Particles and Radiation - Worksheets, Questions and Revision

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A-Level · Physics

AP2 Particles and Radiation

AQA 7408 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The specific charge of a particle is used to compare how strongly different particles respond to electric and magnetic fields.

(a)State what is meant by the specific charge of a particle.(1)

(b)Calculate the specific charge of a proton. Mass of proton = 1.673 x 10^-27 kg, charge of proton = 1.60 x 10^-19 C.(2)

(c)An α particle consists of 2 protons and 2 neutrons bound together, with mass of neutron = 1.675 x 10^-27 kg. Calculate the specific charge of an α particle.(3)

(d)A hydrogen nucleus (a single proton) has a greater specific charge than an α particle, even though the α particle carries twice the charge of a proton. Explain why this is the case, without further calculation.(2)

(Total for Question 1 is 8 marks)

Radioactive decay must obey conservation laws. Nuclide notation ^A_Z X is used throughout, where A is the mass (nucleon) number and Z is the atomic (proton) number.

(a)A nucleus of uranium-238, ^238_92 U, decays by α emission to form a nucleus of thorium. State the mass number and the atomic (proton) number of the thorium nuclide produced.(2)

(b)Carbon-14, ^14_6 C, undergoes β-minus decay to form nitrogen-14. Write a balanced nuclear equation for this decay, including correct nuclide notation for the nitrogen nucleus, the β particle and the antineutrino produced.(3)

(c)State two conservation laws, other than conservation of charge and conservation of nucleon (mass) number, that must be obeyed in this β-minus decay.(2)

(Total for Question 2 is 7 marks)

In the α-scattering experiment, a narrow beam of α particles was directed at a thin gold foil in an evacuated chamber, and the number of α particles scattered was measured at various angles using a rotatable detector screen.

(a)Explain how the results of the α-scattering experiment led scientists to reject the 'plum pudding' model of the atom in favour of the nuclear model.(6)

(b)The radius of a gold nucleus is of order 10^-14 m, while the radius of a gold atom is of order 10^-10 m. Calculate the ratio of the atomic radius to the nuclear radius.(2)

(Total for Question 3 is 8 marks)

Particles can be classified as hadrons or leptons.

(a)State two properties of leptons that distinguish them from hadrons.(2)

(b)Classify each of the following particles as a hadron or a lepton: proton, neutron, electron, muon-neutrino, pion, kaon.(3)

(c)State whether the pion and the kaon named in part (b) are baryons or mesons, and justify your answer in terms of the number of quarks in each.(2)

(Total for Question 4 is 7 marks)

Protons and neutrons are hadrons made up of up (u) and down (d) quarks, with charges of +2/3 e and -1/3 e respectively.

(a)State the quark composition of a proton and of a neutron.(2)

(b)Show that the neutron (quark composition udd) has zero overall charge.(2)

(c)In β-minus decay, a neutron transforms into a proton, an electron and an antineutrino. At the quark level this involves a down quark changing into an up quark. State the exchange particle involved and the fundamental force responsible for this interaction.(2)

(d)A neutral kaon, K0, has quark composition d s-bar (a down quark and a strange antiquark), giving it a strangeness of +1. Use conservation of strangeness to explain why a K0 cannot be created alone in a strong interaction between two protons (each of strangeness zero), but must be produced together with another strange particle.(2)

(Total for Question 5 is 8 marks)

Four proposed particle interactions are given below. For each, deduce whether the interaction is allowed, giving a reason based on the conservation of charge, baryon number, lepton number or strangeness (as appropriate). Where a proposed interaction is not allowed, state the conservation law that is violated.

(a)μ- -> e- + antineutrino_e + neutrino_μ(2)

(b)p -> e+ + pi0 (a proton decaying into a positron and a neutral pion)(2)

(c)n -> p + e- (a neutron decaying into a proton and an electron only, with no antineutrino)(2)

(d)K+ -> μ+ + neutrino_μ (a positive kaon, strangeness +1, decaying into a muon and a muon-neutrino, both of strangeness 0)(2)

(Total for Question 6 is 8 marks)

An electron and a positron, each with negligible kinetic energy, undergo mutual annihilation, producing two identical γ-ray photons travelling in opposite directions. Use the Physics Equations Sheet for E = mc² and E = hf. Mass of electron = mass of positron = 9.11 x 10^-31 kg, c = 3.00 x 10⁸ m/s, h = 6.63 x 10^-34 Js.

(a)Using E = mc², calculate the total energy released in this annihilation.(3)

(b)Calculate the energy of each individual photon produced, and hence use E = hf to find its frequency.(3)

(c)In pair production, a photon converts into an electron-positron pair only in the presence of a nucleus, never in empty space. State and explain the additional condition, besides conservation of energy, that requires a nucleus to be present for pair production to occur.(2)

(Total for Question 7 is 8 marks)

Jamal investigates the photoelectric effect using a clean sodium surface. The work function of sodium is 3.65 x 10^-19 J. Use the Physics Equations Sheet for E = hf. h = 6.63 x 10^-34 Js.

(a)State what is meant by the threshold frequency of a metal surface in the photoelectric effect.(1)

(b)Calculate the threshold frequency of sodium, using E = hf.(2)

(c)Light of frequency 8.00 x 10¹⁴ Hz is incident on the sodium surface. Calculate the maximum kinetic energy of the emitted photoelectrons, using Ek(max) = hf - φ.(3)

(d)The observation that no photoelectrons are emitted below the threshold frequency, however intense the light, cannot be explained by the wave theory of light. Explain why this observation supports the photon (particle) model of electromagnetic radiation instead.(3)

(Total for Question 8 is 9 marks)

Required practical: Priya investigates the relationship between the striking (threshold) voltage of light-emitting diodes (LEDs) and the wavelength of light they emit, in order to determine an approximate value for the Planck constant. Her circuit connects an LED in series with a variable resistor, a milliammeter and a low-voltage d.c. supply, with a voltmeter connected across the LED. She increases the p.d. across the LED slowly from zero until it just begins to emit light (the striking voltage, V). At the striking voltage, assume that all of the electrical energy gained by an electron passing through the LED converts into the energy of a single emitted photon, so that eV = hc/λ, where λ is the wavelength of the emitted light. Her results for four LEDs are: Red, λ = 660 nm, V = 1.62 V; Yellow, λ = 590 nm, V = 1.82 V; Green, λ = 525 nm, V = 2.15 V; Blue, λ = 470 nm, V = 2.45 V. e = 1.60 x 10^-19 C, c = 3.00 x 10⁸ m/s.

(a)Explain why the circuit must include a variable resistor connected in series with the LED, rather than connecting the LED directly across the supply.(2)

(b)Describe how Priya should determine the striking voltage of each LED, and suggest one precaution she should take to improve the precision of this reading.(3)

(c)Show that the equation eV = hc/λ can be rearranged to give V = (hc/e) x (1/λ), and explain how a graph could be used with this equation to determine a value for the Planck constant, h.(3)

(d)Calculate the value of 1/λ, in m^-1, for the green LED (λ = 525 nm).(1)

(e)The graph of V against 1/λ for Priya's four data points has a gradient of 1.26 x 10^-6 V m. Use this gradient to calculate a value for the Planck constant, h.(3)

(f)The accepted value of the Planck constant is 6.63 x 10^-34 Js. Calculate the percentage difference between Priya's value (from part e) and the accepted value.(2)

(g)Suggest one reason, other than random error in judging the striking voltage, why Priya's experimental value of h might differ systematically from the accepted value.(1)

(h)Identify one variable that Priya should keep constant across all four LEDs to ensure a fair comparison of striking voltages.(1)

(Total for Question 9 is 16 marks)

The fundamental forces between particles are mediated by the exchange of virtual particles (gauge bosons).

(a)An electron scatters elastically off a proton by exchanging a virtual particle associated with the electromagnetic force. State the name of this exchange particle and one property of it that distinguishes it from the W and Z bosons.(2)

(b)A neutron decays via the weak interaction: n -> p + e- + antineutrino_e. At the quark level, this occurs when a down quark inside the neutron emits a virtual W- boson and changes into an up quark; the W- boson then decays into an electron and an electron antineutrino. State the quark change that occurs at this interaction vertex, and the exchange particle emitted.(2)

(c)Explain why the range of the weak interaction is much shorter than the range of the electromagnetic force, referring to the exchange particles involved in each.(3)

(Total for Question 10 is 7 marks)

Dr Aisha Khan analyses a wooden artefact to estimate its age using carbon-14 dating. Carbon-14 has a half-life of 5730 years. Use the Physics Equations Sheet for A = λ N, N = N0 e^{-λ t} and λ = ln2 / t(1/2). Take 1 year = 3.156 x 10⁷ s.

(a)Calculate the decay constant, λ, of carbon-14 in s^-1.(3)

(b)A 1.00 g sample of carbon from the artefact has an activity of 0.230 Bq from carbon-14 decay. Calculate the number of carbon-14 nuclei present in the sample, using A = λ N.(2)

(c)A freshly-cut sample of the same mass of wood has an activity of 0.250 Bq. Calculate the age of the artefact, using A = A0 e^{-λ t}.(3)

(d)State one assumption made in using this method to estimate the age of the artefact.(1)

(Total for Question 11 is 9 marks)

A negative pion, π-, with quark composition d u-bar (a down quark and an up antiquark), decays at rest via the weak interaction: π- -> μ- + antineutrino_μ. Use the Physics Equations Sheet for E = mc². c = 3.00 x 10⁸ m/s.

(a)State the quark composition of the π- meson's antiparticle, the π+ meson.(1)

(b)Show that charge is conserved in the decay π- -> μ- + antineutrino_μ.(2)

(c)The rest mass of the π- meson is 2.49 x 10^-28 kg and the rest mass of the muon is 1.88 x 10^-28 kg. Assuming the antineutrino produced has negligible mass, calculate the total kinetic energy released in this decay, using E = mc².(3)

(d)This kinetic energy is shared between the muon and the antineutrino, which move off in opposite directions to conserve momentum. State and explain which of the two particles gains the greater share of the kinetic energy.(2)

(Total for Question 12 is 8 marks)

Mark scheme · AP2 Particles and Radiation

Question 1

(a) B1 the charge of a particle divided by its mass (charge-to-mass ratio) oe
(a) Answer: Specific charge = charge / mass of the particle.
(b) M1 correct substitution: specific charge = 1.60 x 10^-19 / 1.673 x 10^-27
(b) A1 9.56 x 10⁷ C/kg (awrt, cao)
(b) Answer: 9.56 x 10⁷ C/kg
(c) M1 total mass = 2(1.673 x 10^-27) + 2(1.675 x 10^-27) = 6.696 x 10^-27 kg
(c) M1 total charge = 2 x 1.60 x 10^-19 = 3.20 x 10^-19 C, and specific charge = charge/mass correctly substituted
(c) A1 4.78 x 10⁷ C/kg (awrt, ft from their total mass)
(c) Answer: 4.78 x 10⁷ C/kg
(d) B1 the α particle's mass is about 4 times the proton's mass (2 protons + 2 neutrons vs 1 proton), while its charge is only 2 times the proton's charge
(d) B1 since specific charge = charge/mass, the proportionally much larger increase in mass than in charge results in a smaller specific charge for the α particle oe
(d) Answer: The α particle's mass increases by a factor of about 4 while its charge only increases by a factor of 2, so its charge-to-mass ratio (specific charge) is smaller than the proton's.

Question 2

(a) B1 mass number = 234
(a) B1 atomic number = 90
(a) Answer: Mass number 234, atomic number 90 (thorium-234, ^234_90 Th).
(b) B1 correct nitrogen nuclide: ^14_7 N
(b) B1 correct β particle notation: ^0_-1 e (an electron)
(b) B1 correct inclusion of an electron antineutrino on the right-hand side (symbol for antineutrino_e, with mass and charge number both zero)
(b) Answer: ^14_6 C -> ^14_7 N + ^0_-1 e + antineutrino_e
(c) B1 conservation of lepton number
(c) B1 conservation of energy (and momentum) oe
(c) Answer: Conservation of lepton number; conservation of energy and momentum.

Question 3

(a) L3 (5-6): A full, coherent explanation linking all three key observations (most alpha particles undeflected, some deflected through small angles, a very small fraction deflected through large angles including back-scattering) to specific features of the nuclear model (mostly empty space, a small region of concentrated positive charge, and a large concentrated positive charge/mass), with a clear comparison to why the plum pudding model could not account for the large-angle scattering.
(a) L2 (3-4): The answer describes at least two of the key observations and links them to features of the nuclear model, but the explanation is incomplete or does not fully explain why the plum pudding model is inconsistent with the large-angle scattering.
(a) L1 (1-2): A basic statement that most particles passed through with few deflections, with little or no explanation of what this implies about atomic structure.
(a) L0 (0): No relevant content, or the answer does not relate to the question.
(a) Indicative content:
- Most alpha particles passed straight through the foil with little or no deflection, showing that most of the atom is empty space.
- A small proportion of alpha particles were deflected through small angles, showing that there is a concentration of positive charge somewhere in the atom that repels the (positive) alpha particles.
- A very small fraction (about 1 in 8000) of alpha particles were deflected through angles greater than 90 degrees, with some rebounding almost back the way they came.
- This large-angle scattering could only happen if the alpha particles were repelled by an extremely concentrated positive charge and mass, far smaller than the size of the whole atom - the nucleus.
- The plum pudding model, in which positive charge is spread thinly throughout the whole atom, predicts only small deflections everywhere and cannot account for any large-angle scattering, since the electric field from such spread-out charge is never strong enough at any point to reverse an alpha particle's direction.
- The observation of large-angle scattering therefore directly contradicted the plum pudding model and supported a model with a small, dense, positively charged nucleus at the centre of a mostly empty atom.
(a) Answer: The undeflected majority showed the atom is mostly empty space; the small-angle deflections showed a concentration of positive charge; the rare large-angle deflections showed that charge and mass are concentrated in a tiny nucleus, which the plum pudding model (with charge spread throughout the atom) could not explain.
(b) M1 ratio = 10^-10 / 10^-14
(b) A1 1 x 10⁴ (i.e. the atom is about 10000 times larger than its nucleus) cao
(b) Answer: About 1 x 10⁴ (the atom is roughly 10000 times larger than its nucleus).

Question 4

(a) B1 leptons are fundamental particles (not made up of quarks), unlike hadrons, which are composed of quarks
(a) B1 leptons do not experience the strong nuclear force (they interact only via the weak force, and the electromagnetic force if charged), whereas hadrons do experience the strong force
(a) Answer: Leptons are fundamental (not made of quarks) and do not feel the strong nuclear force, unlike hadrons.
(b) B1 proton: hadron and neutron: hadron (both correct)
(b) B1 electron: lepton and muon-neutrino: lepton (both correct)
(b) B1 pion: hadron and kaon: hadron (both correct)
(b) Answer: Hadrons: proton, neutron, pion, kaon. Leptons: electron, muon-neutrino.
(c) B1 both the pion and the kaon are mesons
(c) B1 mesons are made of a quark and an antiquark (two quarks), whereas baryons are made of three quarks; the pion and kaon each contain a quark-antiquark pair, not three quarks
(c) Answer: Both are mesons, as each is made from a quark-antiquark pair (two quarks), not the three quarks that make up a baryon.

Question 5

(a) B1 proton = uud (2 up quarks, 1 down quark)
(a) B1 neutron = udd (1 up quark, 2 down quarks)
(a) Answer: Proton: uud. Neutron: udd.
(b) M1 total charge = (+2/3 e) + (-1/3 e) + (-1/3 e), summing the charges of u, d and d
(b) A1 = 0 (cso, answer given)
(b) Answer: Charge = +2/3 e - 1/3 e - 1/3 e = 0.
(c) B1 exchange particle: W- boson (a virtual W-minus boson)
(c) B1 force: the weak (nuclear) interaction/force
(c) Answer: A (virtual) W- boson is exchanged; the weak nuclear force is responsible.
(d) B1 the total strangeness before the strong interaction is zero, since the two protons each have strangeness zero
(d) B1 strangeness must be conserved in a strong interaction, so a K0 (strangeness +1) can only be produced if it is accompanied by another particle with strangeness -1, keeping the total strangeness at zero (associated production) oe
(d) Answer: Since strangeness is conserved in the strong interaction and the initial protons have total strangeness zero, a K0 (S = +1) must be produced alongside another particle of strangeness -1 so that the total strangeness remains zero.

Question 6

(a) B1 allowed
(a) B1 charge, muon lepton number and electron lepton number are all conserved (e.g. electron lepton number: 0 on the left = +1 - 1 = 0 on the right) oe
(a) Answer: Allowed: charge and both lepton numbers (muon and electron) are conserved.
(b) B1 not allowed
(b) B1 baryon number is not conserved: the proton has baryon number +1, but the positron and pion both have baryon number 0, so total baryon number changes from +1 to 0
(b) Answer: Not allowed: baryon number is violated (+1 on the left, 0 on the right).
(c) B1 not allowed
(c) B1 lepton number is not conserved: the left-hand side has lepton number 0, but the electron alone gives a total of +1 on the right; an antineutrino_e (lepton number -1) is needed to restore the balance to zero
(c) Answer: Not allowed: lepton number is violated; the correct decay requires an antineutrino_e as well.
(d) B1 allowed
(d) B1 even though strangeness changes (from +1 to 0), this is a weak interaction, and strangeness only has to be conserved in strong and electromagnetic interactions, not in weak interactions oe
(d) Answer: Allowed: strangeness need not be conserved in a weak interaction, only in strong and electromagnetic interactions.

Question 7

(a) M1 total mass converted = 2 x 9.11 x 10^-31 = 1.822 x 10^-30 kg
(a) M1 correct substitution: E = 1.822 x 10^-30 x (3.00 x 10⁸)²
(a) A1 1.64 x 10^-13 J (awrt, cao)
(a) Answer: 1.64 x 10^-13 J
(b) M1 energy per photon = half the total energy = 1.64 x 10^-13 / 2 = 8.20 x 10^-14 J (ft from a)
(b) M1 rearranges and substitutes: f = E/h = 8.20 x 10^-14 / 6.63 x 10^-34
(b) A1 1.24 x 10²⁰ Hz (awrt, ft)
(b) Answer: Energy per photon = 8.20 x 10^-14 J; frequency = 1.24 x 10²⁰ Hz.
(c) B1 momentum must also be conserved in the interaction
(c) B1 a single photon in empty space cannot produce an electron-positron pair moving in a way that conserves both energy and momentum simultaneously; a nearby nucleus is needed to absorb/take up the recoil momentum oe
(c) Answer: Momentum, as well as energy, must be conserved; a nucleus is needed nearby to absorb the recoil momentum, which is impossible in empty space with a single incoming photon.

Question 8

(a) B1 the minimum frequency of incident electromagnetic radiation that will just cause photoelectrons to be emitted from the surface oe
(a) Answer: The minimum frequency of light needed to release photoelectrons from the surface.
(b) M1 rearranges and substitutes: f0 = φ/h = 3.65 x 10^-19 / 6.63 x 10^-34
(b) A1 5.51 x 10¹⁴ Hz (awrt, cao)
(b) Answer: 5.51 x 10¹⁴ Hz
(c) M1 hf = 6.63 x 10^-34 x 8.00 x 10¹⁴ = 5.304 x 10^-19 J
(c) M1 Ek(max) = 5.304 x 10^-19 - 3.65 x 10^-19
(c) A1 1.65 x 10^-19 J (awrt, cao)
(c) Answer: 1.65 x 10^-19 J
(d) B1 wave theory predicts that energy is spread continuously across the wavefront, and that increasing the intensity delivers more energy per second regardless of frequency, so given enough time or intensity any frequency should eventually release electrons
(d) B1 experimentally, no electrons are emitted below the threshold frequency no matter how intense the light or how long the surface is exposed
(d) B1 the photon model explains this because each photon carries a fixed, discrete energy hf, and a single electron absorbs the energy of a single photon; if hf is less than the work function, no individual photon can ever release an electron, regardless of the number of photons (intensity)
(d) Answer: Wave theory predicts electrons should eventually be released at any frequency given enough intensity or time, but experimentally none are emitted below f0. The photon model explains this because each photon delivers a fixed energy hf to one electron; if hf < φ, no photon can release an electron no matter how many arrive.

Question 9

(a) B1 an LED has a very low resistance once it starts conducting, so without a series resistance the current would become too large and could damage the LED
(a) B1 the variable resistor allows the p.d. across the LED to be increased gradually from zero, so the striking voltage can be found precisely
(a) Answer: The series resistor limits the current so the LED is not damaged, and allows the p.d. across the LED to be increased gradually so the striking voltage can be measured precisely.
(b) B1 the room should be darkened (or the LED viewed against a dark background/through a tube) so that the first faint light is visible
(b) B1 the p.d. is increased slowly from zero and the voltmeter reading recorded at the instant light is first just seen
(b) B1 the reading is repeated several times (e.g. by the same observer, or by increasing then decreasing the voltage) and averaged, to reduce the effect of human judgement/reaction on the result oe
(b) Answer: Darken the room, increase the p.d. slowly from zero, record the voltmeter reading when light is just first seen, and repeat several times to average out variation in human judgement.
(c) M1 divides both sides of eV = hc/λ by e to give V = hc/(e x λ), which is V = (hc/e) x (1/λ) (cso)
(c) B1 a graph of V (y-axis) against 1/λ (x-axis) gives a straight line through the origin with gradient equal to hc/e
(c) B1 h can then be found from h = gradient x e / c oe
(c) Answer: V = (hc/e) x (1/λ); plotting V against 1/λ gives a straight line of gradient hc/e, from which h = gradient x e/c.
(d) B1 1.90 x 10⁶ m^-1 (awrt, cao)
(d) Answer: 1.90 x 10⁶ m^-1
(e) M1 rearranges h = gradient x e / c
(e) M1 correct substitution: h = 1.26 x 10^-6 x 1.60 x 10^-19 / 3.00 x 10⁸
(e) A1 6.72 x 10^-34 Js (awrt, cao)
(e) Answer: h = 6.72 x 10^-34 Js
(f) M1 percentage difference = |6.72 - 6.63| / 6.63 x 100 (ft from e)
(f) A1 1.36% (awrt, accept 1.4%)
(f) Answer: 1.36% (or awrt 1.4%)
(g) B1 the assumption that all of the electrical energy converts into a single photon's energy ignores energy losses (e.g. as heat/lattice vibrations) inside the LED, so eV may not exactly equal the true photon energy oe (or: resistance of the connecting wires/voltmeter affects the true p.d. across the LED junction)
(g) Answer: The assumption that eV equals the photon energy ignores energy losses to heat within the LED, which introduces a systematic error.
(h) B1 the ambient lighting/level of darkness in the room (or: the same observer/method used to judge 'just visible' light) oe
(h) Answer: The level of room darkness (or the observer/method used to judge when light is 'just visible') must be kept the same for every LED.

Question 10

(a) B1 the (virtual) photon
(a) B1 the photon is massless and uncharged, whereas the W and Z bosons have (large) mass, and the W boson is charged
(a) Answer: The photon; unlike the W and Z bosons, the photon is massless (and uncharged).
(b) B1 a down quark changes into an up quark
(b) B1 a (virtual) W- boson is emitted
(b) Answer: A down quark changes into an up quark, emitting a virtual W- boson.
(c) B1 the range of a force is inversely related to the mass of its exchange particle, because a more massive virtual particle can only exist for a shorter time/travel a shorter distance before violating energy conservation
(c) B1 the W and Z bosons that mediate the weak interaction have a very large mass, giving the weak interaction a very short range (of order 10^-18 m)
(c) B1 the photon that mediates the electromagnetic force is massless, giving the electromagnetic force (in principle) an infinite range
(c) Answer: The range of a force decreases as the mass of its exchange particle increases. The massive W/Z bosons give the weak force a very short range, while the massless photon gives the electromagnetic force an effectively infinite range.

Question 11

(a) M1 converts the half-life to seconds: t(1/2) = 5730 x 3.156 x 10⁷ = 1.808 x 10¹¹ s
(a) M1 correct substitution: λ = ln2 / 1.808 x 10¹¹
(a) A1 3.83 x 10^-12 s^-1 (awrt, cao)
(a) Answer: λ = 3.83 x 10^-12 s^-1
(b) M1 rearranges: N = A / λ = 0.230 / (their value of λ from a)
(b) A1 6.00 x 10¹⁰ nuclei (awrt, ft from a)
(b) Answer: N = 6.00 x 10¹⁰ nuclei
(c) M1 rearranges to make t the subject: t = ln(A0/A) / λ
(c) M1 correct substitution: t = ln(0.250/0.230) / (their value of λ from a)
(c) A1 awrt 690 years (ft from a; accept 2.18 x 10¹⁰ s)
(c) Answer: t = 690 years (2 s.f.)
(d) B1 the proportion (activity) of carbon-14 in the atmosphere, and hence in living wood, has remained constant over time; or the sample has not been contaminated since the tree died oe
(d) Answer: The atmospheric (and hence living-wood) proportion of carbon-14 has remained constant over time, and the sample has not been contaminated.