Nuclear structure — protons, neutrons, nuclear forces: Worked examples

Before we begin, one symbol reminder so nothing is used unearned:

$A$ = mass number = total count of protons + neutrons (the "how many marbles" number).
$Z$ = atomic number = count of protons (the "which element" number).
$N = A - Z$ = count of neutrons.
$R_{0} = 1.2 fm$ where $1 fm = 1 0^{- 15} m$ (a femtometre — a thousandth of a trillionth of a metre).
$ρ$ = density = mass divided by volume, $ρ = \frac{mass}{volume}$ . It answers "how much stuff is packed into each cubic metre?"
$c$ = speed of light $= 3.0 \times 1 0^{8} m/s$ — the fixed conversion rate between mass and energy in $E = m c^{2}$ (see Einstein mass-energy equivalence E=mc^2).
$u$ = atomic mass unit, the mass yardstick where $1 u \cdot c^{2} = 931.5 MeV$ . Read that as: "one unit of mass, if fully converted to energy via $E = m c^{2}$ , gives 931.5 million electron-volts."

The scenario matrix

Every question about nuclear structure falls into one of these cells. The last column names the example that covers it.

#	Case class	What makes it tricky	Covered by
C1	Radius, ordinary $A$	plain plug-in of $R = R_{0} A^{1/3}$	Ex 1
C2	*Radius ratio* of two nuclei**	the $R_{0}$ cancels — ratio-only shortcut	Ex 2
C3	Degenerate input: $A = 1$	single proton, $A^{1/3} = 1$	Ex 3
C4	Density ( $A$ cancels algebraically)	prove $ρ$ is the same for all $A$	Ex 4
C5	Mass defect + binding energy	atomic vs nuclear mass, electron bookkeeping	Ex 5
C6	Binding energy per nucleon	divide by $A$ — the "stability" number	Ex 6
C7	Range of the force (uncertainty)	Yukawa $r \sim ℏ/ (m_{π} c)$ , unit juggling	Ex 7
C8	Coulomb vs strong tug-of-war	compare two forces at one distance	Ex 8
C9	Real-world word problem	"teaspoon of nucleus" mass	Ex 9
C10	Exam twist: unknown $A$ from a ratio	invert the cube-root law	Ex 10

Notice there are no signs or quadrants here (unlike trig/vectors) — nuclear counts $A, Z, N$ are non-negative integers, and $R, ρ, E_{B}$ are always positive. The "edge cases" instead live at $A = 1$ (a lone proton, Ex 3) and in the cancellations (Ex 2, Ex 4). We cover those deliberately.

Ex 1 — C1: Radius of an ordinary nucleus

Worked example Radius of iron

$_{26}^{56} Fe$ Statement: Find the radius of an iron-56 nucleus. Use $R_{0} = 1.2 fm$ .

Forecast: $A = 56$ . The cube root of $56$ is a bit under $4$ (since $4^{3} = 64$ ). So expect $R$ roughly $1.2 \times 3.8 \approx 4.6 fm$ . Guess before reading on.

Step 1 — Write the law. $R = R_{0} A^{1/3}$ . Why this step? This is the only radius formula we have, and it came from "constant density → volume ∝ $A$ → radius ∝ $A^{1/3}$ " (built in the parent note). Nothing else is needed.

Step 2 — Cube-root the mass number. $5 6^{1/3} = 3.826$ . Why this step? The cube root is the whole physics — volume grows linearly with $A$ , but radius is the cube root of volume, so it grows slowly.

Step 3 — Multiply by $R_{0}$ . $R = 1.2 \times 3.826 = 4.59 fm$ . Why this step? $R_{0}$ is the radius of a single-nucleon blob; scaling it by $A^{1/3}$ inflates it to the whole nucleus.

Verify: $4.59 fm = 4.59 \times 1 0^{- 15} m$ — femtometre-scale, as every nucleus should be. It matches our forecast ( $\approx 4.6$ ). ✔

Ex 2 — C2: Radius ratio (the $R_{0}$ cancels)

Worked example How much bigger is

$^{216}$ Rn than $_{13}^{27}$ Al? Statement: Compare the radii of an $A = 216$ nucleus and an $A = 27$ nucleus.

Forecast: This is the parent note's forecast-verify problem. $216/27 = 8$ , and $8 = 2^{3}$ , so the cube root is exactly $2$ . Guess: the big one is twice as wide.

Step 1 — Write both radii. $R_{216} = R_{0} (216)^{1/3}$ and $R_{27} = R_{0} (27)^{1/3}$ . Why this step? Writing them side-by-side reveals the shared $R_{0}$ .

Step 2 — Divide. $\frac{R _{216}}{R _{27}} = \frac{R _{0} ( 216 ) ^{1/3}}{R _{0} ( 27 ) ^{1/3}} = (\frac{216}{27})^{1/3} .$ Why this step? The unknown-ish constant $R_{0}$ cancels — so we never need its numerical value. This is why ratio questions are easier than absolute ones.

Step 3 — Evaluate. $(\frac{216}{27})^{1/3} = 8^{1/3} = 2$ . Why this step? Turning the ratio into a single number $8$ first, then cube-rooting, avoids messy decimals.

Verify: Radius scales as $A^{1/3}$ , so an $8 \times$ heavier nucleus is $8^{1/3} = 2 \times$ wider. Volume ratio would be $8$ , matching "8× the nucleons pack into 8× the volume". ✔

Ex 3 — C3: Degenerate case $A = 1$ (a lone proton)

Worked example Radius of a single proton (hydrogen-1 nucleus)

Statement: What does $R = R_{0} A^{1/3}$ predict for $_{1}^{1} H$ , whose nucleus is a single proton?

Forecast: $A = 1$ , and $1^{1/3} = 1$ , so the formula must just give $R = R_{0}$ itself. Guess: about $1.2 fm$ .

Step 1 — Substitute $A = 1$ . $R = R_{0} \cdot 1^{1/3} = R_{0} = 1.2 fm$ . Why this step? This is the degenerate limit — the smallest possible nucleus. It shows $R_{0}$ literally is the effective radius of one nucleon.

Step 2 — Reality check the meaning. Measured proton "charge radius" is $\approx 0.84 fm$ . Why this step? Our formula is a packing model — it slightly overshoots for a lone particle because a single marble doesn't "pack" against neighbours. The law is meant for many-nucleon nuclei; at $A = 1$ it is only a rough guide.

Verify: The formula behaves sensibly at its smallest input — it returns $R_{0}$ , not zero or infinity — and lands in the right ballpark ( $\sim 1 fm$ ) of the measured value. No blow-up, no nonsense. ✔

Ex 4 — C4: Density — where $A$ cancels algebraically

Worked example Show every nucleus has the same density

Statement: Compute the nuclear density $ρ = \frac{mass}{volume}$ and prove it does not depend on $A$ .

Forecast: Mass grows like $A$ ; volume grows like $R^{3} \propto A$ . The same power of $A$ sits on top and bottom, so the $A$ divides out (it cancels — it does not go to zero) → density is a constant. Guess the number: around $2 \times 1 0^{17} kg/m^{3}$ .

Step 1 — Write mass and volume in terms of $A$ . $mass \approx A m_{n}, V = \frac{4}{3} π R^{3} = \frac{4}{3} π (R_{0} A^{1/3})^{3} = \frac{4}{3} π R_{0}^{3} A .$ Why this step? We deliberately keep $A$ visible in both so we can watch it cancel. Note $(A^{1/3})^{3} = A$ — the cube undoes the cube root.

Step 2 — Form the density. $ρ = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n}}{\frac{4}{3} π R _{0}^{3}} .$ Why this step? The $A$ in numerator and denominator divide out — this is the punchline. Big or small nucleus, same density.

Step 3 — Plug numbers. With $m_{n} = 1.675 \times 1 0^{- 27} kg$ , $R_{0} = 1.2 \times 1 0^{- 15} m$ : $ρ = \frac{1.675 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \approx 2.31 \times 1 0^{17} kg/m^{3} .$ Why this step? Turning symbols into a number lets us feel how absurdly dense nuclear matter is.

Verify: Units: $kg / m^{3}$ . ✔ Value $\sim 2.3 \times 1 0^{17}$ matches the parent note. And crucially — no $A$ survived, confirming the constant-density claim. ✔

The figure below makes the cancellation visual. It plots $ρ$ against $A$ for the whole range of real nuclei: the blue line is dead flat (the same $2.31 \times 1 0^{17} kg/m^{3}$ for hydrogen, iron and lead alike), and the orange dots for $^{2}$ H, $^{56}$ Fe and $^{208}$ Pb all land on that one horizontal line. A flat line is what "independent of $A$ " looks like — if density grew with $A$ , this line would slope upward; it doesn't.

Ex 5 — C5: Mass defect & binding energy (electron bookkeeping)

Worked example Binding energy of helium-4,

$_{2}^{4} He$ Statement: Given atomic masses $m (^{1} H) = 1.007825 u$ , $m_{n} = 1.008665 u$ , $M (^{4} He) = 4.002603 u$ , find the binding energy of the helium-4 nucleus.

Forecast: Helium-4 (an alpha particle) is famously very tightly bound. Guess: the total is a big number, tens of MeV — maybe near $28 MeV$ .

Step 1 — Sum the parts. Helium has $Z = 2$ protons and $N = 2$ neutrons: $parts = 2 m (^{1} H) + 2 m_{n} = 2 (1.007825) + 2 (1.008665) = 4.032980 u .$ Why this step? We use the atomic hydrogen mass (proton + 1 electron) twice, which supplies the 2 electrons that helium's neutral atom also has. So the electron masses cancel when we subtract the neutral helium atom mass. Clean bookkeeping.

Step 2 — Mass defect. $Δ m = 4.032980 - 4.002603 = 0.030377 u .$ Why this step? The bound nucleus is lighter than its parts; the missing mass became binding energy.

Step 3 — Convert to energy. $E_{B} = Δ m \cdot 931.5 = 0.030377 \times 931.5 \approx 28.30 MeV .$ Why this step? The factor $931.5 MeV/u$ is $c^{2}$ dressed in convenient units — it turns "u of mass" into "MeV of energy" via $E = Δ m c^{2}$ , where $c = 3.0 \times 1 0^{8} m/s$ is the speed of light.

Verify: $28.3 MeV$ matches our forecast and the accepted value for $^{4}$ He. Positive (it takes energy to break it apart), and comfortably larger than the deuteron's $2.22 MeV$ from the parent note — helium-4 is far more tightly bound. ✔

Ex 6 — C6: Binding energy per nucleon

Worked example Which is more stable: helium-4 or the deuteron?

Statement: Compare stability using binding energy per nucleon, $E_{B} / A$ . Use $E_{B} (^{4} He) = 28.30 MeV$ and $E_{B} (^{2} H) = 2.22 MeV$ .

Forecast: Total binding energy isn't fair to compare — helium has more nucleons, so of course its total is bigger. The fair measure is per-nucleon. Guess: helium wins big.

Step 1 — Divide by $A$ for each. $\frac{E _{B}}{A}_{^{4} He} = \frac{28.30}{4} = 7.07 MeV/nucleon, \frac{E _{B}}{A}_{^{2} H} = \frac{2.22}{2} = 1.11 MeV/nucleon .$ Why this step? Dividing by $A$ answers "how tightly is a typical single nucleon held?" — that is the real stability yardstick (this is the $y$ -axis of the Mass defect and binding energy curve).

Step 2 — Compare. $7.07 > 1.11$ , so helium-4 is far more tightly bound per nucleon. Why this step? Per-nucleon binding removes the unfair size advantage; now the comparison is apples-to-apples.

Verify: $7.07 MeV/nucleon$ sits right where the binding-energy curve says light-but-magic nuclei like $^{4}$ He sit (near its early peak). The deuteron's $1.11$ is famously low — it is a fragile nucleus. ✔

Ex 7 — C7: Range of the nuclear force (Yukawa + uncertainty)

Worked example Estimate the force range from the pion mass

Statement: A pion has rest energy $m_{π} c^{2} \approx 140 MeV$ . Estimate the range $r \sim ℏ/ (m_{π} c)$ of the force it mediates.

Forecast: The parent note quoted $\approx 1.4 fm$ . Let's see if we can land there from scratch. Guess: about $1 - 2 fm$ .

Step 1 — Rewrite in energy-friendly form. Multiply top and bottom by $c$ : $r \sim \frac{ℏ}{m _{π} c} = \frac{ℏ c}{m _{π} c ^{2}} .$ Why this step? We are given $m_{π} c^{2}$ in MeV, and the constant $ℏ c = 197.3 MeV \cdot fm$ is a famous "handy number." Writing it this way lets MeV cancel and leaves fm — no messy SI conversions. This is the Heisenberg uncertainty principle at work: a heavier borrowed particle can only reach a shorter distance.

Step 2 — Plug in. $r \sim \frac{197.3 MeV \cdot fm}{140 MeV} = 1.41 fm .$ Why this step? The MeV units divide out, leaving fm — exactly a length. The heavier the mediator, the smaller $r$ ; a massless photon ( $m = 0$ ) would give $r = \infty$ , which is why electromagnetism has infinite range.

Verify: $1.41 fm$ matches the parent note's $\approx 1.4 fm$ and the observed "few-fm" cutoff of the strong force. Units checked out to fm. ✔

Ex 8 — C8: Coulomb vs strong tug-of-war at a fixed distance

Worked example Two protons at

$r = 1 fm$ — how strong is the repulsion? Statement: Compute the Coulomb repulsion between two protons separated by $r = 1.0 fm$ . Use $k = 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ , $e = 1.602 \times 1 0^{- 19} C$ . Then comment on why the nucleus survives it.

Forecast: At everyday distances proton–proton repulsion is tiny, but $1 fm$ is microscopically close and force $\propto 1/ r^{2}$ blows up. Guess: a shockingly large force — hundreds of newtons on a single proton.

Step 1 — Write Coulomb's law. From Coulomb's law and electrostatic repulsion: $F = \frac{k e ^{2}}{r ^{2}} .$ Why this step? Both particles carry charge $+ e$ , so the product of charges is $e^{2}$ ; this law gives the repulsive push we must overcome.

Step 2 — Plug numbers ( $r = 1.0 \times 1 0^{- 15} m$ ). $F = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{( 1.0 \times 1 0 ^{- 15} ) ^{2}} \approx 231 N .$ Why this step? Squaring the tiny $r$ makes the denominator $1 0^{- 30}$ — enormous division — so the force is huge despite the tiny charges.

Step 3 — Interpret. $231 N$ acting on a proton of mass $1.67 \times 1 0^{- 27} kg$ is an unimaginable push. Yet the nucleus holds — because the strong force is $\sim 100 \times$ stronger at this range and attractive. Why this step? This is the whole "tug-of-war" made numeric: repulsion is real and large, but it loses.

Verify: $\approx 231 N$ — for comparison that is like ~23 kg of weight pressing on one proton. Force is positive (repulsive) as expected for like charges. The strong force, being $\sim 100 \times$ larger and short-ranged, wins inside $\sim 2 fm$ . ✔

Ex 9 — C9: Real-world word problem

Worked example A teaspoon of pure nuclear matter

Statement: A teaspoon is about $5 cm^{3} = 5 \times 1 0^{- 6} m^{3}$ . If it were filled with pure nuclear matter (density $ρ = 2.31 \times 1 0^{17} kg/m^{3}$ from Ex 4), what would it weigh?

Forecast: Nuclear matter is insanely dense, so even a spoonful should weigh a preposterous amount — the parent note said "billions of tonnes." Guess: around $1 0^{12} kg$ .

Step 1 — Mass = density × volume. $m = ρ V = (2.31 \times 1 0^{17}) (5 \times 1 0^{- 6}) .$ Why this step? Density is mass per volume ( $ρ = mass / volume$ ), so multiplying by volume recovers mass — the definition run backwards.

Step 2 — Multiply. $m = 1.155 \times 1 0^{12} kg \approx 1.16 \times 1 0^{9} tonnes .$ Why this step? $1 0^{17} \times 1 0^{- 6} = 1 0^{11}$ , times the coefficients ( $2.31 \times 5 \approx 11.6$ ) gives $\sim 1.16 \times 1 0^{12} kg$ . Dividing by $1000$ converts kg to tonnes.

Verify: $\sim 1.16$ billion tonnes for a teaspoon — matching the parent note's "billions of tonnes." Units: $(kg/m^{3}) (m^{3}) = kg$ ✔. This is the mass of a small mountain, showing how much empty space a normal atom really is.

Ex 10 — C10: Exam twist — find $A$ from a radius ratio

Worked example Unknown nucleus is 3× the radius of carbon-12

Statement: A nucleus $X$ has a radius exactly $3$ times that of $_{6}^{12} C$ . Find its mass number $A_{X}$ .

Forecast: Radius scales as $A^{1/3}$ , so a $3 \times$ radius means $3^{3} = 27 \times$ the mass number. Guess: $A_{X} = 27 \times 12 = 324$ .

Step 1 — Write the ratio. $\frac{R _{X}}{R _{C}} = (\frac{A _{X}}{12})^{1/3} = 3.$ Why this step? $R_{0}$ cancels in a ratio (as in Ex 2), leaving a clean cube-root equation to invert.

Step 2 — Cube both sides to undo the cube root. $\frac{A _{X}}{12} = 3^{3} = 27.$ Why this step? Cubing is the inverse operation of taking a cube root — it isolates $A_{X} /12$ . This is the "invert the law" move that exam twists love.

Step 3 — Solve. $A_{X} = 27 \times 12 = 324$ . Why this step? Simple multiplication finishes it.

Verify: Check forward: $(324/12)^{1/3} = 2 7^{1/3} = 3$ ✔ — matches the given ratio and our forecast. (Real nuclei top out near $A \approx 250$ , so $324$ is hypothetical — a nice reminder that a formula can be extrapolated past where nature stops.) ✔

Recall Quick self-test

Radius of $^{125}$ Te if $R_{0} = 1.2 fm$ ? ::: $1.2 \times 12 5^{1/3} = 1.2 \times 5 = 6.0 fm$ . Why does the $R_{0}$ vanish in a radius ratio? ::: It multiplies both radii, so it cancels in the division. Binding energy per nucleon of $^{4}$ He? ::: $28.3/4 \approx 7.07 MeV/nucleon$ . Why use $ℏ c = 197.3 MeV\cdotfm$ for the force range? ::: So MeV cancels the pion's rest energy and the answer comes out directly in fm.