A Level Physics 23.1 Mass Defect and Nuclear Binding Energy MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 23.1 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level (A2) topic. It explains why a nucleus weighs less than its separate nucleons, how the missing mass becomes binding energy through E = Δm c2, and why both fission and fusion release energy.
Work through all 30 original practice questions, using the Show Answer toggle on each one to check your answer and see the full working.
Formula recap
- Mass defect: Δm = (Z mp + N mn) − mnucleus — the nucleus is always LIGHTER than its separate nucleons
- Binding energy: E = Δm c2, with c = 3.00 x 108 m/s
- 1 u = 1.66 x 10-27 kg; a mass defect of 1 u is equivalent to 931.5 MeV
- mp = 1.00728 u, mn = 1.00867 u, me = 0.00055 u
- Binding energy per nucleon = total binding energy / nucleon number A — the bigger it is, the more stable the nucleus
Full notes: A Level Physics Formula Sheet
Mass Defect & Binding Energy
Q1. What is meant by the mass defect of a nucleus?
- (A) The mass lost by an atom when it emits an alpha particle
- (B) The difference between the total mass of the separate nucleons and the mass of the nucleus
- (C) The total mass of the neutrons inside the nucleus
- (D) The difference between the mass of the atom and the mass of its electrons
Show Answer
Answer: B – The mass defect Δm is the amount by which the nucleus is lighter than the total mass of its separate protons and neutrons: Δm = (Z mp + N mn) − mnucleus. It is not a loss during a decay; it is a permanent feature of the bound nucleus.
Q2. The binding energy of a nucleus is the energy
- (A) that holds each electron in its orbit around the nucleus
- (B) released when the nucleus undergoes radioactive decay
- (C) required to separate the nucleus completely into its constituent protons and neutrons
- (D) stored in the nucleus because of its electric charge
Show Answer
Answer: C – Binding energy is the minimum energy that must be supplied to break the nucleus apart into separate nucleons at rest, far from each other. By energy conservation it is also the energy released when the nucleus forms from separate nucleons.
Q3. A helium-4 nucleus contains 2 protons and 2 neutrons and has mass 4.00151 u. Using mp = 1.00728 u and mn = 1.00867 u, the mass defect of the helium-4 nucleus is
- (A) 0.03039 u
- (B) 0.3039 u
- (C) 0.003039 u
- (D) 4.03190 u
Show Answer
Answer: A – Total mass of separate nucleons = 2(1.00728) + 2(1.00867) = 2.01456 + 2.01734 = 4.03190 u. Mass defect = 4.03190 − 4.00151 = 0.03039 u. (Option D is the total nucleon mass, before subtracting the nuclear mass.)
Q4. The mass defect of helium-4 is 0.03039 u. Given that 1 u is equivalent to 931.5 MeV, the binding energy of the helium-4 nucleus is
- (A) 2.83 MeV
- (B) 283 MeV
- (C) 7.08 MeV
- (D) 28.3 MeV
Show Answer
Answer: D – Binding energy = 0.03039 x 931.5 = 28.31 MeV, about 28.3 MeV. (Option C, 7.08 MeV, is the binding energy PER nucleon, found in Q12.)
Q5. A nucleus has a mass defect of 0.050 u. Its binding energy is approximately
- (A) 4.66 MeV
- (B) 466 MeV
- (C) 46.6 MeV
- (D) 0.466 MeV
Show Answer
Answer: C – Binding energy = 0.050 x 931.5 = 46.575 MeV, about 46.6 MeV.
Q6. A nucleus has a mass defect of 5.0 x 10-29 kg. Using c = 3.00 x 108 m/s, its binding energy in joules is
- (A) 4.5 x 10-13 J
- (B) 4.5 x 10-12 J
- (C) 1.5 x 10-20 J
- (D) 5.0 x 10-13 J
Show Answer
Answer: B – E = Δm c2 = 5.0 x 10-29 x (3.00 x 108)2 = 5.0 x 10-29 x 9.00 x 1016 = 4.5 x 10-12 J. (Option C uses c instead of c2.)
Q7. The binding energy E of a nucleus whose mass defect is Δm is given by
- (A) E = Δm c2
- (B) E = Δm c
- (C) E = Δm / c2
- (D) E = (1/2) Δm c2
Show Answer
Answer: A – This is Einstein’s mass-energy equivalence applied to the nucleus: the “missing” mass Δm has been converted into the binding energy E = Δm c2 released when the nucleus formed.
Q8. The mass of a helium-4 ATOM is 4.00260 u and the mass of an electron is 0.00055 u. The mass of the helium-4 nucleus alone is
- (A) 4.00260 u
- (B) 4.00205 u
- (C) 4.00315 u
- (D) 4.00150 u
Show Answer
Answer: D – Tabulated atomic masses include the electrons. A helium atom has TWO electrons, so the nuclear mass = 4.00260 − 2(0.00055) = 4.00260 − 0.00110 = 4.00150 u, in agreement with the value 4.00151 u used in Q3. Forgetting to subtract the electrons (or subtracting only one) is a common error in mass defect calculations.
Q9. A mass defect of 2.0 u corresponds to a binding energy of
- (A) 466 MeV
- (B) 1863 MeV
- (C) 931.5 MeV
- (D) 186.3 MeV
Show Answer
Answer: B – Binding energy = 2.0 x 931.5 = 1863 MeV (about 1.86 GeV).
Q10. Which statement about the binding energy of a nucleus is correct?
- (A) It is the energy needed to remove one electron from the atom
- (B) It is equal to the mass of the nucleus multiplied by c2
- (C) It is the minimum energy needed to split the nucleus into separate protons and neutrons
- (D) It is measured in atomic mass units
Show Answer
Answer: C – Binding energy is an energy (measured in joules or MeV, not u), and it involves the mass DEFECT, not the whole nuclear mass. Option A describes an ionisation energy of the atom, which is far smaller.
Binding Energy per Nucleon
Q11. The binding energy per nucleon of a nucleus is calculated by dividing the total binding energy by
- (A) the nucleon number A
- (B) the proton number Z
- (C) the neutron number N
- (D) the mass of the nucleus in u
Show Answer
Answer: A – Binding energy per nucleon = (total binding energy) / A, where A = Z + N is the total number of nucleons. It gives a fair comparison of stability between nuclei of different sizes.
Q12. The binding energy of helium-4 is 28.3 MeV. Its binding energy per nucleon is
- (A) 28.3 MeV
- (B) 7.1 MeV
- (C) 3.5 MeV
- (D) 14.2 MeV
Show Answer
Answer: B – Helium-4 has A = 4 nucleons, so binding energy per nucleon = 28.3 / 4 = 7.08 MeV, about 7.1 MeV per nucleon. This is high for such a light nucleus, which is why the alpha particle is a particularly stable, tightly bound unit.
Q13. Nucleus X has a binding energy per nucleon of 8.0 MeV and nucleus Y has a binding energy per nucleon of 7.2 MeV. This means that nucleus X is
- (A) less stable than Y, because it is heavier
- (B) exactly as stable as Y
- (C) radioactive, while Y is stable
- (D) more stable than Y
Show Answer
Answer: D – The greater the binding energy per nucleon, the more energy is needed to pull each nucleon out of the nucleus, so the nucleus is more tightly bound and more stable. X (8.0 MeV per nucleon) is more stable than Y (7.2 MeV per nucleon).
Q14. On the graph of binding energy per nucleon against nucleon number, the maximum value of about 8.8 MeV per nucleon occurs for nuclei close to
- (A) helium-4
- (B) uranium-235
- (C) iron-56
- (D) hydrogen-1
Show Answer
Answer: C – The curve rises steeply for light nuclei and reaches its peak at iron-56 (about 8.8 MeV per nucleon), the most stable nucleus per nucleon. Beyond iron it falls slowly, to about 7.6 MeV per nucleon for uranium.
Q15. The total binding energy of a deuterium nucleus (hydrogen-2, A = 2) is 2.2 MeV. Its binding energy per nucleon is
- (A) 1.1 MeV
- (B) 2.2 MeV
- (C) 4.4 MeV
- (D) 0.55 MeV
Show Answer
Answer: A – 2.2 / 2 = 1.1 MeV per nucleon. Deuterium is only weakly bound per nucleon, which is why energy is released when light nuclei like deuterium fuse into more tightly bound nuclei.
Q16. A uranium-235 nucleus has a binding energy per nucleon of 7.6 MeV. The total binding energy of the nucleus is about
- (A) 7.6 MeV
- (B) 243 MeV
- (C) 32 MeV
- (D) 1790 MeV
Show Answer
Answer: D – Total binding energy = 7.6 x 235 = 1786 MeV, about 1790 MeV. Although uranium has a huge total binding energy, its binding energy PER nucleon is lower than iron’s, so it can release energy by splitting (see Q23).
Q17. The total binding energy of an oxygen-16 nucleus is 128 MeV. Its binding energy per nucleon is
- (A) 128 MeV
- (B) 8.0 MeV
- (C) 16 MeV
- (D) 2.0 MeV
Show Answer
Answer: B – 128 / 16 = 8.0 MeV per nucleon, close to the iron peak, so oxygen-16 is a very stable nucleus.
Q18. On the binding energy per nucleon curve, as the nucleon number increases beyond about 60, the binding energy per nucleon
- (A) increases rapidly
- (B) stays at its maximum value
- (C) decreases slowly
- (D) falls to zero at A = 100
Show Answer
Answer: C – Past the iron peak the curve falls only gently: even uranium still has about 7.6 MeV per nucleon. The slow fall happens because, as more protons are added, their mutual electrostatic repulsion increasingly offsets the attractive strong nuclear force.
Q19. Carbon-12 has a binding energy per nucleon of 7.7 MeV. The total binding energy of a carbon-12 nucleus is
- (A) 7.7 MeV
- (B) 0.64 MeV
- (C) 84.7 MeV
- (D) 92.4 MeV
Show Answer
Answer: D – Total binding energy = 7.7 x 12 = 92.4 MeV.
Q20. A tritium nucleus (hydrogen-3, A = 3) has a total binding energy of 8.5 MeV. Its binding energy per nucleon is
- (A) 2.8 MeV
- (B) 8.5 MeV
- (C) 25.5 MeV
- (D) 1.4 MeV
Show Answer
Answer: A – 8.5 / 3 = 2.83 MeV, about 2.8 MeV per nucleon. Like deuterium, tritium is a light, weakly bound nucleus used as a fusion fuel (see Q25).
Fission & Fusion
Q21. Nuclear fission is the process in which
- (A) two light nuclei join together to form a heavier nucleus
- (B) a nucleus emits an alpha particle
- (C) a heavy nucleus splits into two nuclei of comparable mass
- (D) a nucleus absorbs an electron from its inner shell
Show Answer
Answer: C – In fission a very heavy nucleus (such as uranium-235, often after absorbing a neutron) splits into two medium-mass nuclei plus neutrons. Option A describes fusion, the opposite journey along the binding energy curve.
Q22. Nuclear fusion is the process in which
- (A) two light nuclei combine to form a more massive nucleus
- (B) a heavy nucleus splits into two lighter nuclei
- (C) a neutron decays into a proton and an electron
- (D) two atoms share a pair of electrons
Show Answer
Answer: A – In fusion, light nuclei such as deuterium and tritium join to form a heavier nucleus. Energy is released because the product is more tightly bound per nucleon than the reactants.
Q23. A uranium-235 nucleus (binding energy per nucleon 7.6 MeV) undergoes fission to form two nuclei with a binding energy per nucleon of 8.4 MeV. The energy released in this fission is about
- (A) 0.8 MeV
- (B) 188 MeV
- (C) 8.4 MeV
- (D) 1974 MeV
Show Answer
Answer: B – Energy released = (increase in binding energy per nucleon) x (number of nucleons) = (8.4 − 7.6) x 235 = 0.8 x 235 = 188 MeV, roughly 200 MeV per fission. (Option D, 1974 MeV, wrongly multiplies the final value 8.4 by 235 without subtracting the initial binding energy.)
Q24. A nucleus of mass number 200 with binding energy per nucleon 7.8 MeV splits into two nuclei of mass number 100, each with binding energy per nucleon 8.3 MeV. The energy released is
- (A) 0.5 MeV
- (B) 50 MeV
- (C) 1660 MeV
- (D) 100 MeV
Show Answer
Answer: D – All 200 nucleons gain (8.3 − 7.8) = 0.5 MeV of binding energy each, so the energy released = 0.5 x 200 = 100 MeV. The nucleon number stays 200 in total; only the binding per nucleon changes.
Q25. A deuterium nucleus (binding energy 2.2 MeV) fuses with a tritium nucleus (binding energy 8.5 MeV) to form a helium-4 nucleus (binding energy 28.3 MeV) and a neutron. The energy released is
- (A) 10.7 MeV
- (B) 17.6 MeV
- (C) 28.3 MeV
- (D) 39.0 MeV
Show Answer
Answer: B – Energy released = (binding energy of products) − (binding energy of reactants) = 28.3 − (2.2 + 8.5) = 28.3 − 10.7 = 17.6 MeV, shared as kinetic energy of the helium nucleus and the neutron. This is the famous deuterium-tritium fusion reaction.
Q26. Both fission and fusion can release energy because in each case
- (A) the products are always lighter nuclei than the reactants
- (B) mass is created during the reaction
- (C) the products have a greater total binding energy than the reactants, so some mass is converted into energy
- (D) extra neutrons are produced
Show Answer
Answer: C – Fission moves heavy nuclei UP the curve towards iron from the right; fusion moves light nuclei UP the curve towards iron from the left. In both, the products sit higher on the binding energy per nucleon curve, total binding energy increases, and the equivalent mass Δm appears as energy through E = Δm c2.
Q27. Fusion reactions require extremely high temperatures, of the order of 108 K, mainly so that
- (A) the nuclei have enough kinetic energy to overcome their electrostatic repulsion and get close enough for the strong nuclear force to act
- (B) the solid fuel can melt into a liquid
- (C) the strong nuclear force becomes stronger
- (D) neutrons can be produced from protons
Show Answer
Answer: A – Nuclei are positively charged and repel each other strongly. Only at very high temperatures do they move fast enough to approach within the very short range of the strong nuclear force, which can then bind them together. This is why fusion powers the Sun but is so hard to sustain on Earth.
Q28. A self-sustaining fission chain reaction is possible because each uranium-235 fission event
- (A) produces a large amount of heat
- (B) absorbs all the neutrons around it
- (C) splits the atoms of the moderator
- (D) releases two or three neutrons, each of which can go on to cause another fission
Show Answer
Answer: D – A typical fission is triggered by one neutron but releases two or three more. If on average at least one of those neutrons causes a further fission, the reaction sustains itself; control rods absorb surplus neutrons to keep the rate steady.
Q29. The energy released by the Sun comes mainly from
- (A) the fission of uranium nuclei
- (B) the fusion of helium nuclei into carbon
- (C) the fusion of hydrogen nuclei into helium
- (D) the fission of iron nuclei
Show Answer
Answer: C – In the Sun’s core, hydrogen nuclei (protons) fuse through a series of steps into helium-4. Each helium nucleus formed has a mass defect of about 0.030 u relative to its ingredients, and that mass appears as the sunlight energy reaching Earth. Iron cannot release energy by fission or fusion because it sits at the peak of the binding energy curve.
Q30. A single fission of uranium-235 releases about 200 MeV of energy. Expressed in joules, this is (1 eV = 1.6 x 10-19 J)
- (A) 1.3 x 10-9 J
- (B) 3.2 x 10-11 J
- (C) 3.2 x 10-9 J
- (D) 5.0 x 10-12 J
Show Answer
Answer: B – 200 MeV = 200 x 106 eV = 2.0 x 108 eV. Energy = 2.0 x 108 x 1.6 x 10-19 = 3.2 x 10-11 J per fission. This is tiny for one nucleus, but a kilogram of uranium-235 contains about 2.6 x 1024 nuclei, which is why nuclear fuel is so energy-dense.
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