IB Physics HL E.3 Radioactive decay Questions

Practise IB Physics HL E.3 by analysing decay chains, activity data, detection statistics and nuclear energy evidence.

271 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Define isotopes by equal proton numbers and different neutron numbers; a nuclide is specified by its proton and neutron constitution.
Recognize isotope evidence from nuclear-mass measurements or mass-spectrometer ion paths; chemical similarity does not imply equal half-lives.
Define nuclear binding energy as the energy needed to completely separate nucleons, or the energy released when they assemble into the nucleus.
Calculate Δm = Zmₚ + (A − Z)mₙ − Mnucleus and use Ebind = Δmc²; distinguish nuclear mass from neutral-atom mass.
Convert mass defect in u to binding energy with 931.5 MeV/u; divide by A only when energy per nucleon is requested.
To infer nuclear mass from binding energy, subtract Ebind/c² from the free-nucleon mass sum and show the binding term explicitly.
For reactions conserving proton and neutron composition, weight each BE/A by A and take total product BE minus total reactant BE for released Q.
For the complete Am-241 decay task, combine alpha-decay bookkeeping, mass loss and binding-energy totals; the supplied data give Q = 7.60 MeV.
Sketch BE/A with a rapid light-nucleus rise, a broad iron–nickel maximum and a gradual heavy-nucleus decline; pass through supplied graph points.
Explain energy release in light fusion or heavy fission through increased BE/A toward the peak; ordinary stellar fusion beyond iron requires energy.
Read BE/A values from the graph, multiply by nucleon counts and form total binding-energy differences before converting the released energy.
Use Q = (Σmreactants − Σmproducts)c² with consistent mass data; check whether electrons cancel before using atomic masses.
For nuclear beta-plus masses, subtract the emitted positron mass as well as the daughter mass; the stated Bi decay releases 3.42 MeV.
If products have greater rest mass, the incoming KE must supply the deficit; subtract that rest-energy increase when finding total product KE.
Keep rest, kinetic and total energy distinct: include qV for an accelerated charged particle and all supplied initial rest energies in a balance.
For symmetric colliding beams at threshold, subtract both incident rest energies from the required energy, then divide the remaining KE equally.
Compare specific nuclear energy as Q per fuel mass; count complete reacting pairs rather than treating each reactant nucleus as one reaction.
Convert a mass-loss rate to power with P = (dm/dt)c², apply the stated efficiency, and convert mass and time units before substituting.
Explain nuclear stability by short-range attraction between nucleons overcoming proton repulsion at nuclear separations.
For a charged particle approaching a nucleus, link Coulomb repulsion to the KE needed to reach the short range of nuclear attraction.
For strong-force evidence, pair an observation with its explanation: stability despite proton repulsion, large binding energy or high-energy scattering.
Use near-constant BE/A above A ≈ 60 as evidence of short-range attraction; compare a supplied 10⁻¹⁵ m range with nucleon separation.
On an N–Z plot, distinguish light stable nuclei near N = Z from heavy stable nuclei with neutron excess; do not use one ratio rule for all nuclei.
For the uranium–lead comparison, explain proton repulsion and strong binding, then use the accepted greater-BE/A argument for lead.
Distinguish random decay, whose individual nucleus or time cannot be predicted, from spontaneous decay, which occurs without external control.
Balance A and signed charge in decay equations: alpha changes (A,Z) by (−4,−2), beta-minus by (0,+1), and beta-plus by (0,−1).
Include an antineutrino with beta-minus and a neutrino with beta-plus; neutral leptons have zero electric charge.
Track every step of a decay route, including neutron capture; on N–Z charts draw arrows in the decay direction using the actual axis orientation.
Describe gamma emission as daughter-nucleus de-excitation with unchanged A and Z; distinguish nuclear transitions from atomic electron transitions.
Use discrete alpha and gamma energies as evidence of nuclear levels; link fixed gamma wavelength or frequency to a fixed level gap with E = hf.
For a gamma wavelength, use the actual excitation gap and λ = hc/ΔE; the supplied 0.48 MeV palladium gap gives about 2.6 pm.
Explain a continuous beta spectrum by variable sharing of fixed released energy with a neutrino or antineutrino, rather than continuous nuclear levels.
Read the supplied available-energy diagram by differences to find the beta endpoint and gamma gap; use the source energy convention consistently.
Compare ionization and penetration separately: alpha strongly ionizes over a short range, while gamma penetrates more with weak direct ionization.
Identify radiation using absorber evidence and equal thicknesses; a missing alpha signal after foil does not prove the source emits no alpha.
Read absorber graphs nonlinearly: find the initial thickness, apply the stated thickness change and read the new transmitted count.
For a stated implanted source, link short particle range to localized energy deposition and reduced exposure outside the target region.
Define half-life using sample activity or the number of remaining radioactive parent nuclei; changing sample mass does not change that half-life.
Use A = λN and T = ln 2/λ with current radioactive N; apply isotope fractions before converting a mixed sample mass to radioactive nuclei.
For a long-lived sample, determine N from mass and measure A, then use λ = A/N; activity and nucleus count suffice without waiting a half-life.
Use inverse seconds for λ when activity is in Bq; convert hours, days or years before calculating λ, A or elapsed time.
In show-that calculations, display the substitution or required precision; repeating the rounded target does not establish the method.
Correct every count reading for background before halving; for readings one half-life apart use B = 2R(T) − R(0).
Add background back when predicting a detector reading, and retain the graph rate unit; a late-time floor can also be offset or a contaminant.
Sketch a smooth exponential through the initial value and successive half-life points; use the actual graph tolerance when reading T.
Track parent remaining and stable daughter formed separately: after n half-lives their fractions are 2⁻ⁿ and 1 − 2⁻ⁿ.
Use the explicit constant-halving model before extrapolating the foam data; a loss of 7/8 leaves 1/8, corresponding to three half-lives.
Apply exponential decay to non-integer times; for the Xe-133 graph, 25 days gives A/A₀ between 0.036 and 0.040.
Use remaining fraction e^(−λt) and decayed fraction 1 − e^(−λt); distinguish a decrease TO 98% from a 98% loss.
For dating, solve t = −ln(N/N₀)/λ or the equivalent activity ratio, and convert any percentage decayed into a remaining fraction first.
In the branched K-40 dating task, use the supplied 11% argon branch to infer the original amount, then date from the remaining K-40 fraction.
Determine λ from the negative gradient of ln(activity) against time or from corresponding A and N readings using λ = A/N.
For two activity readings, calculate λ = ln(A₀/A)/t; convert MBq and GBq to matching units before forming an activity ratio.
Find newly formed stable daughter nuclei from N₀ − N; for alpha decay, correct daughter mass for its different mass number.
For a pure isolated sample use m = m₀e^(−λt); do not import continued parent feeding from a different decay-chain situation.
With fixed energy per decay, power follows activity: use P = P₀e^(−λt) and carry forward the valid launch-power calculation.
Distinguish activity at an instant from decays over an interval: use N₀ − N or area under A(t), with AΔt only when activity is nearly constant.
Compare the given physical half-life with the required exposure time: long T keeps a source steady, while very short T rapidly reduces activity.
Define λ by single-nucleus decay probability per unit time; approximate an interval probability by λΔt only when λΔt is very small.
At equal parent and stable-daughter counts on the supplied graph, half the initial parent remains, so the intersection time is ln 2/λ.

Covered syllabus points

E.3.1—Isotopes E.3.2—Binding energy and mass defect E.3.3—Binding energy curve E.3.4—Mass-energy equivalence E.3.5—Strong nuclear force E.3.6—Random radioactive decay E.3.7—Decay nuclear changes E.3.8—Decay equations E.3.9—Neutrinos and antineutrinos E.3.10—Radiation penetration and ionization E.3.11—Activity, count rate and half-life E.3.12—Half-life changes E.3.13—Background radiation E.3.14 (HL)—Evidence for strong force E.3.15 (HL)—Neutron-proton ratio E.3.16 (HL)—Binding energy above A≈60 E.3.17 (HL)—Discrete nuclear energy levels E.3.18 (HL)—Beta spectrum and neutrino E.3.19 (HL)—Radioactive decay law E.3.20 (HL)—Decay constant meaning E.3.21 (HL)—Activity equation E.3.22 (HL)—Half-life and decay constant

Exam points

Define isotopes using both conditions: the same number of protons and different numbers of neutrons; do not merely name atomic and mass numbers.
Distinguish a nuclide, specified by its proton and neutron constitution, from isotopes, which compare nuclei of the same element.
Identify nuclear-mass measurements or different ion paths in a mass spectrometer as isotope evidence; energy spectra alone test different nuclear ideas.
Isotopes of an element share chemical properties, while their masses, half-lives and decay constants need not be the same.

[Maximum number: 2]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

Outline what is meant by isotopes.

Exam points

Define binding energy as the minimum energy to completely separate a nucleus into nucleons, or the energy released when those nucleons form the nucleus.
Define positive nuclear mass defect as the free proton and neutron mass sum minus the nuclear mass: Δm = Zmₚ + (A − Z)mₙ − M.
Link the mass defect to binding energy using Ebind = Δmc²; explain that formation releases energy or complete separation requires it.
Distinguish units: mass defect may be in u, kg or MeV/c²; binding energy is in energy units, and energy per nucleon requires division by A.
For a nucleus with nuclear mass supplied in u, count protons and neutrons, calculate Δm, convert with 931.5 MeV/u and divide by A if requested.
For a reaction conserving proton and neutron composition, sum total binding energies: Q = ΣBEproducts − ΣBEreactants, weighting each BE/A by A.
When inferring a parent’s binding energy, subtract released Q from the product binding-energy sum, then divide by the parent’s nucleon number.
For an energy-releasing reaction with conserved nucleon composition, total rest mass decreases as total binding energy increases.
Equal nucleon number does not guarantee equal nuclear mass; explicitly compare the parent and daughter masses and account for emitted particles and energy.
When finding nuclear mass from binding energy, show M = Zmₚ + Nmₙ − Ebind/c², convert the units and include the binding-energy term explicitly.

[Maximum number: 3]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

(a)

Outline what is meant by binding energy.

[ 1 ]

(b)

Calculate, in $\mathrm{GeV} \mathrm{c}^{-2}$ , the mass of a nucleus of ${ }_{54}^{131} \mathrm{Xe}$ .

[ 2 ]

Use of Binding energy = mass of separate nucleons - mass of nucleus
$\begin{aligned} \text { « } 1105=54 \times 938+77 \times 940-M c^{2} \text { » } M=122 \text { « } \mathrm{GeVc}^{-2} \text { » } \end{aligned}$

Do not allow BCA.

Award ECF from using binding energy with incorrect sign.

An answer that doesn't include binding energy scores [0].

Exam points

Identify greatest binding stability near the maximum of BE/A, around the iron–nickel region; do not use the largest nucleon number.
Sketch BE/A rising rapidly for light nuclei to a broad maximum, then decreasing gradually for heavier nuclei; it must not remain flat after the peak.
When a point is supplied, pass the curve through the marked cross and put the peak within the stated tolerance; read the actual axes.
Explain energy release in light-nucleus fusion or heavy-nucleus fission by an increase in binding energy per nucleon toward the curve’s peak.
Convert each BE/A value to total BE before comparing reactants and products: multiply by A and by the number of identical nuclei.
For a graph-based Q estimate, read the relevant nuclide values, form the weighted total binding-energy difference and show the unit conversion.
Use products minus reactants for released energy from binding energies with conserved nucleon composition; retain precision until the subtraction.
For the stated isotope comparisons, link larger BE/A to greater stability explicitly; naming the larger value without the required reason is insufficient.
Use the iron–nickel peak to explain why fusion beyond it cannot sustain ordinary stellar energy production: energy must be supplied instead of released.
Combine reaction bookkeeping with the curve: include the incoming neutron, then compare fragment BE/A values by their positions relative to the peak.

[Maximum number: 1]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

$\quad{ }_{54}^{133} \mathrm{Xe}$ is radioactive. Suggest how the binding energy per nucleon for ${ }_{54}^{131} \mathrm{Xe}$ compares with that of ${ }_{54}^{133} \mathrm{Xe}$ .

binding energy per nucleon is a measure of stability so ${ }_{54}^{131} \mathrm{Xe}$ has the greater binding
energy per nucleon

Correct mention of stability required to award the mark.

Marking guidance:

Allow opposite comment for Xe 133.

Exam points

Calculate released energy from Q = (Σmreactants − Σmproducts)c²; convert u using 931.5 MeV/c² and retain the requested final energy unit.
Distinguish atomic and nuclear mass data: neutral-atom electron masses can cancel in balanced reactions, while nuclear beta-plus decay needs positron mass.
For nuclear beta-plus masses, include the emitted positron: the stated Bi-to-Pb decay gives Q = 3.42 MeV, rather than 3.93 MeV.
When product rest mass is larger, explain that incoming KE must supply the extra rest energy; for total product KE subtract that energy deficit.
Distinguish rest energy from kinetic and total energy: the accelerated deuterium has Etotal = mc² + qV.
For supplied decay products, find converted rest mass by subtracting all product rest masses; the stated 770-to-two-140 decay loses 490 MeV/c².
In the stated symmetric electron–positron annihilation, add both initial KEs and both rest energies, divide equally and use photon p = E/c.
For opposite colliding beams at threshold, subtract both incident rest energies from the stated required energy and divide the remaining KE equally.
For a thermal order estimate of the supplied electron–positron pair, compare kT with 2mₑc²; include both masses and report an order of magnitude.
Specific energy is energy released per mass of reactants, so compare Q/mfuel rather than Q per reaction.
For fusion fuel, count reacting pairs from the combined mass per pair; do not count each reactant nucleus as a separate complete reaction.
Convert a mass-loss rate to energy per second using P = (dm/dt)c², then apply the stated efficiency; convert mg and hours before substituting.
Use every particle in the supplied reaction: the stated proton-chain balance includes four protons and two electrons before forming the helium nucleus.
For binding-energy data, multiply BE/A by each nuclide’s A before calculating Q; use the actual supplied reaction and keep sufficient precision.
Keep mass and energy units distinct: multiplying a mass in u by 931.5 gives its value in MeV/c²; multiplying by c² then gives rest energy in MeV.
Explain a decrease in reaction rest mass as energy appearing in the products, such as their kinetic energy; total energy is conserved.

[Maximum number: 2]

The diagram shows a simplified energy-balance model for the Earth surface-atmosphere system.

The following data are given:

\begin{aligned} \text { Average albedo of Earth } & =0.30 \\ \text { Average global temperature of the surface } & =288 \mathrm{~K} \\ \text { Average Earth-Sun distance } & =1.5 \times 10^{11} \mathrm{~m} \end{aligned}

The primary energy source of the Sun is the proton-proton (p-p) chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons. The overall reaction is:

4{ }_{1}^{1} \mathrm{p}+2{ }_{-1}^{0} \mathrm{e} \rightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{0}^{0} v_{e}+4 \gamma

The mass of the helium nucleus is 4.001506 u . Calculate, in MeV , the energy released in the reaction.

ALTERNATIVE 1
$\begin{aligned} \Delta m=4 \times 1.007276+2 \times 0.000549-4.001506 «=0.028696 \mathrm{u} » \text { «0.028696×931.5 }=» 26.7 \mathrm{MeV} \text { ALTERNATIVE } 2 \begin{array}{l} \Delta m=4 \times 938+2 \times 0.511-4.001506 \times 931.5 \ll=25.6 \mathrm{MeV} \mathrm{c}^{-2} \gg 25.6 \mathrm{MeV} \end{array} \end{aligned}$

Marks can only be awarded from one Alternative

Unit is not required.

Particle masses in u and $M e V c^{-2}$ lead to differently rounded answers, working must be checked.

Award [1] max if electrons are omitted. This leads to ALT $125.7 \ll \mathrm{MeV} \gg$ , ALT $224.6 \ll \mathrm{MeV} \gg$

Award ECF if incorrect number of particles used e.g. 1 electron rather than 2.

Exam points

Identify the strong nuclear force as a short-range attractive force between nucleons; a proton can also experience the Coulomb force because it is charged.
Compare both forces in a stable nucleus: nuclear attraction acts between nucleons, while electromagnetic repulsion acts between protons.
State that the strong force is short range and electromagnetism is long range; at nuclear separation, strong attraction dominates proton repulsion.
Relate the stated near-additive nuclear volume to tightly packed nucleons and the dominance of nuclear attraction over electrostatic repulsion.
For a high-speed proton approaching a nucleus, explain Coulomb repulsion, the need to reach nuclear-force range and the KE needed to do so.
For the stated force comparison, order the strengths as strong nuclear, electromagnetic, then gravitational; do not confuse strength with range.

[Maximum number: 1]

A short-range attractive force operates between nucleons.

State the name of this force.

Exam points

Define random decay by the inability to predict which nucleus will decay, or when a particular nucleus will decay.
Define spontaneous decay as occurring without external influence or control; keep this distinct from unpredictability of an individual decay.

[Maximum number: 2]

Radioactive decay is said to be "random" and "spontaneous". Outline what is meant by each of these terms.

Random:
Spontaneous:

Exam points

Explain gamma emission after beta decay as the excited daughter nucleus falling to a lower nuclear energy state; its proton and nucleon numbers stay fixed.
Read the supplied decay-energy diagram by differences: the stated positron maximum is 4.1 − 2.3 = 1.8 MeV and the gamma energy is 2.3 MeV.
Track A and Z separately through each route: alpha reduces A by 4 and Z by 2, while beta-minus preserves A and raises Z by 1.
On an N–Z chart, alpha changes (Z, N) by (−2, −2) and beta-minus by (+1, −1); draw every required arrow in its decay direction.
For the stated neutron-rich nucleus, suggest beta-minus decay, which converts a neutron into a proton.

[Maximum number: 1]

Rhodium-106 $\left({ }_{45}^{106} \mathrm{Rh}\right)$ decays into palladium-106 $\left({ }_{46}^{106} \mathrm{Pd}\right)$ by beta minus $\left(\beta^{-}\right)$ decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the $\beta^{-}$ decay.

Suggest why the $\beta^{-}$ decay is followed by the emission of a gamma ray photon.

Exam points

Balance nucleon number and signed charge on both sides; in alpha decay the daughter has A − 4 and Z − 2, with ⁴₂He or ⁴₂α on the right.
For beta-minus decay keep A unchanged, raise daughter Z by 1 and include an electron or β⁻ plus an antineutrino.
For beta-plus decay keep A unchanged, lower daughter Z by 1 and include a positron plus a neutrino, not an antineutrino.
Keep gamma at A = 0 and Z = 0 in a supplied alpha-plus-gamma equation; it does not alter the daughter’s nuclear numbers.
For a multi-step route add every A/Z change: one alpha and two beta-minus emissions restore Z while reducing A by 4.
Neutron capture increases A by 1; two subsequent beta-minus decays increase Z by 2, giving the stated U-238 to Pu-239 route.
In a supplied capture template with the electron and neutrino already printed, fill A and Z by conservation rather than recalling an extra decay model.
Use the actual answer slots and rubric: some tasks separately credit the daughter, charged beta particle and neutral lepton.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

Radioactive potassium-40 $\left({ }_{19}^{40} \mathrm{~K}\right)$ undergoes $\beta^{-}$ decay to form a nuclide of calcium (Ca).

Write down the complete equation for this radioactive decay.

{ }_{19}^{40} \mathrm{~K} \rightarrow

$\begin{aligned} { }_{19}^{40} \mathrm{~K} \rightarrow{ }_{20}^{40} \mathrm{Ca} +{ }_{-1}^{0} \beta+\bar{\nu} \end{aligned}$

Exam points

Associate beta-plus decay with a positron and a neutrino; distinguish this pair from the electron and antineutrino of beta-minus decay.
State that an electron antineutrino has zero electric charge and is emitted in beta-minus decay; “anti” does not mean negatively charged.

[Maximum number: 1]

Magnesium-27 nuclei $\left({ }_{12}^{27} \mathrm{Mg}\right)$ decay by beta-minus $\left(\beta^{-}\right)$ decay to form nuclei of aluminium-27 (Al).

A Magnesium-27 nucleus can decay by one of two routes:

Route 1: 70\% of the beta particles are emitted with a maximum kinetic energy of 1.76656 MeV , accompanied by a gamma photon of energy 0.84376 MeV .

Route 2: 30 % of the beta particles have a maximum kinetic energy of 1.59587 MeV with a gamma photon of energy 1.01445 MeV .

The final state of the aluminium-27 nucleus is the same for both routes.

Calculate the difference between the magnitudes of the total energy transfers in parts (a) and (b).
(ii) Explain how the difference in part (b)(ii) arises.

$\text { Difference }=2.6175-(1.76656+0.84376)=2.6175-2.61032=0.007195 \text { «MeV» }$

OR

Difference =2.6175-(1.59587+1.01445)=2.61032=0.007195 «MeV»

9.

b

iii

Another particle/«anti» neutrino is emitted «that accounts for this mass / energy»

Exam points

For the stated comparison, order ionizing ability as γ < β < α; do not confuse this with the usual penetration order.
Alpha particles have low penetration and short range: use a very thin window or glass wall, and do not expect them to cross the stated 20 cm air gap.
At equal initial KE, alphas are more massive and slower than betas; their larger charge and stronger ionization cause faster energy loss over distance.
For the stated implanted source, link short radiation range to localized tumour damage and reduced exposure of healthy tissue or people outside the body.
Explain weak gamma response in a thin gas chamber using its lack of charge and weak direct ionization; this does not mean gamma photons cannot be detected.
For absorber graphs, read the thickness producing the stated count first, multiply that thickness, then read the new count; do not scale counts linearly.
Use equal absorber thicknesses when comparing materials: lower corrected transmitted count means stronger absorption in the stated experiment.
Support radiation identification with penetration evidence: alpha does not cross thin foil, while a component passing through lead supports gamma.

[Maximum number: 2]

This question is about the use of radioactive isotopes.

Patients suffering from prostate cancer can experience metastatic bones. As part of the cancer treatment, patients are injected with an isotope of strontium which is readily taken up by the bones.

The table shows the half-life and emission information for strontium-89 (Sr-89) and strontium-90 (Sr-90).

Beta particles have a quality factor of 1 .

State, in terms of range, two reasons why Sr -89 is the more suitable choice of isotope to be used in the treatment of bone cancers.

Exam points

Define half-life as the time for sample activity to halve or for half its initial radioactive parent nuclei to decay; name the sample quantity clearly.
Use A = λN and λ = ln 2/T₁⁄₂; equal activity with different numbers of nuclei implies different half-lives.
Doubling the mass of the same pure nuclide doubles its initial activity but leaves its half-life unchanged.
Count half-lives for parent remaining and daughter formed separately: after three half-lives of an initially pure parent, their ratio is 1:7.
Use the exponential law for non-integer times and compare each isotope using its own half-life; do not replace decay by a linear fall.
Distinguish decays during an interval from activity at an instant: use N₀ − N, the area under A(t), or AΔt only when activity is nearly constant.
Show why a long-lived source is nearly constant over a short experiment by comparing times with T₁⁄₂ or evaluating a small λt.
Convert moles to nuclei with Avogadro’s constant and use A = λN; convert the decay constant to s⁻¹ when the requested rate is per second.
For short half-lives monitor count or activity over time; for extremely long half-lives measure activity and infer N from the sample mass.
When drawing an activity curve, plot the initial value and successive half-life points, then join them with a smooth exponential curve.
Subtract the constant background before halving a detector signal, or halve the net signal and then add the background back for the graph reading.
Use the actual graph’s reading range: Xe-133 gives 5.2–5.4 days, while the two Pa-234 plots give about 70 s after background correction.
Explain falling activity by a decreasing number of radioactive atoms or nuclei; an unspecified statement that “the isotope decreases” may not score.
For the stated short-lived radon context, compare its 55 s half-life with slow production to explain limited accumulation.
For a source intended to remain steady over a stated treatment interval, choose a physical half-life much longer than that interval and explain the stability.
For supplied competing time scales, very fast physical decay can remove activity before a slower process acts; a very long physical half-life makes A steadier.
In the stated isotope choice, combine shorter-lived residual activity with the supplied particle range to explain why radiation remains localized.
In a show-that estimate of decay count, display the method; the supplied order of magnitude alone does not demonstrate how it was obtained.

[Maximum number: 1]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

The graph shows the variation with time of the activity of a pure sample of ${ }_{54}^{133} \mathrm{Xe}$ .
$A / 10^{7} \mathrm{~Bq}$

Estimate the half-life of ${ }_{54}^{133} \mathrm{Xe}$ .

Exam points

After n half-lives the parent fraction is 2⁻ⁿ; for an initially pure parent with a stable daughter, the daughter fraction is 1 − 2⁻ⁿ.
Convert daughter-to-parent ratios before counting half-lives: a daughter/parent ratio of 7 means the parent fraction is 1/8, so three half-lives elapsed.
For samples with different half-lives, count each sample’s halvings separately before taking a ratio.
Convert elapsed time into the same units as the half-life; the supplied Sr-94 graph gives 75 s, so 10 minutes is eight halvings.
Work backwards by doubling for every elapsed half-life, then use A = λN if activity rather than number is requested.
Keep background constant while halving only the source count: the stated 72-to-40 count data imply background 8 s⁻¹ and a total 16 s⁻¹ after three half-lives.
Sketch a smooth decay curve through N₀/2, N₀/4 and N₀/8 at T, 2T and 3T, showing the full requested time interval.
In the supplied foam model, a decrease by 7/8 means 1/8 remains: verify the model’s constant halving time, then extrapolate three half-lives.

[Maximum number: 3]

The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus ( $\beta$ -) decay to form a stable boron (B) nuclide.

The initial number of nuclei in a pure sample of beryllium-10 is $\mathrm{N}_{0}$ . The graph shows how the number of remaining beryllium nuclei in the sample varies with time.

After $4.3 \times 10^{6}$ years,

\frac{\text { number of produced boron nuclei }}{\text { number of remaining beryllium nuclei }}=7 .

Show that the half-life of beryllium-10 is $1.4 \times 10^{6}$ years.

ALTERNATIVE 1

fraction of $\mathrm{Be}=\frac{1}{8}, 12.5\%$ , or 0.125

therefore 3 half-lives have elapsed
$t_{\frac{1}{2}}=\frac{4.3\times10^{6}}{3}=1.43\times10^{6}\ \mathrm{y}\approx1.4\times10^{6}\ \mathrm{y}$

ALTERNATIVE 2

fraction of $\mathrm{Be}=\frac{1}{8}, 12.5\%$ , or 0.125
$\frac{1}{8}=e^{-\lambda(4.3\times10^{6})}\text{, leading to }\lambda=4.836\times10^{-7}\ \mathrm{y}^{-1}$ $\frac{\ln 2}{\lambda}=1.43\times10^{6}\ \mathrm{y}$

Exam points

Correct each detector reading by subtracting the background count rate; halve only the resulting source count.
If the half-life is known, use two readings one half-life apart to find background: B = 2R(T) − R(0).
For a target total count, subtract background before counting half-lives; add it back when predicting what the detector measures.
Read a persistent late-time count with its actual rate unit; the stated Bi-204 background is 600 counts per hour.
A nonzero late-time floor can indicate background radiation; the rubric also allows a counter offset or a longer-lived contaminant.

[Maximum number: 1]

The graph shows the variation with time t of the activity A of a sample of protactinium-234.

Suggest why the activity approaches a non-zero constant value.

Exam points

Give one observed piece of evidence and its force explanation; in the stated rubric the evidence mark is required before the explanation can score.
Use nuclear stability despite proton repulsion as evidence for an attractive force strong enough to hold nucleons together.
Alternatively, link high-energy departures from Rutherford scattering to the nuclear force acting on the approaching alpha particle.
Use the near-constant BE/A above about A = 60 as evidence for short-range attraction, or use large nuclear binding energies as evidence of strong binding.
In the supplied range comparison, an interaction scale around 10⁻¹⁵ m matches adjacent-nucleon separation; use the given scale without deriving a new law.
Recognize that the pattern of neutron number versus proton number for stable nuclides supports the short-range character of nuclear attraction.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

State and explain one piece of evidence that suggests the presence of this force.

ALTERNATIVE 1

Protons repel each other so no nuclei could form

Unless an attractive force exists to overcome this repulsion

ALTERNATIVE 2

Deviation from Rutherford scattering / drop off of scattering at high energies $(\geq 28 \mathrm{MeV})$

This is due to strong nuclear force acting on alpha particle

ALTERNATIVE 3

The binding energy per nucleon is mostly constant above a certain point (A~60)

this is the consequence of an attractive <<short range>> force

ALTERNATIVE 4

Most nuclei have large binding energies

Indicating that a strong force keeps nucleons together

MP1 is required to award MP2

Exam points

On the supplied equal-scale N–Z plot, light stable nuclei lie near N = Z, while heavy stable nuclei require N > Z; do not apply N > Z to every nucleus.
Explain neutron excess in the stated heavy nucleus using proton repulsion and short-range nuclear attraction; extra neutrons add binding without extra charge.
Distinguish a stable-band trend from a universal ratio rule; count N = A − Z for the given nuclide before comparing N and Z.

[Maximum number: 3]

This question is about nuclear reactions.

The daughter nuclei of U-235 undergo radioactive decay until eventually a stable isotope of lead is reached.

Explain why the nuclei of U-235 are unstable whereas the nuclei of the lead are stable.

Exam points

Identify discrete alpha and gamma energies as evidence for nuclear energy levels; the continuous beta-energy spectrum is a different observation.
Link a fixed gamma wavelength or frequency to a fixed photon energy, then link that energy to a difference between nuclear energy levels.
Explain gamma emission as de-excitation of the daughter nucleus from an excited state to a lower or ground state, rather than an atomic electron transition.
For the given palladium diagram, use the 0.48 MeV excitation gap, convert it to joules and calculate λ = hc/ΔE ≈ 2.6 × 10⁻¹² m.
Infer nuclear levels from the stated alternative beta and gamma energy routes; use fixed gamma gaps rather than claiming every beta has one energy.
State explicitly that discrete gamma energies demonstrate nuclear levels; distinguish this evidence from atomic spectra, pair annihilation and tunnelling.

[Maximum number: 2]

Calculate the wavelength of the gamma ray photon in (d)(ii).

Photon energy

$\begin{aligned} E&=0.48\times10^6\times1.6\times10^{-19}=7.68\times10^{-14}\ \mathrm{J}\\ \lambda&=\frac{hc}{E}=\frac{6.63\times10^{-34}\times3\times10^8}{7.68\times10^{-14}}=2.6\times10^{-12}\ \mathrm{m} \end{aligned}$

Exam points

Use the continuous beta-energy spectrum as evidence for an additional neutrino or antineutrino, rather than for continuous nuclear energy levels.
For beta-minus decay, state that an antineutrino is also emitted and shares the released energy with the electron.
Explain the continuum by variable division of a fixed total released energy: different decays give the beta particle different energy shares.
In the supplied rhodium-to-palladium diagram, the available gap is 3.06 MeV; electron energy ranges approximately from zero to that endpoint.
Both the emitted positron and neutrino have continuous energy spectra in the stated beta-plus process.

[Maximum number: 2]

This question is in two parts.

The alpha particles and gamma rays produced in radioactive decay have discrete energy spectra. This suggests that nuclei also possess discrete energy levels. However, beta particles produced in radioactive decay have continuous energy spectra. Describe how the existence of the antineutrino accounts for the continuous nature of beta spectra.

Exam points

Use N/N₀ = e^(−λt) for the fraction remaining; obtain the decayed fraction from 1 − e^(−λt), not from λt unless an approximation is justified.
Solve t = −ln(A/A₀)/λ when activity falls to a stated fraction; keep the time unit consistent with the decay constant.
For dating, convert the percentage decayed into the percentage remaining before using the exponential law; 90% decayed means 10% remains.
Distinguish a decrease TO 98% from a 98% loss: use 0.98 as the remaining fraction in the stated potassium-40 calculation.
Round the C-14 age to two significant figures and give years: 21% remaining with a 5700-year half-life gives 1.3 × 10⁴ years.
On a graph of ln(activity) against time, take λ as the negative gradient; the supplied slope −6 day⁻¹ gives λ = 6 day⁻¹.
Read corresponding A and N values from the paired decay graphs and use λ = A/N, including the inverse-time unit.
For a stable daughter formed after irradiation ends, subtract the remaining parent nuclei from the initial parent count.
When Po-210 becomes Pb-206, multiply the decayed polonium mass by 206/210; 4.2 g of decayed Po produces about 4.1 g of Pb.
For the stated Au-198 to Hg-198 decay, subtract the remaining gold mass from the initial mass and keep mass prefixes consistent.
For a pure isolated sample, use m = m₀e^(−λt); read the isotope half-life from the supplied table before calculating the remaining mass.
With constant energy released per decay, apply P = P₀e^(−λt) to the spacecraft power and carry forward the earlier power estimate.
Convert MBq and GBq to a common unit before forming A/A₀; substitute numerical values into the decay equation, not just the formula.

[Maximum number: 3]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.

Part 1 Kinematics and gravitation

A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

In a pure sample of lead-212 at one instant, $8.0 \times 10^{-3} \mathrm{~kg}$ of the lead-212 is present. Calculate the mass of lead-212 that remains after a period of 35 hours.

use of $\lambda$ from (h)(i);
correct substitution into $N=N_{0} \mathrm{e}^{-\lambda t}$ ;
8.0 to $8.3 \times 10^{-4} \mathrm{~kg}$ ;

Exam points

Define λ as the probability per unit time that one nucleus decays; include both the single-nucleus meaning and the time basis.
Distinguish decay constant from activity: activity counts disintegrations per unit time in the whole sample.
When λΔt is very small, approximate decay probability by λΔt; λ = 10⁻³ s⁻¹ gives a next-second probability of about 1/1000.

[Maximum number: 2]

State what is meant by decay constant.

Exam points

Use A = λN: the decay rate is proportional to the number of radioactive nuclei still present, rather than the number already decayed.
For a long-lived unknown nuclide, A and N determine λ = A/N and hence its half-life; sample mass alone is insufficient without a molar mass.
Outline a laboratory method: determine N from a pure sample mass, measure A with a counter, and calculate λ = A/N.
Convert a pure isotope mass to N using its molar mass and Avogadro constant, then use A = λN to calculate activity or λ.
Convert λ from hour⁻¹ to s⁻¹ before calculating activity in Bq; divide the copper-64 decay constant by 3600.
Use λ = ln(A₀/A)/t for two activity readings; 24 Bq falling to 17 Bq in one week gives about 5.7 × 10⁻⁷ s⁻¹.
Calculate activity after a fractional number of half-lives using A = A₀(1/2)^(t/T); 12 Bq becomes about 4.8 Bq after 4 days when T = 3 days.
For a mixed potassium sample, apply the radioactive isotope percentage before A₀ = λN₀; convert the given half-life to seconds for Bq.
Put MBq and GBq in matching units before using A = A₀e^(−λt), and show a numerical substitution for the required activity interval.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

A sample of potassium has a mass of 5.00 g . The sample contains 0.012 % of potassium-40 by mass. The remainder of the sample is a stable nuclide of potassium. The half-life of potassium-40 is $1.28 \times 10^{9}$ years.

The mass of one mole of potassium is $39.1 \mathrm{~g} \mathrm{~mol}^{-1}$ .

Determine the initial activity of the sample.

$\begin{aligned} N_{0}=\frac{5}{39.1} \times 0.00012 \times 6.02 \times 10^{23}=9.24 \times 10^{18} A_{o}=\lambda N_{o}=\left(1.72 \times 10^{-17}\right)\left(9.24 \times 10^{18}\right)=159 \mathrm{~Bq} \end{aligned}$

Exam points

Use λ = ln 2/T with matching time units; divide by the half-life in hours when λ is required in hour⁻¹.
Convert the half-life from days or years to seconds before calculating λ in s⁻¹.
For the potassium-40 show-that task, display the full time-conversion substitution or give λ to at least three significant figures.
At the intersection of the supplied parent and stable-daughter curves, half the initial parent nuclei remain, so s = ln 2/λ.
Determine a long half-life from the measured activity and number of nuclei using T = N ln 2/A; the mass number is unnecessary.

[Maximum number: 1]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.

Part 1 Kinematics and gravitation

Calculate, in hour ${ }^{-1}$ , the decay constant of lead-212.

$\left.\quad \lambda=\frac{\ln 2}{\mathrm{~T}_{\frac{1}{2}}}=\frac{0.693}{10.6}=\right) 6.5 \times 10^{-2}$ hour $^{-1}$

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