Radioactive sources are described using nuclide notation and classified by the type of radiation they emit.
(a)State what is meant by an isotope.(2)
(b)An atom of radon has the notation ^222_86 Rn. State the number of protons, neutrons and electrons in a neutral atom of this isotope.(3)
(c)State the order of penetrating power of α, β-minus and γ radiation, from most to least penetrating, and state one material that will substantially stop each type.(3)
(d)A radium-226 nucleus (proton number 88) decays by α emission to form a nucleus of radon. Write a balanced nuclear equation for this decay, using correct nuclide notation.(2)
(Total for Question 1 is 10 marks)
2
A cobalt-60 source (proton number 27) used in radiotherapy decays by β-minus emission to an excited nucleus of nickel-60, which subsequently decays to its ground state by emitting γ radiation.
(a)Determine the proton number and nucleon number of the nickel nucleus formed by this β-minus decay, and write a balanced nuclear equation for the decay, including the antineutrino.(3)
(b)Explain why the nickel-60 nucleus produced is described as 'excited', and state how it loses its excess energy.(2)
(c)State why the emission of a γ photon does not change the proton number or nucleon number of the nucleus, and state why γ radiation, unlike α and β particles, is not deflected by electric or magnetic fields.(3)
(Total for Question 2 is 8 marks)
3
Required practical: a student uses a GM tube and counter to investigate the absorption of radiation from a sealed, unknown radioactive source, placing different absorbers between the source and the tube at a fixed distance. The background count rate, measured with no source present, is 22 counts per minute. The student's results are shown below.
Absorber used
Measured count rate / counts per minute
None
850
Paper
845
5 mm aluminium
210
5 cm lead
26
(a)Explain why the student should measure the background count rate before the experiment, and explain how this measurement should be used with the readings in the table.(2)
(b)Determine the corrected count rate for each absorber, and use these values to identify the type(s) of radiation emitted by the source, justifying your answer using the data.(4)
(c)State one precaution, other than repeating readings, that the student should take to improve the reliability of each count rate measurement.(1)
(Total for Question 3 is 7 marks)
4
A sample of technetium-99m used in medical imaging has a half-life of 6.0 hours. A hospital receives a sample with an initial activity of 6.4 x 109 Bq. Use A = λ N and A = A0 exp(-λ t), where λ = ln2 / (half-life).
(a)Show that the decay constant of technetium-99m is 3.2 x 10-5 per second.(3)
(b)Calculate the number of technetium-99m nuclei present in the sample when it is received.(2)
(c)Calculate the activity of the sample 24 hours after it is received.(3)
(d)The random nature of radioactive decay means individual nuclear decays cannot be predicted. Explain what is meant by this, and why activity calculations such as part (c) can still give reliable predictions for a macroscopic sample.(2)
(Total for Question 4 is 10 marks)
5
An archaeological wood sample has a measured activity per unit mass of carbon of 6.9 disintegrations per minute per gram. Living wood has an activity per unit mass of carbon of 15.0 disintegrations per minute per gram. The half-life of carbon-14 is 5730 years. Use A = A0 exp(-λ t), where λ = ln2 / (half-life).
(a)Show that the decay constant of carbon-14 is about 1.21 x 10-4 per year.(2)
(b)Calculate the age of the wood sample.(4)
(c)The sample is later found to have been contaminated with a small amount of modern carbon during excavation. State and explain the effect this contamination would have on the calculated age of the sample compared with its true age.(2)
(Total for Question 5 is 8 marks)
6
Required practical: a student investigates how the corrected count rate, C, from a small γ source varies with distance, x, from a GM tube, to test whether C is inversely proportional to x2. All readings have already been corrected for background. Results:
x / cm
C / counts per second
10
96.0
20
24.1
30
10.7
40
6.0
50
3.8
(a)Explain how the student could use a graph of ln(C) against ln(x) to test whether C = k / x2, and state the gradient expected if this relationship holds.(3)
(b)Two data points give ln(x) = 2.30, ln(C) = 4.564 (at x = 10 cm) and ln(x) = 3.91, ln(C) = 1.335 (at x = 50 cm). Calculate the gradient between these two points and comment on whether it supports the inverse square law.(3)
(c)State two control variables the student should keep constant to ensure this is a valid test.(2)
(d)Suggest one reason why the experimental gradient found in part (b) might differ slightly from the theoretical value of exactly -2.(1)
(Total for Question 6 is 9 marks)
7
Mass of proton = 1.00728 u, mass of neutron = 1.00867 u, mass of a helium-4 nucleus = 4.00151 u. 1 u is equivalent to 931.5 MeV of energy.
(a)Show that the mass defect of a helium-4 nucleus is 0.0304 u.(2)
(b)Calculate the binding energy of the helium-4 nucleus in MeV, and hence the binding energy per nucleon.(4)
(c)State why binding energy per nucleon, rather than total binding energy, is used to compare the stability of different nuclides.(1)
(Total for Question 7 is 7 marks)
8
The graph of binding energy per nucleon (MeV) against nucleon number, A, for stable nuclides rises steeply for light nuclei, reaches a maximum at around A = 56 (near iron), then decreases slowly for heavier nuclei up to A = 238.
Using ideas about binding energy per nucleon, explain why both the fission of a very heavy nucleus (such as uranium-235) and the fusion of very light nuclei (such as isotopes of hydrogen) release energy.
(Total for Question 8 is 6 marks)
9
A possible fission reaction is: n + U-235 -> Ba-141 + Kr-92 + 3n. Masses: U-235 = 235.0439 u, Ba-141 = 140.9144 u, Kr-92 = 91.9262 u, neutron = 1.00867 u. 1 u is equivalent to 931.5 MeV, and 1 MeV = 1.602 x 10-13 J.
(a)Show that the total mass before the reaction (a neutron plus a U-235 nucleus) is 236.0526 u.(1)
(b)The total mass of the products (Ba-141 + Kr-92 + 3 neutrons) is 235.8666 u. Calculate the mass defect for this fission reaction.(2)
(c)Calculate the energy released in this single fission reaction, in MeV and in joules.(4)
(d)A nuclear power station generates 1200 MW of electrical power output at an overall efficiency of 33%. Calculate the number of fission reactions like the one above that must occur per second in the reactor to produce this electrical output.(4)
(Total for Question 9 is 11 marks)
10
In a deuterium-tritium fusion reaction: ^2_1 H + ^3_1 H -> ^4_2 He + ^1_0 n. Masses: deuterium = 2.01410 u, tritium = 3.01605 u, helium-4 = 4.00260 u, neutron = 1.00867 u. 1 u = 931.5 MeV = 1.6605 x 10-27 kg. 1 MeV = 1.602 x 10-13 J.
(a)Calculate the energy released, in MeV, in a single deuterium-tritium fusion reaction.(4)
(b)A future fusion reactor aims to produce a continuous power output of 500 MW using this reaction. Calculate the number of deuterium-tritium fusion reactions required per second, and hence estimate the mass of deuterium fuel consumed per day.(6)
(Total for Question 10 is 10 marks)
11
A thermal nuclear power station reactor uses uranium-235 fuel rods, a graphite moderator, boron control rods, and a pressurised water coolant.
Explain the function of the moderator, the control rods and the coolant in sustaining and controlling a chain reaction, and explain why each is necessary for the reactor to operate safely at a steady power output.
(Total for Question 11 is 6 marks)
12
In an α-particle scattering experiment, α particles of kinetic energy 6.0 MeV are fired directly (head-on) at gold-197 nuclei (proton number 79). At the point of closest approach, all the initial kinetic energy of the α particle has been converted to electric potential energy. Use the equation for electric potential energy between two point charges, Ep = Q1 Q2 / (4 π epsilon0 r), from the Physics Equations Sheet. Elementary charge e = 1.60 x 10-19 C, epsilon0 = 8.85 x 10-12 F/m, 1 eV = 1.60 x 10-19 J.
(a)Show that the kinetic energy of the α particle in joules is 9.6 x 10-13 J.(1)
(b)Calculate the distance of closest approach, r, of the α particle to the gold nucleus, and state what this calculation gives an estimate of.(5)
(c)The measured radius of a gold-197 nucleus is about 7.0 fm (7.0 x 10-15 m). Using R = R0 A1/3, calculate the value of R0 suggested by this measurement, and comment on whether it is consistent with the accepted value of R0 = 1.2 fm.(4)
(Total for Question 12 is 10 marks)
Mark scheme · AP8 Nuclear Physics
Question 1
(a) B1 atoms (nuclei) of the same element, with the same number of protons (same proton/atomic number), oe
(a) B1 with different numbers of neutrons (different nucleon/mass numbers), oe
(a) Answer: Atoms of the same element (same number of protons) with different numbers of neutrons (different mass numbers).
(b) B1 protons = 86
(b) B1 neutrons = 136 (222 - 86)
(b) B1 electrons = 86 (equal to protons, since the atom is neutral)
(c) B1 correct order: γ most penetrating, then β, then α least penetrating
(c) B1 α is stopped by a sheet of paper (or a few cm of air); β is stopped by a few mm of aluminium
(c) B1 γ requires several cm of thick lead (or thick concrete) to substantially reduce its intensity, oe
(c) Answer: Gamma > β > α in penetrating power. Alpha stopped by paper; β stopped by a few mm of aluminium; γ needs thick lead (or concrete) to reduce it substantially.
(d) B1 nucleon number and proton number balance correctly (226 = 222 + 4 and 88 = 86 + 2)
(d) B1 correct nuclide symbols for the products: ^222_86 Rn and ^4_2 He
(d) Answer: ^226_88 Ra -> ^222_86 Rn + ^4_2 He
Question 2
(a) B1 proton number of nickel nucleus = 28
(a) B1 nucleon number of nickel nucleus = 60 (unchanged by β-minus decay)
(a) B1 correct equation: ^60_27 Co -> ^60_28 Ni + ^0_-1 e + antineutrinoe (allow β^- symbol for the electron)
(a) Answer: ^60_27 Co -> ^60_28 Ni + ^0_-1 e + antineutrinoe
(b) B1 the nucleus is produced with excess (internal) energy, i.e. it is not in its lowest possible energy state (ground state), oe
(b) B1 it loses this excess energy by emitting one or more γ-ray photons as it de-excites to the ground state, oe
(b) Answer: The nucleus has excess internal energy above its ground state; it loses this energy by emitting γ-ray photon(s).
(c) B1 a γ photon is electromagnetic radiation (energy) with no mass and no charge, not a particle made of nucleons, so its emission does not alter the proton or nucleon number, oe
(c) B1 γ radiation carries no (electric) charge, so it experiences no force in an electric or magnetic field, oe
(c) B1 α and β particles are charged and are therefore deflected (in opposite senses, since they have opposite-sign charge), unlike γ, oe
(c) Answer: Gamma photons carry no charge and no nucleons, so proton/nucleon numbers are unchanged and γ is undeflected by fields, unlike the charged α and β particles.
Question 3
(a) B1 background radiation (e.g. from cosmic rays and rocks) is present at all times and would be added to (counted along with) the readings from the source, oe
(a) B1 the background count rate (22 counts per minute) must be subtracted from each measured reading to give the corrected count rate due to the source alone, oe
(a) Answer: Background radiation adds to every reading, so 22 counts per minute must be subtracted from each measured value to find the count rate due to the source alone.
(b) B1 corrected count rates calculated: none = 828, paper = 823, aluminium = 188, lead = 4 (counts per minute)
(b) B1 no α radiation is emitted, since the count rate barely changes when paper is added (828 to 823), oe
(b) B1 β radiation is emitted, since the count rate falls greatly when aluminium is added (823 to 188), showing β particles are being absorbed, oe
(b) B1 γ radiation is emitted, since the count rate only falls to (near) background level once thick lead is added (188 to 4), showing γ radiation was passing through the paper and aluminium but is absorbed by lead, oe
(b) Answer: Corrected count rates: 828, 823, 188, 4 counts per minute. The source emits β and γ radiation, but no α.
(c) B1 count for a long time period (e.g. several minutes) at each absorber, to reduce the percentage (statistical) uncertainty caused by the random nature of radioactive decay, oe
(c) Answer: Count for a long time period at each setting, to reduce the statistical uncertainty in the count rate.
Question 4
(a) M1 half-life converted to seconds: 6.0 x 3600 = 21600 s
(a) M1 λ = ln2 / 21600 correctly substituted
(a) A1 λ = 3.2 x 10-5 s-1 (cso, answer given)
(a) Answer: λ = 3.2 x 10-5 s-1
(b) M1 N0 = A0 / λ correctly substituted, ft from (a)
(b) A1 N0 = 2.0 x 1014 (awrt)
(b) Answer: 2.0 x 1014 nuclei
(c) M1 t converted to seconds (86400 s) and substituted with λ into A = A0 exp(-λ t)
(c) M1 exponent evaluated correctly: λ t = 2.77, giving exp(-2.77) = 0.0625 (equivalent to 4 half-lives, factor of 1/16)
(c) A1 A = 4.0 x 108 Bq (awrt, ft from (a))
(c) Answer: 4.0 x 108 Bq
(d) B1 it is impossible to predict which nucleus, or when any particular (individual) nucleus, will decay; each nucleus has the same constant probability of decaying per unit time, oe
(d) B1 because the sample contains a very large number of nuclei, the average (statistical) behaviour of the whole sample follows the exponential decay law reliably, even though individual decays are random, oe
(d) Answer: Individual nuclear decay is random and unpredictable, but with a very large number of nuclei the overall (average) decay rate is predictable and follows the exponential law reliably.
Question 5
(a) M1 λ = ln2 / 5730 correctly substituted
(a) A1 λ = 1.21 x 10-4 per year (cso, answer given)
(b) A1 t = 6.4 x 103 years (awrt, allow 6300 to 6500 years)
(b) Answer: About 6.4 x 103 years (6400 years)
(c) B1 the calculated age would be an underestimate (too young), oe
(c) B1 because the modern carbon has a much higher carbon-14 activity than the old sample, it increases the measured activity above the true value for the sample's actual age, making it appear to have decayed less (and so be younger) than it really is, oe
(c) Answer: The calculated age would be too young (an underestimate), because the modern carbon raises the measured activity above the true value for the sample's real age.
Question 6
(a) B1 taking logs of C = k / xn gives ln(C) = ln(k) - n ln(x), a linear equation, oe
(a) B1 a graph of ln(C) (y-axis) against ln(x) (x-axis) that is a straight line confirms a power-law relationship between C and x, oe
(a) B1 the gradient of this line equals -n, so a gradient of -2 would confirm the inverse square law (n = 2)
(a) Answer: ln(C) = ln(k) - n ln(x), so a straight-line graph of ln(C) against ln(x) with gradient -2 would confirm the inverse square law.
(b) M1 gradient = (change in ln(C)) / (change in ln(x)) correctly substituted
(b) A1 gradient = -2.0 (awrt -2.0)
(b) B1 this is very close to -2, supporting the inverse square law for this γ source, ft from calculated gradient
(b) Answer: Gradient = -2.0, which supports the inverse square law.
(c) B1 the same counting time is used at each distance (e.g. a fixed time such as 60 s), long enough to obtain a reliable count, oe
(c) B1 the same source, GM tube and detector alignment are used throughout, with no other radioactive sources nearby affecting the background, oe
(c) Answer: Same counting time at each distance; same source, GM tube and alignment used throughout.
(d) B1 the background count rate may not have been fully or accurately subtracted, or the source may not behave as a true point source, or there is greater statistical (random) uncertainty in the count rate at larger distances where count rates are low, oe (any one valid reason)
(d) Answer: For example, incomplete background subtraction, the source not being a true point source, or greater statistical uncertainty in the low count rates at large distances.
Question 7
(a) M1 total mass of separate nucleons = 2(1.00728) + 2(1.00867) = 4.0319 u
(a) A1 mass defect = 4.0319 - 4.00151 = 0.0304 u (cso, answer given)
(a) Answer: Mass defect = 0.0304 u
(b) M1 binding energy = mass defect x 931.5 MeV/u, ft from (a)
(b) A1 binding energy = 28.3 MeV (awrt)
(b) M1 binding energy per nucleon = total binding energy / 4
(b) A1 binding energy per nucleon = 7.08 MeV (awrt 7.1)
(b) Answer: Binding energy = 28.3 MeV; binding energy per nucleon = 7.08 MeV
(c) B1 it allows nuclei of different sizes (different numbers of nucleons) to be compared fairly, since it represents the average energy needed to remove one nucleon, oe
(c) Answer: It gives a fair, size-independent measure of stability, as it is the average energy needed to remove one nucleon.
Question 8
Level 3 (5-6): A full, coherent explanation that correctly links both fission and fusion to the shape of the binding energy per nucleon graph, explaining that the products of each process lie closer to the peak (iron) than the reactants, with a clear reference to mass-energy equivalence linking the increase in binding energy per nucleon to the energy released.
Level 2 (3-4): The explanation addresses either fission or fusion in good detail, or addresses both but with less clarity or a minor inaccuracy in linking to the shape of the graph or to mass-energy equivalence.
Level 1 (1-2): Basic, largely descriptive statements are made, for example that energy is released in fission and/or fusion, without a clear explanation of why this follows from the binding energy per nucleon graph.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
Binding energy per nucleon measures the average energy needed to separate one nucleon from the nucleus (or the average energy released per nucleon when the nucleus is assembled from separate nucleons).
The graph rises steeply for light nuclei, reaches a maximum near A = 56 (iron), then falls gradually for heavier nuclei.
In fission, a very heavy nucleus (such as uranium-235), which lies on the falling, right-hand side of the graph, splits into two medium-mass nuclei that lie closer to the peak and so have a higher binding energy per nucleon than the original nucleus.
In fusion, very light nuclei (such as isotopes of hydrogen), which lie on the steeply rising, left-hand side of the graph, combine to form a heavier nucleus that lies closer to the peak and so has a higher binding energy per nucleon than the separate light nuclei.
In both processes, the total binding energy of the products is greater than that of the reactants, meaning the nucleons in the products are, on average, more tightly bound.
By mass-energy equivalence (E = mc^2), this increase in total binding energy corresponds to a decrease in total mass (a mass defect) between reactants and products, which appears as kinetic energy of the products and/or gamma radiation - the energy released.
(b) M1 mass defect = 236.0526 - 235.8666, ft from (a)
(b) A1 0.1860 u (awrt)
(b) Answer: 0.1860 u
(c) M1 energy in MeV = mass defect x 931.5, ft from (b)
(c) A1 173 MeV (awrt 172-174)
(c) M1 conversion to joules: energy x 1.602 x 10-13
(c) A1 2.78 x 10-11 J (awrt, ft)
(c) Answer: About 173 MeV, or 2.78 x 10-11 J
(d) M1 thermal (input) power = electrical power / efficiency = 1200 / 0.33, oe
(d) A1 thermal power = 3.64 x 109 W (awrt)
(d) M1 number of reactions per second = thermal power / energy released per reaction (in J), ft from (c)
(d) A1 1.31 x 1020 reactions per second (awrt)
(d) Answer: About 1.31 x 1020 fission reactions per second
Question 10
(a) M1 total mass of reactants = 2.01410 + 3.01605 = 5.03015 u
(a) M1 total mass of products = 4.00260 + 1.00867 = 5.01127 u, and mass defect = 5.03015 - 5.01127 = 0.01888 u
(a) A1 energy = 0.01888 x 931.5 correctly evaluated
(a) A1 17.6 MeV (awrt)
(a) Answer: 17.6 MeV
(b) M1 energy per reaction converted to joules: 17.6 MeV x 1.602 x 10-13, ft from (a)
(b) M1 number of reactions per second = power / energy per reaction (in J)
(b) A1 1.78 x 1020 reactions per second (awrt)
(b) M1 mass of deuterium per reaction = 2.01410 x 1.6605 x 10-27 kg correctly evaluated
(b) M1 mass per day = (reactions per second) x (mass of deuterium per reaction) x (86400 s), ft
(b) A1 about 0.051 kg (51 g) per day (awrt, allow 50 to 52 g)
(b) Answer: About 1.78 x 1020 reactions per second, consuming about 0.051 kg (51 g) of deuterium per day
Question 11
Level 3 (5-6): A full, coherent explanation of the function of all three components (moderator, control rods, coolant), correctly explaining why fast fission neutrons must be slowed to sustain further fission, how control rods regulate the number of neutrons available to maintain a steady (critical) chain reaction, and how the coolant removes heat to prevent overheating, with a clear link to safe, steady operation.
Level 2 (3-4): The function of at least two of the three components is explained correctly, or all three are mentioned but with less clarity or detail, for example without fully explaining why neutrons need to be slowed or how the reaction rate is kept constant.
Level 1 (1-2): Basic or partial statements are made about the reactor, for example that control rods absorb neutrons or that coolant removes heat, without a clear explanation of how the chain reaction is controlled.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
Fission of a uranium-235 nucleus, following absorption of a slow (thermal) neutron, produces (on average) two or three fast neutrons, together with fission fragments and energy.
Fast neutrons are much less likely to cause further fission in uranium-235 than slow neutrons, so they must be slowed down (thermalised) for the chain reaction to continue.
The moderator (graphite, or the water itself) slows the fast neutrons through repeated elastic collisions, increasing the probability that they go on to cause further fission.
Control rods (e.g. boron or cadmium) absorb neutrons; inserting them further into the core removes more neutrons and reduces the rate of fission, while withdrawing them increases the rate of fission.
By adjusting the position of the control rods, the reactor is kept critical, so that on average exactly one neutron from each fission goes on to cause another fission, giving a constant, controlled rate of energy release rather than a runaway (supercritical) or dying-out (subcritical) reaction.
Control rods can be inserted fully in an emergency to absorb enough neutrons to shut the chain reaction down quickly (an important safety feature).
The coolant (pressurised water) carries heat away from the reactor core to a heat exchanger, generating steam to drive turbines, and prevents the core from overheating.
Question 12
(a) B1 6.0 x 106 x 1.60 x 10-19 = 9.6 x 10-13 J (cso, answer given)
(a) Answer: 9.6 x 10-13 J
(b) M1 charge of α particle = 2e and charge of gold nucleus = 79e correctly identified
(b) M1 Ep = KE equation rearranged to make r the subject: r = (2e)(79e) / (4 π epsilon0 x KE), oe
(b) M1 correct substitution of values into the rearranged equation
(b) A1 r = 3.8 x 10-14 m (awrt 3.7-3.8 x 10-14)
(b) B1 this gives an estimate of an upper limit for the radius of the gold nucleus, since it assumes all the kinetic energy converts to electric potential energy exactly at the nuclear surface, oe
(b) Answer: r = 3.8 x 10-14 m, an estimate of an upper limit for the nuclear radius
(c) B1 this is (very close to) consistent with the accepted value of 1.2 fm, supporting the idea that nuclear density is approximately constant for all nuclei, oe
(c) Answer: R0 = 1.20 fm, consistent with the accepted value of 1.2 fm