Nuclear structure — protons, neutrons, nuclear forces: Exercises

Full derivations of these live in the parent note Nuclear structure — protons, neutrons, nuclear forces. Here we only use them.

Level 1 — Recognition

L1·Q1 — Read the symbol

For $_{20}^{40} Ca$ , state $Z$ , $A$ , and $N$ (number of neutrons).

Recall Solution

WHAT the symbols mean. In $_{Z}^{A} X$ the bottom number is $Z$ = protons, the top number is $A$ = total nucleons.

$Z = 20$ (protons).
$A = 40$ (nucleons).
$N = A - Z = 40 - 20 = 20$ (neutrons). A calcium-40 nucleus is $20$ protons + $20$ neutrons.

L1·Q2 — Spot the relationship

Which pair below are isotopes, and which are isobars? $_{1}^{3} H,_{2}^{3} He,_{1}^{2} H .$

Recall Solution

Isotopes = same $Z$ (same element, different mass). $_{1}^{3} H$ and $_{1}^{2} H$ both have $Z = 1$ → isotopes. Isobars = same $A$ (same mass number, different element). $_{1}^{3} H$ and $_{2}^{3} He$ both have $A = 3$ → isobars.

Level 2 — Application

L2·Q1 — Radius of a nucleus

Find the radius of $_{13}^{27} Al$ .

Recall Solution

WHY this formula. Nuclear matter has constant density, so volume ∝ number of nucleons, giving $R = R_{0} A^{1/3}$ . We only need $A$ . $R = R_{0} A^{1/3} = 1.2 fm \times 2 7^{1/3} .$ Now $2 7^{1/3} = 3$ (since $3^{3} = 27$ ). So $R = 1.2 \times 3 = 3.6 fm = 3.6 \times 1 0^{- 15} m .$

L2·Q2 — Ratio of radii

By what factor is the radius of $^{125} Te$ ( $A = 125$ ) larger than $^{27} Al$ ( $A = 27$ )?

Recall Solution

WHY a ratio is easy. $R_{0}$ cancels, so only $A$ survives: $\frac{R _{125}}{R _{27}} = (\frac{125}{27})^{1/3} = \frac{12 5 ^{1/3}}{2 7 ^{1/3}} = \frac{5}{3} \approx 1.67.$ The tellurium nucleus is about $1.67 \times$ wider — not $125/27 \approx 4.6 \times$ , because radius follows the cube root, not $A$ directly.

L2·Q3 — Binding energy of $_{2}^{4} He$

Given $M (^{4} He) = 4.002603 u$ , find the total binding energy.

Recall Solution

Step 1 — sum of parts. Helium-4 has $Z = 2$ , $N = 2$ . Using atomic masses (so electrons cancel): $2 m_{p} + 2 m_{n} = 2 (1.007825) + 2 (1.008665) = 2.015650 + 2.017330 = 4.032980 u .$ Step 2 — mass defect. The bound nucleus is lighter: $Δ m = 4.032980 - 4.002603 = 0.030377 u .$ Step 3 — convert to energy. $E_{B} = Δ m \times 931.5 MeV$ : $E_{B} = 0.030377 \times 931.5 \approx 28.3 MeV .$ That is the energy to rip a helium nucleus (an alpha particle) into 4 free nucleons.

Level 3 — Analysis

L3·Q1 — Nuclear density is independent of $A$

Show algebraically that $ρ$ does not depend on $A$ , then evaluate it.

Recall Solution

WHY it should cancel. Mass ∝ $A$ (each nucleon ≈ same mass). Volume ∝ $A$ (from $R^{3} \propto A$ ). Ratio → no $A$ . $ρ = \frac{mass}{volume} = \frac{A m _{n}}{\frac{4}{3} π R ^{3}} = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n}}{\frac{4}{3} π R _{0}^{3}} .$ The $A$ divides out — the same for every nucleus. Plug numbers ( $m_{n} \approx 1.67 \times 1 0^{- 27}$ kg, $R_{0} = 1.2 \times 1 0^{- 15}$ m): $ρ = \frac{1.67 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \approx 2.3 \times 1 0^{17} kg/m^{3} .$

L3·Q2 — Coulomb vs strong at contact

Two protons just touch, centre-to-centre separation $r = 2.0 fm$ . Compute the Coulomb repulsion energy in MeV. (This shows the "hill" the strong force must beat.)

Figure — Nuclear structure — protons, neutrons, nuclear forces

Recall Solution

WHY this energy. The electrostatic potential energy of two charges $+ e$ a distance $r$ apart is $U = \frac{k e ^{2}}{r}$ (from Coulomb's law and electrostatic repulsion). We want it in MeV to compare with binding energies (~MeV scale). $U = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{2.0 \times 1 0 ^{- 15}} J .$ Numerator: $(1.602 \times 1 0^{- 19})^{2} = 2.566 \times 1 0^{- 38}$ , times $8.99 \times 1 0^{9} = 2.307 \times 1 0^{- 28}$ . Divide by $2.0 \times 1 0^{- 15}$ : $U = 1.15 \times 1 0^{- 13} J$ . Convert to MeV ( $1 MeV = 1.602 \times 1 0^{- 13} J$ ): $U = \frac{1.15 \times 1 0 ^{- 13}}{1.602 \times 1 0 ^{- 13}} \approx 0.72 MeV .$ So repulsion at contact is under a MeV, while the strong attraction supplies several MeV per bond — the strong force wins, exactly as the parent note claimed.

L3·Q3 — Yukawa range from a mediator mass

The strong force is carried by exchanging a pion of mass $m_{π} c^{2} = 140 MeV$ . Estimate the range $r \sim \frac{ℏ}{m _{π} c}$ and check it lands near a few fm.

Recall Solution

WHY this estimate. From Heisenberg uncertainty principle, a mediator can "borrow" energy $Δ E \sim m_{π} c^{2}$ only for a time $Δ t \sim ℏ/Δ E$ . In that time, moving at ≈ $c$ , it covers $r \sim c Δ t = ℏ/ (m_{π} c)$ . Convert the mass to SI energy: $140 MeV = 140 \times 1.602 \times 1 0^{- 13} = 2.243 \times 1 0^{- 11} J$ . Then $m_{π} c = \frac{m _{π} c ^{2}}{c} = \frac{2.243 \times 1 0 ^{- 11}}{3.0 \times 1 0 ^{8}} = 7.48 \times 1 0^{- 20} kg\cdotm/s$ . $r = \frac{ℏ}{m _{π} c} = \frac{1.055 \times 1 0 ^{- 34}}{7.48 \times 1 0 ^{- 20}} \approx 1.41 \times 1 0^{- 15} m \approx 1.4 fm .$ A heavier mediator would give a shorter range; a massless one (photon) gives infinite range — that is why EM reaches forever but the strong force cuts off at a few fm.

Level 4 — Synthesis

L4·Q1 — Neutrons as glue, not deadweight

Explain quantitatively-flavoured reasoning for why heavy stable nuclei have $N > Z$ . Then check: for $_{82}^{208} Pb$ , is $N > Z$ ? By how much?

Recall Solution

The physics. Coulomb repulsion sums over every proton pair, so it grows like $Z^{2}$ (fast). The strong attraction saturates (each nucleon bonds only to neighbours), so binding grows only like $\sim A$ (slow). As $Z$ climbs, repulsion outpaces attraction. Extra neutrons add strong attraction without adding charge, restoring balance — so heavy stable nuclei drift to $N > Z$ . Check lead-208: $Z = 82$ , $A = 208$ , so $N = A - Z = 208 - 82 = 126.$ Then $N - Z = 126 - 82 = 44$ . Indeed $N > Z$ by $44$ — a big neutron excess, exactly as predicted.

L4·Q2 — Binding energy per nucleon comparison

Compute $E_{B} / A$ for the deuteron $^{2} H$ ( $E_{B} = 2.22 MeV$ ) and for $^{4} He$ ( $E_{B} = 28.3 MeV$ from L2·Q3). Which is more tightly bound, and why does that matter for fusion?

Recall Solution

WHY per-nucleon. Total $E_{B}$ favours big nuclei just for having more parts. To compare stability fairly, divide by $A$ (see Mass defect and binding energy curve).

Deuteron: $E_{B} / A = 2.22/2 = 1.11 MeV per nucleon$ .
Helium-4: $E_{B} / A = 28.3/4 = 7.07 MeV per nucleon$ . Helium-4 is far more tightly bound per nucleon (~ $7$ vs ~ $1$ MeV). Why it matters: fusing light nuclei (like deuterons) toward $^{4} He$ moves nucleons to a deeper binding well, releasing the difference as energy — the basis of Nuclear fission and fusion (stellar fusion).

Level 5 — Mastery

L5·Q1 — From radius to a testable prediction

A nucleus has radius $R = 4.8 fm$ . (a) Estimate its mass number $A$ . (b) Predict which side of $N = Z$ it likely sits, and give a plausible $Z$ . (c) Sanity-check its density.

Recall Solution

(a) Invert the radius law. From $R = R_{0} A^{1/3}$ , $A = (\frac{R}{R _{0}})^{3} = (\frac{4.8}{1.2})^{3} = 4^{3} = 64.$ So $A \approx 64$ . (b) Which side of $N = Z$ ? $A = 64$ is a mid-heavy nucleus, where Coulomb repulsion is significant, so we expect a modest neutron excess $N ≳ Z$ . A real example is $_{30}^{64} Zn$ : $Z = 30$ , $N = 34$ — indeed $N > Z$ by $4$ , matching the "heavier nuclei need spacer neutrons" reasoning. (c) Density check. Density must be the same $\sim 2.3 \times 1 0^{17} kg/m^{3}$ as every nucleus — because $ρ$ is independent of $A$ (L3·Q1). No new computation needed; if you did compute $\frac{A m _{n}}{\frac{4}{3} π R ^{3}}$ you'd recover that number.

L5·Q2 — Full chain: mass defect → energy → mass-energy consistency

For $_{3}^{7} Li$ , given $M (^{7} Li) = 7.016004 u$ : (a) find the mass defect, (b) the total binding energy, (c) $E_{B} / A$ , and (d) convert the mass defect to kilograms and confirm via $E = m c^{2}$ that you get the same energy in joules.

Recall Solution

Lithium-7: $Z = 3$ , $N = 7 - 3 = 4$ . (a) Sum of parts, then defect. $3 m_{p} + 4 m_{n} = 3 (1.007825) + 4 (1.008665) = 3.023475 + 4.034660 = 7.058135 u .$ $Δ m = 7.058135 - 7.016004 = 0.042131 u .$ (b) Binding energy. $E_{B} = 0.042131 \times 931.5 \approx 39.2 MeV .$ (c) Per nucleon. $\frac{E _{B}}{A} = \frac{39.2}{7} \approx 5.6 MeV per nucleon .$ (d) Independent check via $E = m c^{2}$ (see Einstein mass-energy equivalence E=mc^2). Convert defect to kg ( $1 u = 1.6605 \times 1 0^{- 27}$ kg): $Δ m = 0.042131 \times 1.6605 \times 1 0^{- 27} = 6.996 \times 1 0^{- 29} kg .$ Then $E_{B} = Δ m c^{2} = 6.996 \times 1 0^{- 29} \times (3.0 \times 1 0^{8})^{2} = 6.30 \times 1 0^{- 12} J .$ Convert back: $6.30 \times 1 0^{- 12} / (1.602 \times 1 0^{- 13}) \approx 39.3 MeV$ ✔ — matches part (b) within rounding. The two routes agree, confirming the $931.5 MeV/u$ shortcut is just $E = m c^{2}$ in disguise.

Recall Self-test: fill these in from memory

Radius law and value of $R_{0}$ ::: $R = R_{0} A^{1/3}$ , $R_{0} = 1.2$ fm Why nuclear density is constant ::: mass ∝ $A$ and volume ∝ $A$ , so $A$ cancels Mass defect definition ::: $Δ m = [Z m_{p} + N m_{n}] - M_{nucleus}$ Conversion factor ::: $1 u \cdot c^{2} = 931.5$ MeV Why heavy nuclei need $N > Z$ ::: Coulomb grows like $Z^{2}$ , strong force saturates (∝ $A$ ), so extra neutrons add binding without charge