Visual walkthrough — Binding energy — mass defect, BE per nucleon curve

We follow one question the whole way: "If I glue nucleons together, where does the weight go?"

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Step 1 — What is a nucleon, and what is "mass"?

WHAT. A nucleus is a tight clump of two kinds of tiny balls:

• protons (positively charged), we count them with the letter $Z$,
• neutrons (no charge), we count them with the letter $N$.

Together they are called nucleons. The total count is $A=Z+N$, called the mass number.

WHY these letters. We need names before we can weigh anything. $Z$ = "how many protons", $N$ = "how many neutrons", $A$ = "how many balls total". That's all — no physics yet, just labels.

PICTURE. On the left, the balls float free and far apart. On the right, they are snapped together into one clump. Same balls, different arrangement — that difference is the whole story.

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Step 2 — Weigh the loose parts first

WHAT. Put every loose ball on a scale before gluing. One free proton has mass $m_p$; one free neutron has mass $m_n$. Add them all up:

$M_{\text{parts}}=\underset{\text{how many protons}}{\underbrace{Z}}\underset{\text{one proton’s mass}}{\underbrace{m_p}}+\underset{\text{how many neutrons}}{\underbrace{N}}\underset{\text{one neutron’s mass}}{\underbrace{m_n}}$

WHY. To find out if gluing changes the weight, we need the "before" number. $M_{\text{parts}}$ is the total mass of the ingredients laid out separately.

PICTURE. A chalk balance: the left pan holds all the separate balls, and the pointer reads $M_{\text{parts}}$ — the honest sum of every ingredient.

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Step 3 — Weigh the finished nucleus — it's lighter

WHAT. Now glue the balls into the actual nucleus and weigh that: call it $M_{\text{nucleus}}$. Experiment says something surprising:

$M_{\text{nucleus}}<M_{\text{parts}}$

The assembled clump weighs less than the same balls did apart.

WHY it feels wrong but is right. No ball vanished — you can count them, still $Z$ protons and $N$ neutrons. Yet the scale reads lower. The missing amount has a name, the mass defect:

$\Delta m=\underset{\text{parts apart}}{\underbrace{M_{\text{parts}}}}-\underset{\text{clump together}}{\underbrace{M_{\text{nucleus}}}}=[Z{m}_p+N{m}_n]-M_{\text{nucleus}}$

Here $\Delta$ (Greek "delta") just means "the difference in" — it is a subtraction, not a new object.

PICTURE. Two balances side by side. Left pan: the loose balls (heavier, pointer tilts down). Right pan: the glued nucleus (lighter, pointer tilts up). The little gap between the pointers is $\Delta m$.

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Step 4 — Why work + energy conservation force the mass to drop

WHAT. Imagine the reverse job: pull the glued nucleus back apart into free balls. The strong nuclear force fights you — it pulls the balls back. So you must do work, push energy into the system, to separate them.

WHY this pins down the mass. Track the energy honestly:

• Free balls apart have some total energy $E_{\text{parts}}$.
• The bound clump has total energy $E_{\text{nucleus}}$.
• You had to add energy $E_B$ to go from clump → apart.

Energy conservation: $E_{\text{nucleus}}+E_B=E_{\text{parts}}$, so $E_B=E_{\text{parts}}-E_{\text{nucleus}}>0.$ The bound clump sits lower in energy — it lives in an energy well. Lower energy is the reason it is bound at all. This $E_B$ is the binding energy.

PICTURE. A valley on the chalkboard. The free nucleons stand on the high rim; the bound nucleus rests at the bottom of the well. The vertical drop is $E_B$ — the energy released going down, or the energy you must supply to climb back out.

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Step 5 — The bridge: $E=m{c}^2$ turns the energy drop into a mass drop

WHAT. Einstein's relation says energy and mass are the same currency in two units: $E=m{c}^2,$ where $c$ is the speed of light (a fixed conversion number). Any change in energy is automatically a change in mass: $\Delta E=\Delta m{c}^2.$

WHY this is the missing link. Step 3 saw the mass go down by $\Delta m$. Step 4 saw the energy go down by $E_B$ when the clump formed. These are the same drop, seen in two units. Equate them:

$\boxed{E_B=\Delta m{c}^2}$

Term by term:

• $E_B$ = binding energy (the energy that left the system),
• $\Delta m$ = mass defect from Step 3 (the weight that went missing),
• $c^2$ = the exchange rate turning kilograms of missing mass into joules of released energy.

PICTURE. A money-changer's desk: pour "missing mass" $\Delta m$ into the slot, the machine multiplies by $c^2$, and out comes energy $E_B$. Same value, different currency.

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Step 6 — Edge case: the deuteron (only 2 balls)

WHAT. The smallest bound nucleus is the deuteron ${}_1^2$H: one proton, one neutron. Let's run the machine with the tiniest possible input.

$\Delta m=\underset{m_p,u}{\underbrace{(1.007825}}+\underset{m_n,u}{\underbrace{1.008665)}}-\underset{M{(}^2\text{H}),u}{\underbrace{2.014102}}=0.002388u$ $E_B=0.002388\times 931.5=2.224\text{MeV}$

WHY it matters as an edge case. Two balls means only one bond. There is a real mass defect (the well exists), but it is shallow: only $2.224$ MeV, and per nucleon just $2.224/2=1.11$ MeV. This is the far-left, barely-bound end of the curve — the smallest non-trivial case, and it still obeys $E_B=\Delta m{c}^2$.

PICTURE. A very shallow chalk well holding just a proton and a neutron, drop labelled $2.224$ MeV — tiny compared with the deep wells to come.

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Step 7 — Fair comparison: divide by $A$ to get depth per ball

WHAT. Total $E_B$ always rises with $A$ (more balls, more bonds), so it can't rank stability. Instead ask: "how deep is the well per nucleon?" $\overline{E_B}=\frac{E_B}{A}(\text{energy to remove, on average, }one\text{ ball})$

Run helium-4 ($Z=2,N=2,M=4.002602u$): $\Delta m=2(1.007825)+2(1.008665)-4.002602=0.030378u$ $E_B=0.030378\times 931.5=28.30\text{MeV},\overline{E_B}=\frac{28.30}{4}=7.07\text{MeV}.$

WHY. Deuteron scored $1.11$ MeV/nucleon; helium-4 scores $7.07$ MeV/nucleon — a much deeper well per ball. Dividing by $A$ is the fair judge that reveals medium nuclei as the winners, peaking near ${}^{56}$Fe at $\approx 8.8$ MeV.

PICTURE. Two wells drawn to the same "per ball" ruler: deuteron shallow ($1.11$), helium deep ($7.07$) — same physics, but now compared fairly per nucleon.

Recall Why both fusion and fission release energy (in one line)

Both push nucleons toward the deepest per-ball well near iron; the extra depth gained comes out as energy — fusion climbing the steep left, fission climbing the gentle right, and the released amount is the Q-value.

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The one-picture summary

Everything above compresses into a single chain: loose balls (heavy) → glue them → clump (lighter) → the missing mass $\Delta m$ becomes released energy $E_B=\Delta m{c}^2$ → divide by $A$ to compare depth, giving the iron-peaked curve.

Recall Feynman retelling — the whole walkthrough in plain words

Lay out a pile of magnet-balls on a scale and note the weight — that's your parts list. Now let them snap together into a clump; they release a little click of energy as they lock. Weigh the clump: it's a hair lighter. Nothing fell off — count them, all still there — but a sliver of weight ($\Delta m$) turned into the energy that flew out as the click. Einstein's exchange rate $c^2$ tells you exactly how much energy that sliver was worth ($E_B$). To break the clump you'd have to hand all that energy back, which is why the clump is "bound": it's parked at the bottom of an energy valley. Some clumps park deeper per ball than others — medium-sized ones near iron are deepest — so tiny clumps love to join up and giant clumps love to split, each sliding toward the comfy iron-deep valley and spitting out the difference as energy.

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Connections

• Mass–energy equivalence ($E=mc^2$)
• Atomic mass unit (u)
• Strong nuclear force
• Nuclear fusion
• Nuclear fission
• Q-value of nuclear reactions
• Stability of nuclei & N-Z curve