ExaminerPrep › Question bank › IAL Physics › 5.3 Binding energy, fission and fusion
Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
40 free, high-quality past-paper-style questions on Binding energy, fission and fusion for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which statement describes the binding energy of a nucleus?
| Answer | Mark | Notes |
|---|---|---|
| Binding energy is the energy needed to separate a nucleus completely into its individual protons and neutrons. The same amount of energy is released when the nucleus forms from free nucleons. | ✓ 1 | Answer D |
Answer: D
Common error: Binding energy is not energy stored in the nucleus waiting to come out. It has to be supplied to pull the nucleus apart.
The mass deficit of a nucleus is 0.0240 u. 1 u = 1.66 × 10−27 kg.
What is the binding energy of the nucleus?
| Answer | Mark | Notes |
|---|---|---|
| Convert to kilograms first: 0.0240 × 1.66 × 10−27 = 3.98 × 10−29 kg. Then ΔE = c2Δm = 9.00 × 1016 × 3.98 × 10−29 = 3.59 × 10−12 J. | ✓ 1 | Answer C |
Answer: C
Exam tip: ΔE = c2Δm gives joules only when the mass is in kilograms. Convert from u first.
A gamma-ray photon has an energy of 4.80 MeV. 1 u is equivalent to 931.5 MeV. 1 u = 1.66 × 10−27 kg.
What mass is equivalent to the energy of this photon?
| Answer | Mark | Notes |
|---|---|---|
| Mass in u = energy in MeV ÷ 931.5 = 4.80 ÷ 931.5 = 5.15 × 10−3 u. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check the size of the answer: 4.80 MeV is much less than 931.5 MeV, so the mass must be much less than 1 u.
Which nucleus has the greatest binding energy per nucleon?
| Answer | Mark | Notes |
|---|---|---|
| The binding energy per nucleon curve has its peak near iron-56 (about 8.8 MeV). Helium-4 is about 7.1 MeV, lead-208 about 7.9 MeV and uranium-235 about 7.6 MeV. | ✓ 1 | Answer B |
Answer: B
Common error: Uranium-235 has the largest total binding energy of these, because it has the most nucleons, but the smallest binding energy per nucleon apart from helium.
The mass deficit of a carbon-12 nucleus is 0.0989 u. 1 u = 1.66 × 10−27 kg.
State what is meant by the binding energy of a nucleus.
Calculate the mass deficit in kg.
Calculate the binding energy of the nucleus in J.
Calculate the binding energy per nucleon in MeV.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy needed to separate the nucleus completely into its individual protons and neutrons | ✓ 1 | |
| Part (b) | ||
| 0.0989 × 1.66 × 10−27 = 1.64 × 10−28 kg | ✓ 1 | |
| Part (c) | ||
| (3.00 × 108)2 × 1.64 × 10−28 = 1.48 × 10−11 J | ✓ 1 | |
| Part (d) | ||
| 1.48 × 10−11 ÷ 1.60 × 10−13 = 92.3 MeV; ÷ 12 = 7.70 MeV | ✓ 1 | |
Exam tip: 1 MeV = 1.60 × 10−13 J. Divide a joule value by this to get MeV.
In an experimental fusion reactor, a mixture of deuterium and tritium is heated to about 108 K and held by magnetic fields so that it does not touch the walls.
State why two nuclei repel each other.
Explain why the mixture must be at a very high temperature for fusion to take place.
Explain why the mixture must also have a high density.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| both are positively charged, so there is an electrostatic repulsion | ✓ 1 | |
| Part (b) | ||
| nuclei must come within the short range of the strong force, about 10−15 m, so they need a very large kinetic energy to overcome the repulsion | ✓ 1 | |
| the mean kinetic energy of the nuclei is proportional to the absolute temperature | ✓ 1 | |
| Part (c) | ||
| more nuclei per unit volume gives a higher collision rate, so more fusions per second (and enough power output) | ✓ 1 | |
Exam tip: Name the force in each step: electrostatic repulsion keeps nuclei apart; the strong force acts only once they are very close.
The graph shows how the binding energy per nucleon of a nucleus varies with its nucleon number A. Four nuclei P, Q, R and S are marked on the curve.
Which nucleus has the greatest total binding energy?
| Answer | Mark | Notes |
|---|---|---|
| Total binding energy = binding energy per nucleon × A. From the graph: P ≈ 8.1 × 20 ≈ 160 MeV, Q ≈ 8.8 × 56 ≈ 490 MeV, R ≈ 8.5 × 120 ≈ 1020 MeV, S ≈ 7.6 × 235 ≈ 1790 MeV. Nucleus S has the greatest total binding energy, even though its value per nucleon is lower than that of Q or R. | ✓ 1 | Answer D |
Answer: D
Common error: Picking the highest point on the curve. The peak shows the greatest binding energy per nucleon (the most stable nucleus), not the greatest total.
The figure represents the protons and neutrons in a nucleus X.
Mass of a proton = 1.00728 u, mass of a neutron = 1.00867 u, mass of nucleus X = 9.00999 u. 1 u is equivalent to 931.5 MeV.
What is the binding energy of nucleus X?
| Answer | Mark | Notes |
|---|---|---|
| The figure shows 4 protons and 5 neutrons (plus the two nucleons in the key, which are not part of X). Δm = 4 × 1.00728 + 5 × 1.00867 − 9.00999 = 9.07247 − 9.00999 = 0.06248 u. Binding energy = 0.06248 × 931.5 = 58.2 MeV. | ✓ 1 | Answer C |
Answer: C
Exam tip: Count protons and neutrons separately and check that they add up to the nucleon number before you start the mass calculation.
The graph shows how the electrostatic potential energy of two protons depends on their separation r.
Two protons, initially very far apart, move directly towards each other with equal initial kinetic energies. They stop momentarily at the separation marked P. Ignore all forces except the electrostatic force.
What was the initial kinetic energy of each proton?
| Answer | Mark | Notes |
|---|---|---|
| At P the potential energy is 0.60 MeV. Both protons stop there, so their total initial kinetic energy is 0.60 MeV and each had 0.30 MeV. | ✓ 1 | Answer B |
Answer: B
Common error: Giving all the potential energy to one proton. When both particles move towards each other, the potential energy at closest approach equals their combined kinetic energy.
The energy level diagram shows the energy of two protons and three neutrons in three arrangements: all separate, as a hydrogen-2 nucleus and a hydrogen-3 nucleus, and as a helium-4 nucleus and a free neutron.
What is the energy released when a hydrogen-2 nucleus and a hydrogen-3 nucleus fuse to form a helium-4 nucleus and a neutron?
| Answer | Mark | Notes |
|---|---|---|
| The energy released is the drop from the 2H + 3H level to the 4He + n level: 28.3 − 10.7 = 17.6 MeV. (The levels are the total binding energies below the separate nucleons.) | ✓ 1 | Answer B |
Answer: B
Exam tip: On an energy level diagram the energy released is the gap between the starting and final levels, not the depth of either level.
The bar chart compares the total mass of seven separate protons and seven separate neutrons with the mass of a nitrogen-14 nucleus. 1 u is equivalent to 931.5 MeV.
State what is meant by the mass deficit of a nucleus.
Use the figure to calculate the mass deficit of the nitrogen-14 nucleus.
Calculate the binding energy of the nitrogen-14 nucleus in MeV.
Calculate the binding energy per nucleon of nitrogen-14.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the difference between the total mass of the separate protons and neutrons and the mass of the nucleus | ✓ 1 | |
| Part (b) | ||
| 14.11159 − 13.99923 = 0.11236 u = 0.1124 u | ✓ 1 | |
| Part (c) | ||
| 0.11236 × 931.5 = 104.7 MeV = 105 MeV | ✓ 1 | |
| Part (d) | ||
| 104.7 ÷ 14 = 7.48 MeV | ✓ 1 | |
Exam tip: When a bar chart has a broken or zoomed axis, use the labelled values, not the bar heights: here the bars exaggerate the difference enormously.
A student calculates the mass deficit and the binding energy of seven light nuclei and plots the graph shown. The points lie on a straight line through the origin.
Determine the gradient of the line, in MeV u−1.
Show that this gradient is equivalent to c2 when it is converted to SI units. 1 u = 1.66 × 10−27 kg. 1 MeV = 1.60 × 10−13 J.
A calcium-40 nucleus has a binding energy of 342 MeV. Use the graph to find its mass deficit.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| large triangle, e.g. from (0, 0) to (0.40 u, 373 MeV) | ✓ 1 | |
| gradient = 373 ÷ 0.40 = 930 MeV u−1 (allow 920–945) | ✓ 1 | |
| Part (b) | ||
| 930 MeV = 930 × 1.60 × 10−13 = 1.49 × 10−10 J | ✓ 1 | |
| 1.49 × 10−10 J ÷ 1.66 × 10−27 kg = 8.96 × 1016 J kg−1, which is close to c2 = (3.00 × 108)2 = 9.00 × 1016 m2 s−2 (J kg−1 = m2 s−2) | ✓ 1 | |
| Part (c) | ||
| reading the line at 342 MeV: 0.367 u (allow 0.36–0.37 u) | ✓ 1 | |
Exam tip: For a gradient, draw the triangle as large as the line allows and read both corners at gridlines. Small triangles magnify reading errors.
A lithium-7 nucleus contains 3 protons and 4 neutrons. Mass of a proton = 1.00728 u, mass of a neutron = 1.00867 u, mass of a lithium-7 nucleus = 7.01436 u.
What is the mass deficit of the lithium-7 nucleus?
| Answer | Mark | Notes |
|---|---|---|
| Δm = 3 × 1.00728 + 4 × 1.00867 − 7.01436 = 7.05652 − 7.01436 = 0.04216 u. | ✓ 1 | Answer C |
Answer: C
Exam tip: Write Z and N = A − Z before you start, and keep all five decimal places until after the subtraction.
The mass deficit of a neon-20 nucleus (10 protons, 10 neutrons) is 0.1725 u. 1 u is equivalent to 931.5 MeV.
What is the binding energy per nucleon of neon-20?
| Answer | Mark | Notes |
|---|---|---|
| Total binding energy = 0.1725 × 931.5 = 160.7 MeV. Per nucleon: 160.7 ÷ 20 = 8.03 MeV. | ✓ 1 | Answer A |
Answer: A
Common error: “Per nucleon” means dividing by the nucleon number A = 20, not by the proton number.
When a uranium-235 nucleus undergoes fission, energy is released.
Which statement explains why?
| Answer | Mark | Notes |
|---|---|---|
| The products are closer to the peak of the curve, so their nucleons are more tightly bound: binding energy per nucleon increases. With the same number of nucleons, the total binding energy increases, the products have less mass and the difference is released. | ✓ 1 | Answer A |
Answer: A
Exam tip: Energy is released when nucleons end up more tightly bound. Moving towards the iron peak, from either side, releases energy.
In the core of a star, two helium-3 nuclei fuse: 3He + 3He → 4He + 21H.
Binding energy per nucleon: helium-3 = 2.57 MeV; helium-4 = 7.07 MeV. A free proton has no binding energy.
What is the energy released in this reaction?
| Answer | Mark | Notes |
|---|---|---|
| Before: 2 × 3 × 2.57 = 15.42 MeV. After: 4 × 7.07 = 28.28 MeV (the two protons add nothing). Energy released = 28.28 − 15.42 = 12.9 MeV. | ✓ 1 | Answer B |
Answer: B
Exam tip: Multiply each binding energy per nucleon by its nucleon number to get total binding energies, then subtract: released = after − before.
A fusion reactor needs its fuel to be at a very high density as well as a very high temperature.
Why is a high density needed?
| Answer | Mark | Notes |
|---|---|---|
| More nuclei per unit volume means more collisions per second, so more fusion reactions per second and a useful power output. Temperature, not density, sets the mean kinetic energy. | ✓ 1 | Answer D |
Answer: D
Exam tip: Temperature decides whether a collision can lead to fusion; density decides how often collisions happen.
An oxygen-16 nucleus contains 8 protons and 8 neutrons. Mass of a proton = 1.007276 u, mass of a neutron = 1.008665 u, mass of an oxygen-16 nucleus = 15.990523 u. 1 u = 1.66 × 10−27 kg.
Calculate the mass deficit of an oxygen-16 nucleus in u.
Calculate the binding energy per nucleon of oxygen-16 in MeV.
Explain why the mass of the oxygen-16 nucleus is less than the total mass of its separate nucleons.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8 × 1.007276 + 8 × 1.008665 = 16.127528 u | ✓ 1 | |
| Δm = 16.127528 − 15.990523 = 0.137005 u = 0.1370 u | ✓ 1 | |
| Part (b) | ||
| Δm = 0.137005 × 1.66 × 10−27 = 2.27 × 10−28 kg | ✓ 1 | |
| ΔE = c2Δm = 2.05 × 10−11 J = 128 MeV | ✓ 1 | |
| ÷ 16 = 8.00 MeV | ✓ 1 | |
| (allow 0.137005 × 931.5 = 127.6 MeV, giving 7.98 MeV) | — | |
| Part (c) | ||
| energy is released when the nucleons bind together / work must be done to separate them | ✓ 1 | |
| so the nucleus has less energy than the separate nucleons, and by ΔE = c2Δm less energy means less mass | ✓ 1 | |
Common error: Rounding the masses before subtracting. The deficit is in the third decimal place, so 3 s.f. masses give nonsense.
The graph of binding energy per nucleon against nucleon number rises steeply from hydrogen to a maximum near iron-56 and then falls gradually towards uranium.
Explain, with reference to this graph, why energy is released both when heavy nuclei undergo fission and when light nuclei undergo fusion. You may refer to mass and energy in your answer.
| Answer | Mark | Notes |
|---|---|---|
| Indicative content: | — | |
| binding energy per nucleon is a measure of how tightly bound / stable a nucleus is; the peak near iron-56 is the most stable | ✓ 1 | |
| in fission a heavy nucleus splits into two middle-mass nuclei, which are closer to the peak | ✓ 1 | |
| in fusion two light nuclei form a heavier nucleus, also closer to the peak | ✓ 1 | |
| in both cases the binding energy per nucleon of the products is greater than that of the reactants | ✓ 1 | |
| nucleon number is conserved, so the total binding energy of the products is greater | ✓ 1 | |
| so the products have less mass than the reactants, and the mass difference Δm is released as energy ΔE = c2Δm (as kinetic energy of the products) | ✓ 1 |
Exam tip: In a six-mark explanation, each separate physics point earns credit. Build a chain: graph → position of products → binding energy per nucleon → total binding energy → mass → energy.
The table gives binding energies per nucleon for some nuclei.
2H: 1.11 MeV · 4He: 7.07 MeV · 56Fe: 8.79 MeV · 93Rb: 8.48 MeV · 140Cs: 8.31 MeV · 235U: 7.59 MeV
State which nucleus in the table is the most stable.
One fission of uranium-235 is 235U + n → 140Cs + 93Rb + 3n. Calculate the energy released.
Two deuterium nuclei can fuse to form a helium-4 nucleus. Calculate the energy released.
Explain why two iron-56 nuclei would not release energy by fusing.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 56Fe — it has the greatest binding energy per nucleon | ✓ 1 | |
| Part (b) | ||
| before: 235 × 7.59 = 1784 MeV (the absorbed neutron has no binding energy) | ✓ 1 | |
| after: 140 × 8.31 + 93 × 8.48 = 1163 + 789 = 1952 MeV | ✓ 1 | |
| energy released = 1952 − 1784 = 168 MeV | ✓ 1 | |
| Part (c) | ||
| before: 2 × 2 × 1.11 = 4.44 MeV; after: 4 × 7.07 = 28.28 MeV | ✓ 1 | |
| released = 23.8 MeV | ✓ 1 | |
| Part (d) | ||
| the product would be beyond the peak, with a lower binding energy per nucleon than iron-56, so energy would have to be supplied | ✓ 1 | |
Common error: Subtracting binding energies per nucleon directly. Multiply each by its nucleon number first: totals, not averages, are conserved in the energy book-keeping.
A small research reactor produces a thermal power of 25 MW from the fission of uranium-235. Each fission releases 3.1 × 10−11 J. The mass of a uranium-235 atom is 235 u. 1 u = 1.66 × 10−27 kg.
Calculate the number of fissions per second.
Calculate the mass of uranium-235 used in one day.
Calculate the total decrease in mass, in kg, of the reacting material in one day.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| N = P/E = 25 × 106 ÷ 3.1 × 10−11 | ✓ 1 | |
| = 8.1 × 1017 s−1 | ✓ 1 | |
| Part (b) | ||
| fissions per day = 8.06 × 1017 × 86 400 = 6.97 × 1022 | ✓ 1 | |
| mass = 6.97 × 1022 × 235 × 1.66 × 10−27 = 0.027 kg | ✓ 1 | |
| Part (c) | ||
| energy per day = 25 × 106 × 86 400 = 2.16 × 1012 J | ✓ 1 | |
| Δm = ΔE/c2 = 2.4 × 10−5 kg | ✓ 1 | |
Exam tip: Power is energy per second. Dividing the power by the energy of one event gives the number of events per second.
The bar chart shows the binding energy per nucleon of some light nuclei.
A lithium-6 nucleus fuses with a hydrogen-2 nucleus to form two helium-4 nuclei: 6Li + 2H → 2 4He.
What is the energy released in this reaction?
| Answer | Mark | Notes |
|---|---|---|
| Total binding energy before: 6 × 5.33 + 2 × 1.11 = 31.98 + 2.22 = 34.20 MeV. After: 2 × 4 × 7.07 = 56.56 MeV. Energy released = 56.56 − 34.20 = 22.4 MeV. | ✓ 1 | Answer B |
Answer: B
Exam tip: Turn every binding energy per nucleon into a total (multiply by A) before you compare the two sides of the equation.
The graph shows how the kinetic energies of the nuclei in a plasma are distributed at two temperatures, T1 and T2. The number of nuclei is the same at both temperatures. EF is the kinetic energy a nucleus needs for fusion to be likely.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| The T2 curve peaks at a higher energy and spreads further, so T2 is the higher temperature. The shaded area beyond EF is much larger for T2: more nuclei have enough energy to fuse. Only a small fraction exceed EF, and EF depends on the charges of the nuclei, not on temperature. | ✓ 1 | Answer A |
Answer: A
Exam tip: On an energy distribution, the area under the curve beyond a given energy represents the number of particles above that energy. A higher temperature moves the peak to higher energy and fattens the tail.
The graph shows how the total binding energy B of a nucleus varies with nucleon number A. P is a point on the curve. The figure also shows the tangent to the curve at P and the straight line from the origin O to P.
Which quantity is equal to the binding energy per nucleon of the nucleus at P?
| Answer | Mark | Notes |
|---|---|---|
| Binding energy per nucleon = B/A. At P this is (B at P − 0) ÷ (A at P − 0), which is the gradient of the line from O to P. The tangent gives the extra binding energy for one extra nucleon, which is smaller here because the curve bends downwards. | ✓ 1 | Answer C |
Answer: C
Common error: Using the tangent whenever a gradient is mentioned. A ratio y/x is the gradient of the line from the origin; the tangent gives the rate of change dy/dx.
The graph shows the binding energy per nucleon curve. Each numbered arrow represents a nuclear change. An arrow pointing towards larger A shows two identical nuclei fusing; an arrow pointing towards smaller A shows one nucleus splitting into two identical nuclei.
Which two changes would release energy?
| Answer | Mark | Notes |
|---|---|---|
| Energy is released when the binding energy per nucleon increases (the nucleon number is unchanged, so the total binding energy increases). Arrow 2 (two light nuclei fusing) and arrow 3 (a heavy nucleus splitting) both move up the curve towards the peak. Arrows 1 and 4 move down the curve, away from the peak, so energy would have to be supplied. Answer: 2 and 3. | ✓ 1 | Answer C |
Answer: C
Exam tip: Ignore whether a change is called fusion or fission: look only at whether the products are higher on the curve than the starting nuclei.
The graph shows the mass deficit Δm of a range of nuclei plotted against nucleon number A. Nucleus X, which contains 26 protons, is labelled. 1 u is equivalent to 931.5 MeV.
What is the binding energy per nucleon of nucleus X?
| Answer | Mark | Notes |
|---|---|---|
| Binding energy = 0.528 × 931.5 = 491.8 MeV. Per nucleon: 491.8 ÷ 56 = 8.78 MeV. | ✓ 1 | Answer A |
Answer: A
Exam tip: Per nucleon means divide by the nucleon number A (protons + neutrons), not by the proton number.
The graph shows the binding energy per nucleon curve. Values for four nuclei are labelled, in MeV.
Calculate the energy released in the reaction 2H + 3H → 4He + n.
A 235U nucleus absorbs a neutron and splits into two nuclei of equal nucleon number, A = 117, and two free neutrons. Use the curve to estimate the energy released.
Compare the energy released per nucleon taking part in each reaction, and use the shape of the curve to explain the difference.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| before: 2 × 1.1 + 3 × 2.8 = 10.6 MeV; after: 4 × 7.1 = 28.4 MeV (a free neutron has no binding energy) | ✓ 1 | |
| released = 28.4 − 10.6 = 17.8 MeV | ✓ 1 | |
| Part (b) | ||
| reading the curve at A = 117: about 8.55 MeV (allow 8.5–8.6 MeV) | ✓ 1 | |
| after: 234 × 8.55 = 2000 MeV; before: 235 × 7.6 = 1786 MeV (the free neutrons have no binding energy) | ✓ 1 | |
| released ≈ 2000 − 1786 = 210 MeV (allow 200–230 MeV) | ✓ 1 | |
| Part (c) | ||
| fusion: 17.8 ÷ 5 ≈ 3.6 MeV per nucleon; fission: 210 ÷ 236 ≈ 0.9 MeV per nucleon, so fusion releases about 4 times as much energy per nucleon (allow 3–5 times) | ✓ 1 | |
| the curve is much steeper for light nuclei, so fusing light nuclei increases the binding energy per nucleon by much more than splitting a heavy nucleus, where the curve is nearly flat | ✓ 1 | |
Common error: Subtracting the binding energies per nucleon directly. Each value must be multiplied by its nucleon number to give a total binding energy first; free neutrons add nothing.
The diagram shows one fission event in a reactor.
Atomic masses: 235U 235.043930 u, 141Ba 140.914403 u, 92Kr 91.926173 u. Mass of a neutron = 1.008665 u. 1 u = 1.66 × 10−27 kg. 1 u is equivalent to 931.5 MeV. 1 MeV = 1.60 × 10−13 J.
Use the diagram to show that both nucleon number and proton number are conserved in this event.
Calculate the decrease in mass in this event.
Calculate the energy released, in MeV, and state the main form in which this energy appears immediately after the event.
Calculate the energy released per kilogram of uranium-235 if all of it undergoes fission in this way.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| nucleon number: 235 + 1 = 141 + 92 + 3 = 236; proton number: 92 = 56 + 36 | ✓ 1 | |
| Part (b) | ||
| before: 235.043930 + 1.008665 = 236.052595 u; after: 140.914403 + 91.926173 + 3 × 1.008665 = 235.866571 u | ✓ 1 | |
| Δm = 0.186024 u = 0.1860 u | ✓ 1 | |
| Part (c) | ||
| 0.186024 × 931.5 = 173.3 MeV = 173 MeV | ✓ 1 | |
| mostly as kinetic energy of the two fission fragments | ✓ 1 | |
| Part (d) | ||
| 173.3 × 1.60 × 10−13 = 2.77 × 10−11 J per fission; mass of one atom = 235.04 × 1.66 × 10−27 = 3.90 × 10−25 kg | ✓ 1 | |
| 2.77 × 10−11 ÷ 3.90 × 10−25 = 7.1 × 1013 J kg−1 | ✓ 1 | |
Exam tip: Atomic masses can be used directly here because the electrons balance: 92 electrons before, and 56 + 36 = 92 after. Keep all six decimal places until the subtraction is done.
The graph shows the binding energy per nucleon of some light nuclei. In the core of a red giant star, carbon-12 nuclei fuse with helium-4 nuclei: 12C + 4He → 16O + γ. 1 MeV = 1.60 × 10−13 J.
Use the labelled values to calculate the energy released in this reaction.
Calculate the decrease in mass in this reaction, in kg.
This reaction needs a core temperature of about 108 K, much higher than is needed for the fusion of hydrogen-1 nuclei. Explain why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| before: 12 × 7.68 + 4 × 7.07 = 92.16 + 28.28 = 120.44 MeV | ✓ 1 | |
| after: 16 × 7.98 = 127.68 MeV | ✓ 1 | |
| released = 127.68 − 120.44 = 7.24 MeV | ✓ 1 | |
| Part (b) | ||
| 7.24 × 1.60 × 10−13 = 1.16 × 10−12 J | ✓ 1 | |
| Δm = ΔE/c2 = 1.16 × 10−12 ÷ (3.00 × 108)2 = 1.29 × 10−29 kg | ✓ 1 | |
| Part (c) | ||
| the nuclei have charges +6e and +2e, so the electrostatic repulsion at a given separation is 12 times that between two protons | ✓ 1 | |
| much more kinetic energy is needed to bring them within range of the strong force, and the mean kinetic energy of the nuclei is proportional to the temperature | ✓ 1 | |
Common error: Forgetting to multiply the product's binding energy per nucleon by its own nucleon number (16), or leaving out the helium-4 nucleus from the total before.
The diagram shows a fusion reaction between a hydrogen-2 nucleus and a helium-3 nucleus, with the binding energy B of each nucleus. 1 MeV = 1.60 × 10−13 J.
Explain why the binding energy of the 1H nucleus is zero.
Calculate the energy released in the reaction.
Calculate the decrease in mass in this reaction, in kg.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it is a single proton: there are no other nucleons bound to it, so no energy is needed to separate it into nucleons | ✓ 1 | |
| Part (b) | ||
| total binding energy before = 2.22 + 7.72 = 9.94 MeV; after = 28.30 MeV | ✓ 1 | |
| released = 28.30 − 9.94 = 18.36 MeV = 18.4 MeV | ✓ 1 | |
| Part (c) | ||
| 18.36 × 1.60 × 10−13 = 2.94 × 10−12 J | ✓ 1 | |
| Δm = 2.94 × 10−12 ÷ (3.00 × 108)2 = 3.26 × 10−29 kg | ✓ 1 | |
Exam tip: Energy released = total binding energy of the products − total binding energy of the reactants. A larger binding energy means less mass, so the mass decrease matches the energy released.
For two nuclei to fuse, they must approach to a separation r. Assume that r is the same for a pair of protons and for a pair of helium-4 nuclei, and that each nucleus needs a mean kinetic energy equal to half the electrostatic potential energy at that separation.
What is (temperature needed for helium-4 fusion) ÷ (temperature needed for proton fusion)?
| Answer | Mark | Notes |
|---|---|---|
| The potential energy is kQ1Q2/r. Each helium-4 nucleus has charge 2e, so Q1Q2 is 4 times larger. Mean kinetic energy ∝ T, so the temperature is 4 times larger. | ✓ 1 | Answer C |
Answer: C
Exam tip: Write the general expression first and see which quantities change: here only Q1Q2 does, and T ∝ Q1Q2.
A student uses published mass-spectrometer data to find the binding energy of the deuteron (one proton and one neutron). The masses, each with an absolute uncertainty of ±0.000002 u, are: proton 1.007276 u, neutron 1.008665 u, deuteron 2.013553 u. 1 u is equivalent to 931.5 MeV.
Calculate the mass deficit of the deuteron.
Calculate the binding energy of the deuteron in MeV.
Determine the absolute uncertainty in the mass deficit.
Calculate the percentage uncertainty in the binding energy.
The percentage uncertainty in each measured mass is no more than 0.0002%. Explain why the percentage uncertainty in the binding energy is so much larger.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.007276 + 1.008665 − 2.013553 = 0.002388 u | ✓ 1 | |
| Part (b) | ||
| 0.002388 × 931.5 = 2.22 MeV (2.224 MeV) | ✓ 1 | |
| Part (c) | ||
| for sums and differences, absolute uncertainties add | ✓ 1 | |
| 3 × 0.000002 = ±0.000006 u | ✓ 1 | |
| Part (d) | ||
| 0.000006 ÷ 0.002388 × 100 = 0.25% | ✓ 1 | |
| Part (e) | ||
| the mass deficit is a small difference between two nearly equal masses, so the absolute uncertainties are large compared with the difference itself | ✓ 1 | |
Exam tip: When quantities are added or subtracted, add their absolute uncertainties; when they are multiplied or divided, add their percentage uncertainties. Multiplying by the constant 931.5 does not change the percentage uncertainty.
Two deuterium nuclei can fuse to form a helium-3 nucleus and a neutron: 2H + 2H → 3He + n.
Atomic masses: 2H 2.014102 u, 3He 3.016029 u. Mass of a neutron = 1.008665 u. 1 u = 1.66 × 10−27 kg.
Calculate the decrease in mass in this reaction.
Calculate the energy released, in J.
The two deuterium nuclei have negligible momentum before the reaction. The masses of the helium-3 nucleus and the neutron are in the ratio 3 : 1. Show that the neutron carries about three-quarters of the energy released, and calculate its kinetic energy in MeV.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| before 2 × 2.014102 = 4.028204 u; after 3.016029 + 1.008665 = 4.024694 u | ✓ 1 | |
| Δm = 0.003510 u | ✓ 1 | |
| Part (b) | ||
| 0.003510 × 1.66 × 10−27 = 5.83 × 10−30 kg | ✓ 1 | |
| × (3.00 × 108)2 = 5.24 × 10−13 J | ✓ 1 | |
| Part (c) | ||
| equal and opposite momenta: Ek = p2/2m, so kinetic energy ∝ 1/m and the neutron gets 3/(3 + 1) = ¾ | ✓ 1 | |
| ¾ × 5.24 × 10−13 = 3.93 × 10−13 J = 2.46 MeV | ✓ 1 | |
Exam tip: Atomic masses can be used directly when the number of electrons is the same on both sides — here two deuterium atoms have two electrons in total, the same as one helium-3 atom, and the neutron has none.
The temperature in the core of a star is 1.5 × 107 K. The core contains hydrogen nuclei (protons, charge +1.60 × 10−19 C). k = 8.99 × 109 N m2 C−2.
Calculate the mean kinetic energy of a proton in the core.
Two protons, each with this kinetic energy, approach each other head-on. Calculate their distance of closest approach.
The strong force acts only over distances of about 10−15 m. Use your answer to (b) to explain why fusion in the core still takes place, but only slowly.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3⁄2kBT = 1.5 × 1.38 × 10−23 × 1.5 × 107 = 3.1 × 10−16 J | ✓ 1 | |
| Part (b) | ||
| total kinetic energy 6.21 × 10−16 J = ke2/r | ✓ 1 | |
| r = 8.99 × 109 × (1.60 × 10−19)2 ÷ 6.21 × 10−16 = 3.7 × 10−13 m | ✓ 1 | |
| Part (c) | ||
| protons with the mean kinetic energy stop several hundred times too far apart to fuse, so most collisions do not lead to fusion | ✓ 1 | |
| the protons have a spread of speeds and a very small fraction are much faster than average; the core is so dense that this fraction still gives enough fusions per second (allow reference to tunnelling) | ✓ 1 | |
Common error: Forgetting that both protons are moving. At closest approach both have stopped, so the potential energy equals the total kinetic energy of the pair.
The Sun radiates power at 3.85 × 1026 W. This power comes from a series of fusion reactions whose overall effect is to convert four protons into one helium-4 nucleus. Each helium-4 nucleus formed releases 4.3 × 10−12 J. Mass of a proton = 1.67 × 10−27 kg.
Calculate the number of helium-4 nuclei formed per second.
Calculate the mass of protons used per second.
Calculate the mass lost by the Sun each second as a result of the energy it radiates.
Calculate the percentage of the protons' mass that is converted to energy, and explain, in terms of binding energy, where this energy comes from.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3.85 × 1026 ÷ 4.3 × 10−12 = 9.0 × 1037 s−1 (8.95 × 1037) | ✓ 1 | |
| Part (b) | ||
| 4 protons per helium nucleus: 4 × 8.95 × 1037 = 3.58 × 1038 protons per second | ✓ 1 | |
| × 1.67 × 10−27 = 6.0 × 1011 kg s−1 | ✓ 1 | |
| Part (c) | ||
| Δm = P/c2 per second = 3.85 × 1026 ÷ (3.00 × 108)2 | ✓ 1 | |
| = 4.3 × 109 kg s−1 | ✓ 1 | |
| Part (d) | ||
| 4.28 × 109 ÷ 5.98 × 1011 × 100 = 0.72% | ✓ 1 | |
| helium-4 has a much higher binding energy per nucleon than free protons (which have none), so the products have less mass than the reactants; the difference is released | ✓ 1 | |
Exam tip: Every joule radiated corresponds to a mass of 1/c2 kg. For a power, Δm per second = P/c2.
The graph shows the mass per nucleon of a range of nuclei plotted against nucleon number. The values for helium-4 and carbon-12 are labelled.
In the core of a star, three helium-4 nuclei fuse to form one carbon-12 nucleus. 1 u is equivalent to 931.5 MeV.
What is the energy released in this reaction?
| Answer | Mark | Notes |
|---|---|---|
| The 12 nucleons each lose 1.000376 − 0.999726 = 0.000650 u. Mass decrease = 12 × 0.000650 = 0.00780 u. Energy released = 0.00780 × 931.5 = 7.27 MeV. | ✓ 1 | Answer D |
Answer: D
Exam tip: Mass per nucleon must be multiplied by the number of nucleons involved. Nucleon number is conserved, so the mass lost is (difference per nucleon) × (total nucleons).
The graph shows how the potential energy of a deuterium nucleus (charge +e) and a tritium nucleus (charge +e) varies with their separation r. The strong force acts at separations below 3.0 fm. Point B marks the top of the energy barrier.
kB = 1.38 × 10−23 J K−1. 1 MeV = 1.60 × 10−13 J.
Name the force responsible for the shape of the curve for r greater than 3.0 fm, and explain why the potential energy increases as r decreases.
Use the graph to determine the height of the barrier at B, in J.
The two nuclei approach head-on with equal kinetic energies. Calculate the temperature at which the mean kinetic energy of the nuclei would be just enough for them to reach B.
Deuterium–tritium fusion takes place in plasmas at about 1.5 × 108 K. Suggest why fusion can occur at this lower temperature.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| electrostatic (Coulomb) repulsion between the two positive nuclei; work must be done against it to push them closer together | ✓ 1 | |
| Part (b) | ||
| reading at B: 0.48 MeV (allow 0.47–0.49 MeV) | ✓ 1 | |
| 0.48 × 1.60 × 10−13 = 7.7 × 10−14 J (allow 7.5–7.8 × 10−14 J) | ✓ 1 | |
| Part (c) | ||
| each nucleus needs half the barrier height: 3.84 × 10−14 J | ✓ 1 | |
| 3⁄2kBT = 3.84 × 10−14, so T = 3.84 × 10−14 ÷ (1.5 × 1.38 × 10−23) = 1.9 × 109 K (allow 1.8–1.9 × 109 K) | ✓ 1 | |
| Part (d) | ||
| the nuclei have a spread of kinetic energies, so a small fraction have far more than the mean / nuclei can tunnel through the barrier without reaching its top (either) | ✓ 1 | |
Exam tip: When two nuclei move towards each other with equal kinetic energies, each needs only half of the potential energy at closest approach. Set 3⁄2kBT equal to that half.
The flow diagram shows a chain of fusion reactions that converts hydrogen-1 into helium-4 in the core of a star, with the energy released in each reaction.
Atomic masses: 1H 1.007825 u, 4He 4.002603 u. 1 u is equivalent to 931.5 MeV. 1 u = 1.66 × 10−27 kg. 1 MeV = 1.60 × 10−13 J.
Use the diagram to show that the overall effect of the chain is to convert four 1H nuclei into one 4He nucleus.
Calculate the total energy released for each helium-4 nucleus formed.
Use the atomic masses to show that your answer to (b) is consistent with the decrease in mass.
Calculate the energy released per kilogram of hydrogen-1 converted into helium-4.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| six 1H nuclei are used (two in each first reaction, one in each second reaction) and two are released in the last reaction, so 6 − 2 = 4 are used up for one 4He | ✓ 1 | |
| Part (b) | ||
| the first two reactions each happen twice: 2 × 1.44 + 2 × 5.49 + 12.86 | ✓ 1 | |
| = 26.72 MeV = 26.7 MeV | ✓ 1 | |
| Part (c) | ||
| Δm = 4 × 1.007825 − 4.002603 = 0.028697 u | ✓ 1 | |
| 0.028697 × 931.5 = 26.73 MeV, which agrees with (b) | ✓ 1 | |
| Part (d) | ||
| 26.72 × 1.60 × 10−13 = 4.28 × 10−12 J; mass of four 1H atoms = 4 × 1.007825 × 1.66 × 10−27 = 6.69 × 10−27 kg | ✓ 1 | |
| 4.28 × 10−12 ÷ 6.69 × 10−27 = 6.4 × 1014 J kg−1 | ✓ 1 | |
Exam tip: Atomic masses work here because four hydrogen atoms bring two more electrons than one helium atom, and those two electrons annihilate with the two positrons; the diagram energies include that annihilation energy.
The diagram shows a gamma-ray photon splitting a beryllium-9 nucleus into a beryllium-8 nucleus and a free neutron.
Atomic masses: 9Be 9.012183 u, 8Be 8.005305 u. Mass of a neutron = 1.008665 u. 1 u is equivalent to 931.5 MeV. 1 MeV = 1.60 × 10−13 J. h = 6.63 × 10−34 J s.
Explain why energy must be supplied for this process to happen.
Calculate the minimum photon energy, in MeV, that can cause this process.
Calculate the minimum frequency of a photon that can cause this process.
Estimate the total kinetic energy of the products for the photon shown in the diagram.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the products have a greater total mass than the beryllium-9 nucleus, so energy must be supplied (work must be done against the strong force to remove the neutron) | ✓ 1 | |
| Part (b) | ||
| increase in mass = 8.005305 + 1.008665 − 9.012183 = 0.001787 u | ✓ 1 | |
| 0.001787 × 931.5 = 1.665 MeV = 1.66 MeV | ✓ 1 | |
| Part (c) | ||
| 1.665 × 1.60 × 10−13 = 2.66 × 10−13 J | ✓ 1 | |
| f = E/h = 2.66 × 10−13 ÷ 6.63 × 10−34 = 4.02 × 1020 Hz | ✓ 1 | |
| Part (d) | ||
| 2.00 − 1.66 = 0.34 MeV, shared between the neutron and the beryllium-8 nucleus | ✓ 1 | |
Exam tip: Atomic masses can be used because both sides contain 4 electrons. The minimum photon energy is the energy equivalent of the mass gained; any extra photon energy becomes kinetic energy of the products.
In one type of fusion experiment, a small spherical pellet of fuel is compressed and heated by laser beams, as shown in the diagram. The pellet contains 1.0 mg of fuel made of equal numbers of deuterium (2H) and tritium (3H) atoms. Each fusion of a deuterium nucleus with a tritium nucleus releases 17.6 MeV.
Atomic masses: 2H 2.014102 u, 3H 3.016049 u. 1 u = 1.66 × 10−27 kg. 1 MeV = 1.60 × 10−13 J.
Use the diagram to calculate the factor by which the density of the fuel increases.
Explain why the fuel must be both very hot and very dense for a useful amount of energy to be released.
Calculate the number of deuterium–tritium pairs in the pellet.
In one shot, 20% of the pairs fuse. Calculate the energy released.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| density ∝ 1/volume ∝ 1/r3: (1.0 ÷ 0.050)3 = 8.0 × 103 | ✓ 1 | |
| Part (b) | ||
| high temperature: the nuclei need very large kinetic energies to overcome the electrostatic repulsion and come within range of the strong force, and mean kinetic energy is proportional to temperature | ✓ 1 | |
| high density: more nuclei per unit volume, so collisions, and therefore fusions, happen at a much greater rate (before the pellet flies apart) | ✓ 1 | |
| Part (c) | ||
| mass of one pair = (2.014102 + 3.016049) × 1.66 × 10−27 = 8.35 × 10−27 kg | ✓ 1 | |
| N = 1.0 × 10−6 ÷ 8.35 × 10−27 = 1.2 × 1020 | ✓ 1 | |
| Part (d) | ||
| 0.20 × 1.2 × 1020 = 2.4 × 1019 fusions | ✓ 1 | |
| 2.4 × 1019 × 17.6 × 1.60 × 10−13 = 6.7 × 107 J (allow 6.7–6.8 × 107 J) | ✓ 1 | |
Common error: Using the ratio of the radii (20) as the density factor. Density depends on volume, and volume scales with the cube of the radius.
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