HSC Exam Practice

Sample response. (1) Short range: effective only over ~1–3 fm; it drops to essentially zero beyond nuclear dimensions. (2) Attractive: it attracts all nucleon pairs (p–p, n–n, p–n). (3) Charge-independent: acts with roughly equal strength regardless of whether nucleons are charged. (4) Saturated: each nucleon only interacts with its immediate neighbours, not with all other nucleons; this keeps nuclear density approximately constant. (5) Repulsive at very short distances (<0.5 fm): prevents nucleons from collapsing into each other.

Marking notes. 1 mark per distinct, correct property. Accept any four from: short range, attractive, charge-independent, saturated, repulsive at very short distances.

Sample response. The mass defect (Δm) is the difference between the total mass of all the separated nucleons (Zm_p + Nm_n) and the actual mass of the nucleus: Δm = Zm_p + Nm_n − m_nucleus. The binding energy (E_b) is the energy required to completely disassemble the nucleus into its constituent nucleons, or equivalently, the energy released when those nucleons combine. It equals the energy equivalent of the mass defect: E_b = Δm × c².

Marking notes. 1 mark for defining mass defect with a correct formula or description; 1 mark for defining binding energy as the energy to separate the nucleus into nucleons; 1 mark for stating E_b = Δm × c² (accept E = mc² with context).

Sample response. Electrons orbit at distances of approximately 10⁻¹⁰ m from the nucleus — roughly 100,000 times greater than the nuclear radius (~10⁻¹⁵ m). The strong nuclear force has a range of only ~1–3 fm; at the orbital distance of electrons it is effectively zero. Therefore the strong force exerts no force on electrons.

Marking notes. 1 mark for stating the strong force is short-range (~1–3 fm); 1 mark for stating electrons orbit at distances far exceeding this range, so the force cannot reach them.

Sample response. The binding-energy-per-nucleon curve rises steeply from near zero for hydrogen (A = 1) through a rapid increase for light nuclei, reaches a broad maximum near A = 56 at approximately 8.8 MeV/nucleon, then gradually decreases for heavier nuclei as Coulomb repulsion between the many protons reduces average binding. The nucleus at the maximum is iron-56 (⁵⁶Fe), the most tightly bound and most stable nucleus in nature.

Marking notes. 1 mark for describing the curve rising steeply for light nuclei then decreasing for heavy nuclei; 1 mark for identifying iron-56 at the peak; 1 mark for stating the approximate peak value (~8.8 MeV/nucleon).

Sample response. In very heavy nuclei, there are a large number of protons. The Coulomb force between protons is long-range and acts between every proton pair in the nucleus; its cumulative repulsive effect grows with Z² [1]. The strong nuclear force, by contrast, is short-range and saturates — each nucleon only interacts with its immediate neighbours and does not contribute proportionally more binding as Z increases [1]. For A > ~200, the accumulated Coulomb repulsion exceeds the short-range strong force binding, making the nucleus energetically unstable and prone to alpha or beta decay [1].

Marking notes. 1 mark for explaining that Coulomb repulsion accumulates with proton number and is long-range; 1 mark for contrasting this with the short-range, saturating strong force; 1 mark for concluding Coulomb wins in heavy nuclei, causing instability.

Sample response. Energy is released in any nuclear reaction where the products have a higher binding energy per nucleon than the reactants [1]. For fusion: light nuclei (A < 56) lie on the left of the binding-energy peak and have relatively low E_b/A. Fusing them produces a nucleus closer to iron-56 with higher E_b/A; the increase in binding energy is released as kinetic energy [1]. For fission: heavy nuclei (A > 56) lie on the right of the peak with lower E_b/A than iron-56. Splitting them produces medium-mass fragments with higher E_b/A; again the difference is released as energy [1].

(a) Mass defect [2 marks]. Δm = 2(1.007276) + 2(1.008665) − 4.001506 = 2.014552 + 2.017330 − 4.001506 = 0.030376 u [1 for correct working; 1 for correct value 0.030376 u].

(b) Binding energy [2 marks]. E_b = 0.030376 × 931.5 = 28.30 MeV [1]. E_b = 28.30 × 1.602 × 10⁻¹³ = 4.53 × 10⁻¹² J [1].

(c) Binding energy per nucleon and comparison [2 marks]. E_b/A = 28.30/4 = 7.07 MeV/nucleon [1]. This is less than iron-56 (~8.79 MeV/nucleon), so helium-4 is less tightly bound than iron-56 and is less stable; however, it is more stable than very light nuclei such as hydrogen. Helium-4 lies on the steeply rising part of the binding-energy curve [1].

(d) Assumption [1 mark]. The calculation uses nuclear masses (mass of protons and neutrons alone, excluding electrons). If atomic masses were used, the electron masses on both sides of the equation would need to cancel correctly. Accept: the calculation ignores electron binding energy, which is negligible compared to nuclear binding energies.

Sample Band 6 response. The strong nuclear force is the fundamental agent responsible for nuclear stability and binding. Its four key properties determine how nuclei behave: it is short-range (~1–3 fm), attractive between all nucleon pairs (charge-independent), saturating (each nucleon acts only with immediate neighbours), and repulsive below ~0.5 fm to prevent nuclear collapse. Because the strong force is attractive and stronger than the Coulomb force at nuclear distances (~100×), it overcomes the mutual repulsion between protons in small and medium nuclei and binds them into stable configurations. The energy released when nucleons combine — the binding energy — corresponds to the mass defect via E_b = Δmc². The binding-energy-per-nucleon curve is a direct expression of how tightly bound different nuclei are. It rises steeply for light nuclei, peaks at iron-56 (~8.8 MeV/nucleon), and declines gradually for heavier nuclei. Iron-56 represents the thermodynamic “valley” of nuclear stability: it has the greatest binding per nucleon, and any nuclear reaction that moves products toward this peak releases energy. For heavy nuclei, the saturation of the strong force means that adding protons increases Coulomb repulsion cumulatively (long-range, acting between all proton pairs) without a proportional increase in strong-force binding. This accounts for the decline in E_b/A for A > 56 and the instability of very heavy nuclei, which undergo alpha or beta decay to move toward lower Z. The binding-energy curve thus directly determines which nuclear reactions release energy. Fusion of light nuclei (left of the peak) releases energy by increasing E_b/A. Fission of heavy nuclei (right of the peak) does the same. In both cases, the energy released equals the difference in total binding energy between products and reactants, by mass–energy equivalence. This shows that the strong force, through its specific properties of short range and saturation, is ultimately responsible for both the stability of matter and the energy available in nuclear reactions.

Marking criteria (7 marks). 1 = correctly identifies at least two properties of the strong force with explanations of their physical consequences for nuclear stability. 1 = defines mass defect and binding energy correctly, with E_b = Δmc². 1 = describes the binding-energy-per-nucleon curve (steep rise, peak at Fe-56, gradual fall for heavy nuclei). 1 = explains why E_b/A decreases for heavy nuclei (Coulomb accumulates; strong force saturates). 1 = explains fusion energy release using E_b/A argument. 1 = explains fission energy release using E_b/A argument. 1 = response is cohesive, uses precise scientific terminology throughout (binding energy, mass defect, Coulomb repulsion, saturation, nucleon), and integrates the strong force properties with the energy argument throughout.