Chapter 4 Quiz — The Semi-Empirical Mass Formula

Instructions: Select the best answer for each question. Answers are provided at the end.

Q1. The semi-empirical mass formula is called "semi-empirical" because:

(a) Half of its terms are derived from theory and half from experiment (b) The functional form of each term is motivated by physics, but the coefficients are fit to experimental data (c) It works for approximately half of all known nuclei (d) It was developed jointly by theorists and experimentalists

Q2. The volume term $a_V A$ is the dominant (largest) term in the SEMF. Physically, the volume term reflects:

(a) The fact that nuclei are held together by gravity (b) The saturation of the nuclear force — each nucleon interacts with a fixed number of neighbors (c) The total Coulomb repulsion among all protons (d) The energy stored in the nuclear electric field

Q3. Without the Coulomb term, the binding energy per nucleon $B/A$ would:

(a) Decrease monotonically with $A$ (b) Be approximately constant for all $A$ (approaching $a_V$ minus surface corrections) (c) Increase as $A^2$ (d) Be zero for all nuclei

Q4. The surface term $-a_S A^{2/3}$ is MOST important (has the largest relative effect on $B/A$) for:

(a) Light nuclei, where the surface-to-volume ratio is large (b) Heavy nuclei, where the surface area is large in absolute terms (c) Nuclei with $N = Z$ only (d) Deformed nuclei only

Q5. The Coulomb coefficient $a_C$ can be calculated from first principles using electrostatics. The calculation gives $a_C \approx 0.72$ MeV. The empirically fitted value is approximately 0.71 MeV. This close agreement:

(a) Is a coincidence with no physical significance (b) Confirms that the nucleus is well-approximated as a uniformly charged sphere (c) Proves that the nuclear force is electromagnetic in origin (d) Shows that quantum effects are negligible for the Coulomb interaction

Q6. The asymmetry term penalizes nuclei with $N \neq Z$. Its physical origin is:

(a) The electromagnetic repulsion between excess neutrons (b) The Pauli exclusion principle — placing extra nucleons of one type forces them into higher energy levels (c) The fact that the strong force is weaker between neutrons than between protons (d) The gravitational interaction between nucleons

Q7. Light stable nuclei (e.g., $^4$He, $^{12}$C, $^{16}$O) tend to have $N = Z$. Heavy stable nuclei (e.g., $^{208}$Pb) have $N > Z$. This trend is caused by the competition between:

(a) The volume term (favoring large $A$) and the surface term (favoring small $A$) (b) The asymmetry term (favoring $N = Z$) and the Coulomb term (favoring fewer protons) (c) The pairing term (favoring even-even) and the asymmetry term (favoring $N = Z$) (d) The strong force (favoring many nucleons) and gravity (favoring few nucleons)

Q8. Of the 254 known stable nuclides, approximately how many are even-even ($Z$ even, $N$ even)?

(a) About 30 (roughly one-quarter) (b) About 65 (roughly one-quarter) (c) About 166 (roughly two-thirds) (d) About 250 (nearly all)

Q9. For odd-$A$ isobars (fixed $A$, varying $Z$), the SEMF predicts the nuclear mass varies as:

(a) A single parabola in $Z$, with one most-stable isobar (b) Two parabolas (one for even-$Z$, one for odd-$Z$), with potentially two stable isobars (c) A linear function of $Z$ (d) A cubic function of $Z$

Q10. The neutron drip line is:

(a) The locus of nuclei with the most neutrons that are still stable against beta decay (b) The locus of nuclei where the neutron separation energy $S_n$ goes to zero — adding one more neutron gives an unbound system (c) The line $N = Z$ on the chart of nuclides (d) The maximum neutron number observed in a laboratory, limited by technology rather than physics

Q11. The fissility parameter $x = a_C Z^2 / (2 a_S A)$ determines the stability of a nucleus against fission. For $^{238}$U, $x \approx 0.71$. A nucleus is predicted to undergo instantaneous spontaneous fission when:

(a) $x < 0$ (b) $x = 0.5$ (c) $x \geq 1$ (d) $x = Z/A$

Q12. The most significant systematic deviations between the SEMF and experimental binding energies occur at:

(a) Nuclei with $A$ divisible by 4 (b) Nuclei at or near magic numbers ($N$ or $Z = 2, 8, 20, 28, 50, 82, 126$) (c) Nuclei with odd mass number (d) Nuclei with $Z > 92$

Q13. The SEMF predicts that $B/A$ peaks near $A \approx 56$–$62$. The peak exists because:

(a) Of a coincidence between the values of the five SEMF parameters (b) The surface term (which decreases $B/A$ for small $A$) and the Coulomb term (which decreases $B/A$ for large $A$) create a maximum at intermediate $A$ (c) Shell closure effects at $^{56}$Fe (d) The pairing term is largest at $A \approx 60$

Q14. $^{208}$Pb is doubly magic ($Z = 82$, $N = 126$). Compared to the SEMF prediction for nearby non-magic nuclei, $^{208}$Pb is:

(a) Less tightly bound than predicted (negative residual) (b) More tightly bound than predicted (positive residual) (c) Exactly as predicted (zero residual) (d) Unbound according to the SEMF

Q15. Why does the proton drip line lie closer to the valley of stability than the neutron drip line?

(a) Protons are heavier than neutrons (b) The Coulomb barrier partially confines protons even when $S_p < 0$, but there is no equivalent barrier for neutrons (c) There are more stable even-$Z$ nuclei than stable odd-$Z$ nuclei (d) The neutron drip line is an artifact of the SEMF and does not exist in nature

Q16. A nuclear physicist claims: "The SEMF is useless because it fails for light nuclei and at magic numbers." The best response is:

(a) The claim is correct — the SEMF should not be used for any purpose (b) The SEMF's failures at magic numbers motivated the shell model, and for medium-to-heavy nuclei away from magic numbers, it remains an accurate and physically insightful baseline (c) The SEMF is exact for all nuclei with $A > 100$ (d) The failures can be fixed by using more decimal places for the coefficients

Q17. The nuclear symmetry energy (closely related to $a_{\text{sym}}$) plays a critical role in:

(a) The design of nuclear weapons only (b) Neutron star structure — it determines the proton fraction in dense nuclear matter (c) Alpha decay half-lives only (d) The color of gamma rays emitted by nuclei

Q18. Consider two nuclei: $^{48}$Ca ($Z = 20$, $N = 28$) and $^{48}$Ti ($Z = 22$, $N = 26$). Both have $A = 48$. Based on the SEMF:

(a) $^{48}$Ca should be more bound because it has fewer protons (lower Coulomb energy) (b) $^{48}$Ti should be more bound because it is closer to $N = Z$ (c) The SEMF predicts equal binding because they have the same $A$ (d) Neither can be properly analyzed with the SEMF because $A < 50$

Answers

Q Answer Explanation
1 (b) The physics determines the form; experiment determines the coefficients.
2 (b) Saturation means each nucleon contributes a roughly constant binding energy.
3 (b) Without Coulomb, only volume and surface terms remain; $B/A \to a_V - a_S A^{-1/3}$, which approaches $a_V$ from below.
4 (a) The surface correction to $B/A$ scales as $A^{-1/3}$, largest for small $A$.
5 (b) The agreement validates the uniform charge sphere model for the nuclear Coulomb energy.
6 (b) Pauli exclusion forces excess nucleons into higher shells, costing kinetic energy.
7 (b) The asymmetry term pulls toward $N = Z$; the Coulomb term pushes toward fewer protons (more neutrons).
8 (c) 166 of 254 stable nuclides are even-even, reflecting the strong pairing effect.
9 (a) For odd $A$, the pairing term is zero and there is a single parabola; the most stable isobar is unique.
10 (b) The drip line is where the last neutron becomes unbound ($S_n = 0$).
11 (c) $x = 1$ means Coulomb energy equals twice the surface energy, and the fission barrier vanishes.
12 (b) Magic numbers show extra binding not captured by the smooth SEMF.
13 (b) The peak in $B/A$ arises from the competition between surface (dominant at low $A$) and Coulomb (dominant at high $A$) corrections.
14 (b) Doubly magic nuclei are MORE bound than the SEMF predicts, due to shell closure effects.
15 (b) The Coulomb barrier provides a centrifugal-like confinement for protons, allowing proton-unbound nuclei to exist with finite lifetimes. Neutrons face no such barrier.
16 (b) The SEMF is an excellent macroscopic baseline; its systematic failures point to the microscopic physics that refines the picture.
17 (b) The symmetry energy controls the equation of state of neutron-rich matter, which determines neutron star properties.
18 (b) $^{48}$Ti ($N - Z = 4$) has a smaller asymmetry penalty than $^{48}$Ca ($N - Z = 8$), and the Coulomb difference for $\Delta Z = 2$ is smaller than the asymmetry gain. However, experimentally $^{48}$Ca is MORE bound because of the double magic nature ($Z = 20$, $N = 28$) — a shell effect the SEMF misses. This question tests both SEMF reasoning and its limitations.