count rate. After subtracting the background, the count rate falls from 640 counts s⁻¹ at t = 0 to 80 counts s⁻¹ at t = 120 min. What is the half-life of the nuclide?
Answer and reasoning
69 min — A student who assumes a steady fall finds when a straight line from 640 to 80 would pass 320: 120 × 320/560 = 69 min. The fall is not linear: it is fastest at first, so the first halving takes less time than this.
15 min — A student who takes the factor of 8 to be the number of half-lives gets 120/8 = 15 min. A factor of 8 = 2³ is three halvings.
30 min — A student who counts the four values 640, 320, 160 and 80 as four half-lives gets 120/4 = 30 min. There are only three halvings between them.
40 min — 640 → 320 → 160 → 80 counts s⁻¹ is three halvings, so 120 min is three half-lives and T½ = 40 min.
Working 640/80 = 8 = 2³, so the corrected count rate has halved three times: 640 → 320 → 160 → 80 counts s⁻¹. Three half-lives = 120 min, so T½ = 120 min/3 = 40 min.
15 Carbon-14 (half-life 5730 years) is used to date wood. Equal masses of carbon from a living tree and from an ancient wooden tool give background-corrected count rates of 96 and 12 counts per minute. Assuming that the proportion of carbon-14 in living wood has not changed, how old is the tool?
Answer and reasoning
2.29 × 10⁴ y — A student who counts the four values 96, 48, 24 and 12 as four half-lives gets 4 × 5730 = 2.29 × 10⁴ years. There are only three halvings between them.
1.00 × 10⁴ y — A student who assumes a steady fall reasons that the count rate drops by 48 counts per minute every 5730 years, so a drop of 84 takes 84/48 × 5730 = 1.00 × 10⁴ years. The count rate does not fall linearly; it halves every 5730 years.
4.58 × 10⁴ y — A student who takes the factor of 8 to be the number of half-lives gets 8 × 5730 = 4.58 × 10⁴ years. A factor of 8 = 2³ means three half-lives.
1.72 × 10⁴ y — 96 → 48 → 24 → 12 is three halvings, so the carbon in the tool has been decaying for three half-lives: 3 × 5730 = 1.72 × 10⁴ years. Carbon-14 suits this use because its half-life is comparable with the ages being measured.
Working 96/12 = 8 = 2³, so three half-lives have passed: 96 → 48 → 24 → 12 counts per minute. Age = 3 × 5730 y = 17 190 y ≈ 1.72 × 10⁴ y.
16 A detector placed near a source of half-life 2.0 days records 250 counts min⁻¹. With the source removed, it records a background of 50 counts min⁻¹. What will the detector record with the source in the same place 6.0 days later?
Answer and reasoning
75 counts min⁻¹ — Only the source's contribution halves. Corrected rate now: 250 − 50 = 200 counts min⁻¹. After 3 half-lives: 200/8 = 25 counts min⁻¹. The detector still records the background too: 25 + 50 = 75 counts min⁻¹.
31 counts min⁻¹ — A student who treats the background as part of the source halves the whole reading three times: 250/8 = 31 counts min⁻¹. Background comes from other sources and stays at about 50 counts min⁻¹.
25 counts min⁻¹ — A student who forgets that the detector always records background gives only the source's contribution, 200/8 = 25 counts min⁻¹. The background of 50 counts min⁻¹ must be added back.
50 counts min⁻¹ — A student who thinks half-life is half the time to decay completely expects the source to be exhausted after 4.0 days, leaving only background. After 6.0 days (three half-lives) the source still gives 25 counts min⁻¹.
Working Corrected count rate now = 250 − 50 = 200 counts min⁻¹. 6.0 days = 3 half-lives, so the corrected count rate becomes 200/2³ = 25 counts min⁻¹. Measured count rate = 25 + 50 = 75 counts min⁻¹.
17 The stable isotopes of carbon are ¹²₆C and ¹³₆C. Carbon-14 (¹⁴₆C) is radioactive. Which decay does carbon-14 undergo, and why? HL
Answer and reasoning
β⁺ decay, as this turns one of its surplus neutrons into a proton — A student who reverses the two kinds of β decay picks this. β⁺ decay turns a PROTON into a neutron, which would raise N/Z further. It is β⁻ decay that turns a neutron into a proton.
β⁻ decay, as a neutron changes into a proton and lowers its N/Z ratio — With 8 neutrons and 6 protons (N/Z = 1.33), carbon-14 is neutron-rich compared with the stable isotopes (N/Z = 1.00 and 1.17). β⁻ decay turns a neutron into a proton, giving ¹⁴₇N (N/Z = 1), which is stable.
α decay, as the α particle carries two of its surplus neutrons away with it — A student who sees α emission as a way to lose neutrons picks this. An α particle also removes two protons: ¹⁰₄Be would have N/Z = 1.5, further from stability. α decay is typical of very heavy nuclides.
γ emission, which lets it shed its surplus neutrons as energy — A student who thinks γ emission changes the nucleus picks this. A γ photon has no nucleon number; γ emission changes only the energy of a nucleus, not its numbers of neutrons and protons.
18 Almost all the α particles emitted by radium-226 have one of two energies, 4.78 MeV or 4.60 MeV. The daughter nucleus, radon-222, emits γ photons of energy 0.19 MeV. (The α energies differ by slightly less than 0.19 MeV because the recoiling radon nucleus takes a small share of the energy.) Which conclusion do these observations support? HL
Answer and reasoning
The radium sample is a mixture of two isotopes, each emitting α particles of one energy. — A student who thinks each nuclide has only one α energy picks this. One nuclide can emit α particles of different discrete energies; the γ photon energy matching their difference shows that both decays end in radon-222, one of them in an excited state.
The γ photons come from electrons of the radon atom falling between atomic energy levels. — A student who knows only atomic energy levels picks this. The γ energy matches the difference between the α energies, which is set by the energy left in the radon NUCLEUS; the γ photon comes from a nuclear transition.
Radon-222 has discrete energy levels, one of them 0.19 MeV above its ground state. — Some decays leave the radon nucleus in an excited state, so their α particles carry about 0.19 MeV less energy; the nucleus then drops to its ground state, emitting a 0.19 MeV γ photon. Only discrete energies occur, so the nuclear energy levels are discrete.
The γ photons are emitted as each radon-222 nucleus changes into a different nuclide. — A student who thinks γ emission changes the nuclide picks this. γ emission leaves Z and A unchanged; the excited radon-222 nucleus becomes radon-222 in its ground state.
Working Difference in α energies = 4.78 MeV − 4.60 MeV = 0.18 MeV, which matches the 0.19 MeV γ photon once the small recoil energy of the radon nucleus is allowed for. So one α decay mode leaves radon-222 in an excited state about 0.19 MeV above its ground state, from which it emits the γ photon: evidence for discrete nuclear energy levels.
19 The activity per gram of carbon in a sample of bone is 23% of that in living tissue. Carbon-14 has a half-life of 5730 years. Assuming the proportion of carbon-14 in living tissue has not changed, what is the age of the bone? HL
Answer and reasoning
8.4 × 10³ y — A student who uses λ = 1/T½ gets t = 5730 × ln(1/0.23) = 8.4 × 10³ years. The decay constant is λ = ln 2/T½; without the factor 0.693 the age comes out too small.
1.2 × 10⁴ y — With λ = ln 2/5730 y = 1.21 × 10⁻⁴ y⁻¹, 0.23 = e^(−λt) gives t = ln(1/0.23)/λ = 1.2 × 10⁴ years: between two and three half-lives, as expected for a fall to 23%.
5.3 × 10³ y — A student who presses 'log' instead of 'ln' gets t = log₁₀(1/0.23)/λ = 5.3 × 10³ years, too small by a factor of 2.30. Only ln undoes e^(−λt).
2.5 × 10⁴ y — A student who takes the number of half-lives to be the factor 1/0.23 = 4.35 gets 4.35 × 5730 = 2.5 × 10⁴ years. A fall by a factor of 4.35 is only 2.12 half-lives, because 2^2.12 = 4.35.
Working A = A₀e^(−λt) with A/A₀ = 0.23. λ = ln 2/T½ = 0.693/5730 y = 1.210 × 10⁻⁴ y⁻¹. t = ln(A₀/A)/λ = ln(1/0.23)/(1.210 × 10⁻⁴ y⁻¹) = 1.470/(1.210 × 10⁻⁴ y⁻¹) = 1.21 × 10⁴ y ≈ 1.2 × 10⁴ y.
20 A nuclide has a decay constant λ = 0.10 s⁻¹. What is the probability that a particular undecayed nucleus of this nuclide decays within the next 5.0 s? HL
Answer and reasoning
0.50 — A student who treats λ as exactly the probability per unit time multiplies: λt = 0.10 × 5.0 = 0.50. λt approximates the probability only when λt is small; the exact value is 1 − e^(−λt) = 0.39.
0.61 — A student who takes e^(−λt) as the probability of decay gets e^(−0.50) = 0.61. That is the probability that the nucleus SURVIVES; the probability that it decays is 1 − 0.61 = 0.39.
0.39 — The probability of surviving 5.0 s is e^(−λt) = e^(−0.50) = 0.61, so the probability of decaying is 1 − 0.61 = 0.39. Here λt = 0.50 is not small, so λt is not a good approximation.
0.10 — A student who takes λ itself as the probability of decay, whatever the time interval, answers 0.10. λ has the unit s⁻¹; it must be combined with the time: 1 − e^(−λt) = 0.39 for 5.0 s.
Working Probability of surviving for t = 5.0 s: e^(−λt) = e^(−0.10 s⁻¹ × 5.0 s) = e^(−0.50) = 0.607. Probability of decaying = 1 − 0.607 = 0.393 ≈ 0.39. (λt = 0.50 is not small, so the approximation P ≈ λt = 0.50 is poor.)
21 Iodine-131 has a half-life of 8.0 days and a molar mass of 131 g mol⁻¹. What is the activity of 1.0 μg of iodine-131? Take N_A = 6.02 × 10²³ mol⁻¹ and 1 day = 8.64 × 10⁴ s. HL
Answer and reasoning
6.6 × 10⁹ Bq — A student who uses λ = 1/T½ gets (4.60 × 10¹⁵)/(6.91 × 10⁵ s) = 6.6 × 10⁹ Bq. The decay constant is λ = ln 2/T½, which is 0.693 times smaller.
3.3 × 10⁹ Bq — A student who assumes the nuclei decay at a steady rate spreads half of them, N/2, evenly over one half-life: (2.30 × 10¹⁵)/(6.91 × 10⁵ s) = 3.3 × 10⁹ Bq. The rate is not steady; at this instant A = λN = 4.6 × 10⁹ Bq.
4.6 × 10⁶ Bq — A student who converts the mass to 1.0 × 10⁻⁹ kg but keeps the molar mass as 131 g mol⁻¹ finds 1000 times too few nuclei, 4.60 × 10¹², and an activity of 4.6 × 10⁶ Bq. Both masses must be in the same unit.
22 Samples X and Y each contain 4.0 × 10¹⁸ undecayed nuclei. The nuclide in X has a half-life of 2.0 h; the nuclide in Y has a half-life of 8.0 h. Which statement about their activities at this instant is correct? HL
Answer and reasoning
Y has four times the activity of X, as its nuclei remain radioactive for longer. — A student who links a long half-life with being 'more radioactive' picks this. A longer half-life means a smaller decay constant, so for equal N the activity is SMALLER; Y will simply keep emitting for longer.
X has four times the activity of Y, as its decay constant is four times larger. — A = λN with λ = ln 2/T½. N is the same, and λ for X is ln 2/2.0 h, four times λ for Y (ln 2/8.0 h), so X's activity is four times Y's.
They have equal activities, as they contain equal numbers of nuclei still undecayed. — A student who thinks activity depends only on the number of nuclei picks this. A = λN: equal N gives equal activity only for nuclides with the same decay constant.
Their activities cannot be compared, as the decay of each nucleus is random. — A student who takes 'random' to mean 'unpredictable in every respect' picks this. With 4.0 × 10¹⁸ nuclei the decay rate is highly predictable: A = λN for each sample.
Working A = λN and λ = ln 2/T½. λ_X = 0.693/2.0 h = 0.347 h⁻¹; λ_Y = 0.693/8.0 h = 0.0866 h⁻¹. With equal N, A_X/A_Y = λ_X/λ_Y = 8.0/2.0 = 4.
23 To determine the half-life of a nuclide, a student plots a graph of ln(corrected count rate) against time. The graph is a straight line with a gradient of −0.0231 min⁻¹. What is the half-life of the nuclide? HL
Answer and reasoning
30 min — ln(count rate) = ln(C₀) − λt, so λ = −gradient = 0.0231 min⁻¹ and T½ = ln 2/λ = 0.693/0.0231 min⁻¹ = 30 min.
43 min — A student who takes T½ = 1/λ gets 1/0.0231 = 43 min. The half-life is ln 2/λ, which includes the factor 0.693.
22 min — A student who treats λt as the fraction decayed sets λT½ = 0.5, giving 0.5/0.0231 = 22 min. The fraction decayed is 1 − e^(−λt), which reaches 0.5 when λt = ln 2 = 0.693.
13 min — A student who uses log₁₀ 2 instead of ln 2 gets 0.301/0.0231 = 13 min. The graph uses natural logarithms, so T½ = ln 2/λ.
Working For C = C₀e^(−λt), ln C = ln C₀ − λt, so the gradient is −λ and λ = 0.0231 min⁻¹. T½ = ln 2/λ = 0.693/0.0231 min⁻¹ = 30.0 min ≈ 30 min.
⁴⁰₁₈Ar and ⁴⁰₁₉K — A student who reads the upper number as the proton number sees '40 protons' in both and takes the lower numbers 18 and 19 as differing neutron numbers. The upper number is the nucleon number A. These nuclides have Z = 18 and Z = 19, so they are different elements (argon and potassium) that happen to share A = 40.
³¹₁₅P and ³²₁₆S — A student who has the definition reversed picks two nuclides with the same number of neutrons (31 − 15 = 16 and 32 − 16 = 16) and different numbers of protons. Isotopes share the PROTON number; Z = 15 and Z = 16 are phosphorus and sulfur, different elements.
²³₁₁Na and ²³₁₁Na⁺ — A student who confuses isotopes with ions picks an atom and its ion. Both have the same nucleus (Z = 11, A = 23); they differ only in the number of electrons, so they are the same nuclide, not two isotopes.
³⁵₁₇Cl and ³⁷₁₇Cl — Both nuclides have proton number Z = 17, so both are chlorine. They differ in nucleon number (35 and 37), so chlorine-35 has 18 neutrons and chlorine-37 has 20: the same element, different numbers of neutrons, which is what isotopes are.
25 In the fusion reaction ²₁H + ²₁H → ³₂He + ¹₀n the rest masses are 1875.61 MeV c⁻² for ²₁H, 2808.39 MeV c⁻² for ³₂He and 939.57 MeV c⁻² for the neutron. Take c = 3.00 × 10⁸ m s⁻¹. How much energy is released in one reaction?
Answer and reasoning
939.57 MeV — A student who takes the energy released to be the energy equivalent of the emitted particle's mass answers with the neutron's rest energy. The neutron is not converted into energy; it is a product that still exists. Only the DECREASE in total rest mass, 3.26 MeV c⁻², becomes kinetic energy.
3.04 × 10³ MeV — A student who multiplies every mass difference by 931.5 gets 3.26 × 931.5 = 3.04 × 10³ MeV. The factor 931.5 MeV converts a mass in u; these masses are already in MeV c⁻², so 3.26 MeV c⁻² is 3.26 MeV of energy with no further conversion.
3.26 MeV — Total rest mass before = 2 × 1875.61 = 3751.22 MeV c⁻²; after = 2808.39 + 939.57 = 3747.96 MeV c⁻². The decrease is 3.26 MeV c⁻², and E = Δmc² = 3.26 MeV c⁻² × c² = 3.26 MeV: the c² cancels, so a mass in MeV c⁻² gives its energy in MeV directly.
2.93 × 10¹⁷ MeV — A student who multiplies by c² numerically gets 3.26 × (3.00 × 10⁸)² = 2.93 × 10¹⁷. In the unit MeV c⁻² the c² is part of the unit and cancels: 3.26 MeV c⁻² × c² = 3.26 MeV. The value of c in m s⁻¹ is needed only when the mass is in kg.
Working Rest mass of reactants = 2 × 1875.61 = 3751.22 MeV c⁻². Rest mass of products = 2808.39 + 939.57 = 3747.96 MeV c⁻². Δm = 3751.22 − 3747.96 = 3.26 MeV c⁻². E = Δmc² = 3.26 MeV c⁻² × c² = 3.26 MeV (the c² cancels; c = 3.00 × 10⁸ m s⁻¹ is not used numerically). Accepted value 3.27 MeV. Check: 3.26 MeV = 3.26 × 1.60 × 10⁻¹³ J = 5.22 × 10⁻¹³ J, consistent with Δm = 3.26/931.5 u = 0.003500 u = 5.81 × 10⁻³⁰ kg and E = 5.81 × 10⁻³⁰ × (3.00 × 10⁸)² = 5.23 × 10⁻¹³ J.
26 A helium-4 nucleus contains two protons and two neutrons. Between which pairs of particles does the strong nuclear force act?
Answer and reasoning
Only between a proton and a neutron; two protons or two neutrons do not attract each other. — A student who sees neutrons as the only 'glue' picks this. The strong force acts between any two nucleons, including two protons and two neutrons; the neutrons are not the sole source of attraction.
Between every pair of nucleons: proton–proton, neutron–neutron and proton–neutron. — The strong nuclear force is an attraction between nucleons that does not depend on their charge. It acts between the two protons, between the two neutrons and between each proton–neutron pair, and it is what holds the nucleus together against the electric repulsion of the protons.
Between every pair except the two protons, which repel each other as they carry like charges. — A student who thinks the strong force depends on charge, like the electric force, excludes the proton pair. The two protons DO repel electrically, but the strong force between them is attractive and larger at nuclear separations; the strong force takes no account of charge.
Between every pair of nucleons, and between each nucleon and the atom's two electrons. — A student who thinks the strong force holds the whole atom together includes the electrons. Electrons do not feel the strong force at all: they are held in the atom by the electric attraction of the nucleus, at separations far greater than the force's range.
27 A nucleus emits a β⁺ particle, and the daughter nucleus then emits a γ photon. How do the proton number Z and neutron number N of the final nucleus compare with those of the original nucleus?
Answer and reasoning
Z is one more, and N is one less. — A student who has the two β decays the wrong way round picks this. It is β⁻ decay that turns a neutron into a proton. In β⁺ decay the emitted particle is positive, so the nucleus loses one unit of positive charge: a proton becomes a neutron.
Z is one less; N is unchanged. — A student who thinks the β⁺ particle carries away a nucleon (a proton) leaves N unchanged and lowers A. A positron has nucleon number zero; the proton is not ejected but CHANGES into a neutron, so N rises by one and A stays the same.
Z is one less and N is one more. — In β⁺ decay a proton in the nucleus changes into a neutron, emitting a positron and a neutrino: Z falls by one and N rises by one (A unchanged). γ emission carries away energy only, leaving Z and N as they are.
Neither Z nor N has changed. — A student who thinks β particles come from outside the nucleus, like orbital electrons, leaves the nucleus unchanged. The positron is created inside the nucleus when a proton becomes a neutron, so Z falls by one and N rises by one; only the γ emission leaves Z and N unchanged.
Working β⁺ decay: p → n + e⁺ + ν, so Z → Z − 1, N → N + 1, A unchanged. γ emission: the excited daughter drops to a lower energy state; Z, N and A are all unchanged. Overall: Z − 1, N + 1.
28 Fluorine-18, used in PET scanning, decays by β⁺ emission: ¹⁸₉F → ¹⁸₈O + e⁺ + X. What is the particle X?
Answer and reasoning
an antineutrino, ν̄, from the nucleus — A student who pairs the antineutrino with the positron 'because both are antimatter' picks this. It is the other way round: β⁻ decay emits an electron with an ANTIneutrino, and β⁺ decay emits a positron with a neutrino.
a neutron, n, from the nucleus — A student who confuses the neutrino with the neutron picks this. A neutron has nucleon number 1, which would break the balance of nucleon number (18 ≠ 18 + 1). In β⁺ decay a proton becomes a neutron that STAYS in the nucleus; the light neutral particle emitted is a neutrino.
a γ photon emitted by the nucleus — A student who takes any uncharged emission to be a photon picks this. A γ photon is emitted when an excited nucleus loses energy; it is not part of the β⁺ process itself. The particle created alongside the positron is a neutrino.
a neutrino, ν, from the nucleus — β⁺ decay is p → n + e⁺ + ν: the positron is accompanied by a neutrino. The neutrino has zero charge and nucleon number zero, so both sides of the equation balance (charge 9 = 8 + 1 + 0; nucleon number 18 = 18 + 0 + 0).
29 With no radioactive source near it, a detector still records a count rate. Which statement about this background count rate is correct?
Answer and reasoning
It is the same in every minute, so a single one-minute reading gives its value exactly and no repeat readings are needed. — A student who thinks of background as a steady value picks this. Background counts come from random decays and random cosmic-ray arrivals, so they fluctuate from minute to minute; a reliable value needs a long counting time or the average of several readings.
It is radiation that has soaked into the detector from sources used earlier, and it fades once the detector is left unused. — A student who thinks radiation can be stored in an object picks this. Radiation is not a substance that collects in a detector; the background is present in any new detector, anywhere, because cosmic rays and natural radioactive nuclides surround us.
It comes from cosmic rays and from radioactive nuclides in rock, air and buildings and is subtracted from each reading. — Background radiation arrives from cosmic rays and from natural radioactive nuclides (for example radon in the air and potassium-40 in rocks and buildings). A detector records it whether or not a source is present, so it is measured separately and subtracted from each reading to give the source's corrected count rate.
It needs to be subtracted only from the first reading taken with a source; later readings then show the source alone. — A student who thinks subtraction 'removes' the background from the detector picks this. The detector goes on recording background in every reading; the subtraction is a correction the experimenter applies to each reading, not a change to the detector.
30 The deuteron, ²₁H, is a stable nucleus made of one proton and one neutron; its binding energy is 2.2 MeV. The neutron carries no electric charge. Which conclusion do these facts support? HL
Answer and reasoning
An attractive force other than the electric force acts between a proton and a neutron. — The neutron is uncharged, so no electric force can act between it and the proton, yet the two are bound with 2.2 MeV. Some other attraction must act between nucleons: this is the strong nuclear force, which acts between nucleons whether or not they are charged.
The gravitational attraction between the two nucleons is strong enough to bind them together. — A student who thinks nucleons are close enough for gravity to matter picks this. For a proton and a neutron about 2 × 10⁻¹⁵ m apart the gravitational potential energy is of the order of 10⁻³⁰ eV, some 10³⁶ times smaller than the 2.2 MeV binding energy.
The proton's electric field attracts the neutron, as it attracts the electron in a hydrogen atom. — A student who thinks the binding force is electric in nature picks this. The electric force acts only on charged particles; the neutron has zero charge, so the proton's electric field exerts no force on it at all. A different force must be responsible.
Pulling the proton and neutron apart from each other would release 2.2 MeV of energy. — A student who thinks binding energy is released when a nucleus is broken up picks this. The binding energy is the energy that must be SUPPLIED to separate the nucleons; 2.2 MeV was released when the deuteron formed, and the same amount is needed to pull it apart.
31 Light stable nuclides have roughly equal numbers of neutrons and protons, but heavy stable nuclides have many more neutrons than protons: ²⁰⁸₈₂Pb has N/Z ≈ 1.5. Why do heavy nuclei need this excess of neutrons to be stable? HL
Answer and reasoning
Extra neutrons add strong-force attraction, which acts only between near neighbours, without adding electric repulsion, which acts between every pair of protons. — The electric repulsion is long-range and acts between every pair of protons, so it grows rapidly as Z increases. The strong force is short-range, so each nucleon is attracted only by its near neighbours. Adding neutrons adds attraction with no extra repulsion, which is what a large nucleus needs to stay bound.
The neutrons sit between the protons and shield them from one another's electric repulsion, so more protons need more neutrons between them. — A student who pictures neutrons as insulating spacers picks this. Neutrons do not shield electric forces: the repulsion between two protons is unchanged by what lies between them. Neutrons help by supplying extra strong-force attraction, not by blocking repulsion.
The strong force acts only between a proton and a neutron, so each proton must be surrounded by more than one neutron to be held in place. — A student who thinks neutrons are the only 'glue' picks this. The strong force acts between all pairs of nucleons, including two protons. Extra neutrons are needed because they add attraction without adding repulsion, not because protons cannot attract each other.
The extra neutrons add mass, and the greater gravitational attraction between all the nucleons then holds the larger nucleus together. — A student who thinks gravity plays a part in binding nuclei picks this. Gravitational attraction between nucleons is about 10³⁶ times weaker than the electric repulsion between two protons and has no role in nuclear stability; the extra binding comes from the strong force.
32 The binding energy per nucleon peaks at 8.8 MeV for ⁵⁶₂₆Fe and is roughly constant above a nucleon number of about 60, falling slowly to 7.9 MeV for ²⁰⁸₈₂Pb. Which statement is correct? HL
Answer and reasoning
The total binding energy of lead-208 is about 3.3 times that of iron-56, though each nucleon in it is less tightly bound. — Total binding energy = (binding energy per nucleon) × A. For lead-208: 7.9 MeV × 208 = 1.6 × 10³ MeV; for iron-56: 8.8 MeV × 56 = 4.9 × 10² MeV; the ratio is 3.3. Because E_b/A is roughly constant, the total E_b is roughly proportional to A, even though the per-nucleon value is slightly lower for lead.
The total binding energy of iron-56 is greater than that of lead-208, since its binding energy per nucleon is greater. — A student who uses binding energy and binding energy per nucleon interchangeably picks this. Iron-56 has the larger value PER NUCLEON, but lead-208 has almost four times as many nucleons, so its total binding energy (about 1600 MeV) far exceeds iron-56's (about 490 MeV).
The total binding energy of lead-208 is about 14 times that of iron-56, since every nucleon in it attracts every other nucleon. — A student who assumes the strong force has a long range expects the number of attracting pairs, and so the total binding energy, to grow roughly as A², giving (208/56)² ≈ 14. The near-constant E_b/A shows the total grows in proportion to A, giving a ratio of only about 3.3: evidence that each nucleon is bound only to its near neighbours.
A lead-208 nucleus stores about 1.6 × 10³ MeV of binding energy, which it releases when it takes part in a nuclear reaction. — A student who thinks binding energy is energy stored in the nucleus picks this. 1.6 × 10³ MeV is the energy that would have to be SUPPLIED to separate all 208 nucleons; a reaction releases only the difference between the total binding energies of the products and the reactants.
Working Total binding energy E_b = (E_b/A) × A. Lead-208: 7.9 MeV × 208 = 1643 MeV ≈ 1.6 × 10³ MeV. Iron-56: 8.8 MeV × 56 = 492.8 MeV ≈ 4.9 × 10² MeV. Ratio = 1643/492.8 = 3.33 ≈ 3.3. (If E_b grew as A², the ratio would be (208/56)² = 13.8 ≈ 14.)
33 Thorium-228 decays by α emission to radium-224, which has an excited state 0.08 MeV above its ground state. The α particles emitted when radium-224 is formed in its ground state have kinetic energy 5.42 MeV. Ignoring the recoil of the radium nucleus, which statement about the α particles emitted when radium-224 is formed in its excited state is correct? HL
Answer and reasoning
They have kinetic energy 5.50 MeV, as they carry the extra 0.08 MeV that the excited radium nucleus possesses. — A student who attaches the 'extra' excitation energy to the α particle adds 0.08 MeV. The energy released is fixed; energy kept by the daughter as excitation is NOT available to the α particle, so these α particles have 0.08 MeV LESS, not more.
They also have kinetic energy 5.42 MeV, because a given nuclide emits α particles of a single energy only. — A student who thinks each nuclide has a single α energy picks this. One nuclide can emit α particles of several discrete energies, one for each energy level of the daughter in which it can be left; the energy kept by the excited daughter reduces the α energy to 5.34 MeV.
They have kinetic energies anywhere between 5.34 MeV and 5.42 MeV, as the nucleus can be left with any energy up to 0.08 MeV. — A student who thinks nuclear energy is continuous picks this. The radium nucleus can be left only in its ground state or in the discrete excited state 0.08 MeV above it, so the α particles have only the two energies 5.42 MeV and 5.34 MeV, not a continuous range.
They have kinetic energy 5.34 MeV, and a 0.08 MeV γ photon is emitted as the radium nucleus falls to its ground state. — The energy released in the decay is fixed. If the radium nucleus keeps 0.08 MeV as excitation energy, the α particle receives 5.42 − 0.08 = 5.34 MeV. The excited nucleus then drops to its ground state, emitting the 0.08 MeV as a γ photon: two discrete α energies and one discrete γ energy, as discrete nuclear energy levels require.
Working Energy released is the same in every decay of thorium-228 to radium-224. Ground-state decay: E_α = 5.42 MeV (recoil ignored). Decay to the excited state: the daughter retains 0.08 MeV, so E_α = 5.42 − 0.08 = 5.34 MeV. The excited radium-224 nucleus then emits a γ photon of energy 0.08 MeV. (Measured values: α energies 5.423 MeV and 5.340 MeV; γ energy 0.084 MeV.)
34 The α particles from a given decay to the daughter's ground state all have the same kinetic energy, whereas the β⁻ particles from a given decay have a continuous range of kinetic energies up to a maximum. Which statement correctly explains this difference? HL
Answer and reasoning
α decay produces two bodies, so conservation of momentum fixes the α particle's share of the energy; β⁻ decay produces three, so the electron's share can vary. — With two products, conservation of momentum and energy together fix how the released energy is divided, so every α particle has the same energy. With three products (daughter, electron and antineutrino) the energy can be divided in many ways, so the electron's energy varies from zero up to the maximum, and the antineutrino takes the rest.
Both kinds of particle leave the nucleus with a single energy, but β⁻ particles lose varying amounts of energy in collisions before they escape the source. — A student who looks for an ordinary cause of the spread picks this. The spread is present at emission, and α particles, being far more ionizing, would lose MORE energy in collisions than β⁻ particles, yet their spectrum is discrete. The continuous β⁻ spectrum needs a third particle.
Every α decay of a nuclide releases exactly the same energy, but the energy released varies from one β⁻ decay to the next, so the electron's energy varies with it. — A student who thinks nuclear energy levels are continuous picks this. The daughter nucleus of a β⁻ decay, like that of an α decay, can be left only in discrete energy states, so the energy released is the same in every decay; the variation lies in how it is shared with the antineutrino.
The β⁻ particle shares the released energy with a γ photon emitted at the same instant, whereas an α particle takes all of the energy. — A student who takes the third particle to be a photon picks this. γ photons are emitted later, by an excited daughter nucleus, with discrete energies; they are not produced in every decay. The particle that shares the energy in every β⁻ decay is the antineutrino.
35 A sample contains 8.0 × 10²⁰ nuclei of iodine-131, which has a half-life of 8.0 days. How many iodine-131 nuclei remain after 20 days? HL
Answer and reasoning
3.2 × 10²⁰ — A student who divides by the number of half-lives gets 8.0 × 10²⁰/2.5 = 3.2 × 10²⁰. Each half-life halves the number, so 2.5 half-lives divide it by 2^2.5 = 5.66, not by 2.5.
1.4 × 10²⁰ — λ = ln 2/T½ = 0.693/8.0 days = 0.0866 day⁻¹. N = N₀e^(−λt) = 8.0 × 10²⁰ × e^(−0.0866 × 20) = 8.0 × 10²⁰ × e^(−1.73) = 8.0 × 10²⁰ × 0.177 = 1.4 × 10²⁰. This lies between the values for 2 half-lives (2.0 × 10²⁰) and 3 half-lives (1.0 × 10²⁰), as it should for 2.5 half-lives.
3.8 × 10²⁰ — A student who uses log₁₀ 2 = 0.301 instead of ln 2 = 0.693 gets λ = 0.0376 day⁻¹ and N = 8.0 × 10²⁰ × e^(−0.753) = 3.8 × 10²⁰. The decay law uses e, so its inverse is the natural logarithm: λ = ln 2/T½.
2.0 × 10²⁰ — A student who rounds 20 days down to 2 whole half-lives (16 days) gets 8.0 × 10²⁰/4 = 2.0 × 10²⁰. The decay law N = N₀e^(−λt) applies at any time; after 2.5 half-lives the number is 8.0 × 10²⁰/2^2.5 = 1.4 × 10²⁰.
36 Technetium-99m, used in medical imaging, has a half-life of 6.0 h. What is its decay constant? HL
Answer and reasoning
4.6 × 10⁻⁵ s⁻¹ — A student who uses λ = 1/T½ gets 1/(2.16 × 10⁴ s) = 4.6 × 10⁻⁵ s⁻¹. The half-life is the time for N to fall to N₀/2, so e^(−λT½) = ½ and λ = ln 2/T½; the factor 0.693 cannot be omitted.
1.4 × 10⁻⁵ s⁻¹ — A student who uses log₁₀ 2 = 0.301 instead of ln 2 = 0.693 gets 0.301/(2.16 × 10⁴ s) = 1.4 × 10⁻⁵ s⁻¹. The decay law is exponential in e, so it is the natural logarithm that appears: λ = ln 2/T½.
1.2 × 10⁻¹ s⁻¹ — A student who divides ln 2 by 6.0 without converting the hours gets 0.12, which is λ in h⁻¹, not s⁻¹. λ has the unit of 1/time, so its value depends on the time unit: 0.12 h⁻¹ = 0.12/3600 s⁻¹ = 3.2 × 10⁻⁵ s⁻¹.
That was your twenty minutes. Real practice on E.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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