Nuclear structure — protons, neutrons, nuclear forces: Visual walkthrough

Step 1 — Meet the players (no equations yet)

WHAT. Before any formula, we name the three characters and draw them:

A proton: a tiny ball carrying a positive electric charge (we write that charge as $+ e$ , where $e$ is just the size of one elementary charge, $e \approx 1.6 \times 1 0^{- 19}$ coulombs — a coulomb is the unit we measure charge in).
A neutron: the same size ball, but no charge at all.
The pair of them living inside a sphere — the nucleus.

WHY. Every claim later ("protons repel", "neutrons add glue") is a claim about these objects. If we don't picture them first, the symbols $p$ , $n$ , $+ e$ mean nothing.

PICTURE. Two magenta protons (each stamped $+$ ) and one violet neutron (blank), all the same size, sitting in a peach nucleus. Notice: same size = same volume each. That single visual fact powers Step 2.

Figure — Nuclear structure — protons, neutrons, nuclear forces

Step 2 — Nucleons pack like marbles → volume grows with $A$

WHAT. Scattering experiments (see Rutherford scattering experiment) show nuclear matter has almost constant density: pack in one more nucleon and the ball grows by one nucleon's worth of room — no squeezing, no gaps. So each nucleon owns a fixed little volume, call it $v_{0}$ .

WHY this idea. If every nucleon takes the same room $v_{0}$ , then $A$ of them take $V = A v_{0} .$ Here $V$ = total volume of the nucleus, $A$ = how many nucleons, $v_{0}$ = room per nucleon. Read it left to right: "total room = (number of marbles) × (room per marble)". That's the whole physical assumption, and it's just marbles-in-a-bag.

PICTURE. Three nuclei side by side: $A = 1$ , $A = 8$ , $A = 27$ marbles. The bag visibly grows — but each marble stays the same size. The orange bracket labels one $v_{0}$ .

Step 3 — A sphere's volume forces a cube root → the radius law

WHAT. The nucleus is (roughly) a ball. The volume of a ball of radius $R$ is a fixed geometry fact: $V = \frac{4}{3} π R^{3} .$

$R$ = radius (how far from centre to edge),
$\frac{4}{3} π$ = a pure number ( $\approx 4.19$ ) that geometry hands us for any sphere,
$R^{3}$ = radius cubed, because volume is a 3-D (length × length × length) quantity.

WHY. We now have two expressions for the same volume — the marble count from Step 2 and the sphere formula from geometry. Setting equal things equal is how we solve for $R$ : $sphere geometry \frac{4}{3} π R^{3} = marble count A v_{0} .$

Now unwrap for $R$ , one algebra move at a time: $R^{3} = \frac{3 v _{0}}{4 π} A cube root both sides R = call this R_{0} (\frac{3 v _{0}}{4 π})^{1/3} A^{1/3} .$

WHY the cube root (and not something else)? Volume grew linearly with $A$ , but volume is $R^{3}$ . To get $R$ back out of $R^{3}$ you must undo a cube — that's exactly what the power $1/3$ (the cube root) does. The exponent $\frac{1}{3}$ is not chosen; it is forced by "volume ∝ length $^{3}$ ".

PICTURE. A quarter-circle showing $R$ from centre to edge, with the algebra chain drawn as a flow: $V = A v_{0} \to R^{3} \to R$ . The lump of constants gets boxed and renamed $R_{0}$ .

8^{1/3}

equals what?

2

— so an

A = 8

nucleus has twice the radius of an

A = 1

one.

Why is the exponent exactly

1/3

Volume

\propto A

but volume

\propto R^{3}

, so

R \propto A^{1/3}

Step 4 — Edge check: does density really stay constant? (degenerate-limit test)

WHAT. We claimed constant density in Step 2, then derived $R$ from it. Let's verify the loop closes by computing density from the finished radius law.

WHY. A good derivation must survive its own consequences. If putting $R = R_{0} A^{1/3}$ back in changed the density with $A$ , our starting assumption would be self-contradictory. So we test the extreme cases: tiny $A$ vs huge $A$ .

Density is mass ÷ volume. Mass $\approx A m_{n}$ ( $A$ nucleons, each of mass $\approx m_{n}$ ), volume $= \frac{4}{3} π R^{3}$ : $ρ = \frac{A m _{n}}{\frac{4}{3} π R ^{3}} = \frac{A m _{n}}{\frac{4}{3} π ( R _{0} A ^{1/3} ) ^{3}} = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n}}{\frac{4}{3} π R _{0}^{3}} .$

Numerator $A m_{n}$ : total mass grows with $A$ .
Denominator: $(A^{1/3})^{3} = A$ , so volume also grows with $A$ .
The two $A$ 's cancel → $ρ$ has no $A$ left in it.

PICTURE. A flat magenta line of density vs $A$ : whether $A = 2$ or $A = 238$ , the height is the same $\approx 2.3 \times 1 0^{17} kg/m^{3}$ . The cancelling $A$ 's are shown crossing out.

Step 5 — Now the forces: draw the two that fight

WHAT. Radius done. New question: with all those $+$ charges crammed into $1 0^{- 15}$ m, why doesn't the nucleus explode? Draw the two influences on one proton:

Coulomb repulsion — from Coulomb's law and electrostatic repulsion, like charges push apart with a force $\propto 1/ r^{2}$ (weaker as they separate, but never quite zero: long range).
Strong nuclear force — attractive, and stronger, but only felt by touching/near neighbours: short range.

WHY draw them together. "Why doesn't it fly apart?" is literally the question which arrow wins? You cannot answer it without seeing both arrows on the same proton at the same distance.

PICTURE. One proton with two arrows: an orange repulsion arrow pointing away, and a big magenta strong-force arrow pointing toward a neighbour. At close range the magenta arrow dwarfs the orange one.

Step 6 — The force-vs-distance graph: where each force wins

WHAT. Plot the net nuclear force felt between two nucleons as distance $r$ changes. Three zones appear:

$r < 0.5$ fm: repulsive core (positive spike) — a hard wall that stops the nucleus collapsing.
$r \approx 1$ fm: deep attractive well (large negative) — this is where nucleons sit and bind.
$r > 2 - 3$ fm: force $\to 0$ — beyond here only the gentle Coulomb push survives.

WHY this shape. The attraction must switch off with distance, otherwise nuclei would swell without limit and every nucleon would bind to every other. The short range is explained by Heisenberg uncertainty principle: the force is carried by exchanging a heavy particle (a pion, mass $m_{π}$ ). A heavy borrowed particle can only exist for a flash of time $Δ t \sim ℏ/Δ E$ , so it travels only $r \sim \frac{ℏ}{m _{π} c} \approx 1.4 fm,$

$ℏ$ = the reduced Planck constant (nature's smallest "action" unit),
$m_{π} c$ = the pion's mass times light-speed (heavier mediator → smaller $r$ ). A massless mediator (the photon) gives $r \to \infty$ — that's why electromagnetism is long-range and the strong force is not.

PICTURE. Force $F$ vs $r$ : violet curve spiking up (repulsive) below $0.5$ fm, plunging into a magenta attractive well near $1$ fm, flattening to zero past $3$ fm; a faint orange $1/ r^{2}$ Coulomb tail continues past it.

Step 7 — Cash it out in numbers: the deuteron binding energy

WHAT. The strong force's grip shows up as missing mass. Weigh the parts separately, weigh them bound together — the bound version is lighter. The lost mass became the binding energy (via Einstein mass-energy equivalence E=mc^2).

WHY lighter, not heavier? Binding releases energy; that escaping energy carries mass away ( $E = m c^{2}$ ). To break the bond you must pay that energy back in. So the mass gap directly measures how tightly it's held. This ties to the Mass defect and binding energy curve.

Take the deuteron $_{1}^{2} H$ (1 proton + 1 neutron): $Δ m = [m_{p} + m_{n}] - M (^{2} H)$

$m_{p} = 1.007825$ u (hydrogen-atom mass; the electron cancels later),
$m_{n} = 1.008665$ u,
$M (^{2} H) = 2.014102$ u,
"u" = atomic mass unit, and $1 u \cdot c^{2} = 931.5$ MeV.

Step by step: $parts: 1.007825 + 1.008665 = 2.016490 u$ $Δ m = 2.016490 - 2.014102 = 0.002388 u$ $E_{B} = Δ m c^{2} = 0.002388 \times 931.5 \approx 2.22 MeV .$

PICTURE. A balance scale: left pan holds separate $p + n$ (higher, heavier); right pan holds the bound deuteron (lower... lighter, so the scale tips toward the parts). The tiny difference bar is labelled $Δ m$ , arrow converting it to $2.22$ MeV.

The one-picture summary

WHAT. Everything on one canvas: marbles → sphere → radius law → constant density on one side; the two-force tug-of-war → force well → missing mass → binding energy on the other. Two questions, two answers, one nucleus.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a bag of magnetic marbles. Each marble takes up the same little space, so pouring in more marbles just makes a bigger bag — same "stuff-per-space" no matter how big the bag. Because it's a ball and a ball's volume is (radius)³, the radius only grows as the cube root of how many marbles you have: quadruple... no — eight-fold the marbles, double the radius. That's $R = R_{0} A^{1/3}$ , and the density comes out the same for every nucleus. Now the marbles with "+" stickers (protons) push each other away — but every marble has super-strong velcro that only grabs the marbles touching it. Up close the velcro crushes the pushing; far away the velcro can't reach and only the push is left. That velcro is short-range because it works by tossing a heavy little ball (the pion) back and forth, and a heavy ball can't be thrown far. Finally, when the velcro grabs, the clump ends up weighing a hair less than the loose marbles — and that missing weight is exactly the energy locked into the bond. For a proton-plus-neutron, that missing weight is worth $2.22$ MeV. That's the whole story.

Self-test

Radius ratio of

A = 64

A = 8

nucleus?

(64/8)^{1/3} = 8^{1/3} = 2

Why is nuclear density the same for all nuclei?

Mass

\propto A

and volume

\propto A

cancel in

ρ = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A}

What makes the strong force short-ranged?

Its mediator (the pion) is massive, so it can only travel

\sim ℏ/ (m_{π} c) \approx 1.4

fm.

Is a bound nucleus heavier or lighter than its parts, and why?

Lighter — binding released energy, and that energy carried mass away (

E = m c^{2}