Binding energy — mass defect, BE per nucleon curve: Worked examples

Everything here uses atomic masses (masses of neutral atoms, which include the electrons) so that the electron masses cancel — exactly the fix the parent note stressed. Constants we reuse:

Here $Z$ = protons, $N = A - Z$ = neutrons, $A$ = total nucleons — all as in the parent. Every symbol was earned there; this page only applies them.

The scenario matrix

Every problem on binding energy is one (or a mix) of these case classes. The table lists each, and which worked example below covers it.

Cell	Case class	What makes it different	Covered by
A	Smallest real nucleus	Only 2 nucleons → far down the left slope	Ex 1 (deuteron)
B	Local-peak light nucleus	Doubly-magic, unusually high $\overline{E_{B}}$	Ex 2 ( $^{4}$ He)
C	Iron-region maximum	Top of the curve, the reference stability	Ex 3 ( $^{56}$ Fe)
D	Heavy nucleus	Large total $E_{B}$ but low $\overline{E_{B}}$	Ex 4 ( $^{235}$ U)
E	Reaction — energy released	$Q > 0$ , products more bound (fusion)	Ex 5 (D–T)
F	Reaction — energy absorbed	$Q < 0$ , must supply energy (photodisintegration)	Ex 6 (break deuteron)
G	Degenerate / zero input	1 nucleon: no partner → $E_{B} = 0$	Ex 7 ( $^{1}$ H, free $n$ )
H	Limiting behaviour	Why per-nucleon flattens then falls	Ex 8 (compare Fe vs U)
I	Word / exam twist	Reactor energy, hidden per-nucleon trap	Ex 9 (fission power)

We hit A–I in nine examples. Signs are handled explicitly in E, F, and G, so the "which way does energy flow" question is never left ambiguous.

Case A — the smallest nucleus (deuteron)

Worked example Example 1 — Binding energy of

$_{1}^{2}$ H Data: $M (^{2} H) = 2.014102 u$ . Find $E_{B}$ and $\overline{E_{B}}$ .

Forecast: Only 2 nucleons, so guess — will $\overline{E_{B}}$ be near the iron value of $8.8$ MeV, or far below it? (Think about the left slope of the curve before reading on.)

Count nucleons. $Z = 1$ , $N = 1$ , $A = 2$ . Why this step? Every formula needs $Z$ and $N$ before anything else — they pick how many protons and neutrons to add up.
Mass defect. $Δ m = [1 (1.007825) + 1 (1.008665)] - 2.014102 = 0.002388 u$ Why this step? $Δ m$ is the "missing mass" — the sole source of binding energy.
Convert to energy. $E_{B} = 0.002388 \times 931.5 = 2.224 MeV$ Why this step? $E = m c^{2}$ turns the missing mass into the glue energy; $931.5$ is the ready-made $u \to$ MeV bridge.
Per nucleon. $\overline{E_{B}} = 2.224/2 = 1.11 MeV$ . Why this step? Per-nucleon is the fair stability yardstick.

Verify: $\overline{E_{B}} = 1.11$ MeV is tiny compared to $8.8$ MeV — consistent with the deuteron sitting at the very bottom-left of the curve. Units: $u \times (MeV / u) = MeV$ . ✓ Your forecast should have been "far below iron."

Case B — a local-peak light nucleus

Worked example Example 2 — Helium-4,

$M (^{4} He) = 4.002602 u$ Find $\overline{E_{B}}$ and explain why it beats its neighbours.

Forecast: $^{4}$ He has only 4 nucleons — will it be low like the deuteron, or will "doubly magic" push it surprisingly high?

Count. $Z = 2$ , $N = 2$ , $A = 4$ . Why? Sets how many $m_{^{1} H}$ and $m_{n}$ to add.
Mass defect. $Δ m = 2 (1.007825) + 2 (1.008665) - 4.002602 = 0.030378 u$ Why? Same source-of-energy logic; electrons cancel because we used atomic masses throughout.
Energy. $E_{B} = 0.030378 \times 931.5 = 28.30 MeV$ . Why? $E = m c^{2}$ converts the missing mass into glue energy; $931.5$ is the $u \to$ MeV bridge.
Per nucleon. $\overline{E_{B}} = 28.30/4 = 7.07 MeV$ . Why? Fair comparison per particle.

Verify: $7.07$ MeV is far above the deuteron's $1.11$ despite both being "light." That is the famous local spike at $^{4}$ He — the closed proton shell and closed neutron shell (doubly magic) bind it hard. ✓

Case C — the iron maximum

Worked example Example 3 — Iron-56,

$M (^{56} Fe) = 55.934939 u$ , $Z = 26$ Find $\overline{E_{B}}$ and confirm it is the curve's peak.

Forecast: The parent quoted $\approx 8.8$ MeV for the peak. Will our computed number land there?

Count. $Z = 26$ , $N = 56 - 26 = 30$ , $A = 56$ . Why? $N = A - Z$ ; you must split correctly or the neutron count is wrong.
Mass defect. $Δ m = [26 (1.007825) + 30 (1.008665)] - 55.934939 = 0.528461 u$ Why? 26 hydrogen-atom masses + 30 neutron masses, minus the actual atom.
Energy. $E_{B} = 0.528461 \times 931.5 = 492.3 MeV$ . Why? $E = m c^{2}$ turns the mass defect into binding energy; the $931.5$ factor takes us from $u$ to MeV.
Per nucleon. $\overline{E_{B}} = 492.3/56 = 8.79 MeV$ . Why? Dividing by $A$ gives the fair per-particle glue strength — the only fair way to compare stability across nuclei.

Verify: $8.79 \approx 8.8$ MeV — bang on the quoted peak. Total $E_{B}$ here ( $492$ MeV) is enormous, yet it's the per-nucleon value that crowns iron as most stable. ✓

Case D — a heavy nucleus (the total-vs-per-nucleon trap)

Worked example Example 4 — Uranium-235,

$M (^{235} U) = 235.043924 u$ , $Z = 92$ Find total $E_{B}$ and $\overline{E_{B}}$ . Which is larger, its total or iron's total?

Forecast: $^{235}$ U is huge. Its total binding energy will surely dwarf iron's — but will its per-nucleon value be higher or lower than iron's $8.79$ ?

Count. $Z = 92$ , $N = 235 - 92 = 143$ , $A = 235$ . Why? Big $Z$ means lots of Coulomb repulsion later — but first just count.
Mass defect. $Δ m = [92 (1.007825) + 143 (1.008665)] - 235.043924 = 1.867521 u$ Why? Sum of 92 hydrogen-atom masses and 143 neutron masses, minus the actual atom — the missing mass that became binding energy.
Energy. $E_{B} = 1.867521 \times 931.5 = 1739.6 MeV$ . Why? $E = m c^{2}$ converts the mass defect into glue energy; $931.5$ carries us from $u$ to MeV.
Per nucleon. $\overline{E_{B}} = 1739.6/235 = 7.40 MeV$ . Why? Only the per-nucleon value is a fair stability measure — total $E_{B}$ always grows with $A$ and would mislead.

Verify: Total $E_{B} \approx 1740$ MeV beats iron's $492$ MeV — yet per nucleon $7.40 < 8.79$ . This is exactly the parent's "don't use total for stability" warning: $^{235}$ U is less tightly bound per particle, so it can split (fission) and climb toward the peak. ✓

The next figure places all four nuclei (A–D) on the curve so you can see the story.

Figure — Binding energy — mass defect, BE per nucleon curve

Notice: the deuteron (Case A) is at rock bottom, $^{4}$ He (B) spikes early, $^{56}$ Fe (C) sits at the crest, and $^{235}$ U (D) has slid back down the right side.

Case E — a reaction that RELEASES energy (sign $Q > 0$ )

Worked example Example 5 — D–T fusion,

$^{2} H +^{3} H \to^{4} He + n$ Given total $E_{B}$ values: $E_{B} (^{2} H) = 2.22$ , $E_{B} (^{3} H) = 8.48$ , $E_{B} (^{4} He) = 28.30$ MeV. The neutron is a single nucleon, $E_{B} = 0$ . Find $Q$ .

Forecast: The Sun runs on this — so $Q$ must be positive. But how big, and does the free neutron count in the bookkeeping?

Total BE of reactants. $E_{B} (reactants) = E_{B} (^{2} H) + E_{B} (^{3} H) = 2.22 + 8.48 = 10.70 MeV$ Why? The $Q$ -value formula compares how tightly bound the "before" and "after" states are.
Total BE of products. $E_{B} (products) = E_{B} (^{4} He) + E_{B} (n) = 28.30 + 0 = 28.30 MeV$ Why? A lone neutron has no partner to bind to, so contributes $0$ — the correct way to include it.
Energy released — apply the $Q$ -value formula. $Q = E_{B} (products) - E_{B} (reactants) = 28.30 - 10.70 = 17.6 MeV$ Why this step? If the products are more bound, that extra glue energy is released as kinetic energy of the fragments.

Verify: $Q > 0$ ✓ — matches the Sun powering itself. Sign rule confirmed: products higher on the curve ⇒ energy out. ✓

Case F — a reaction that ABSORBS energy (sign $Q < 0$ )

Worked example Example 6 — Photodisintegration: split the deuteron,

$γ +^{2} H \to p + n$ A gamma photon breaks a deuteron into a free proton and a free neutron. What minimum photon energy is needed?

Forecast: We are un-gluing a bound nucleus — do we get energy out or must we put energy in? Predict the sign of $Q$ before computing.

Identify the states. Before: bound deuteron ( $E_{B} = 2.224$ MeV). After: a free proton and free neutron ( $E_{B} = 0$ each). Why? Breaking a bound state is the reverse of Example 5's logic.
Apply the $Q$ -value formula. $Q = E_{B} (products) - E_{B} (reactants) = 0 - 2.224 = - 2.224 MeV$ Why this step? Products are less bound, so $Q$ is negative — the reaction cannot happen for free.
Minimum photon energy. The photon must supply at least $∣ Q ∣$ : $E_{γ}^{m i n} = 2.224 MeV$ Why? You must add back exactly the binding energy that held the deuteron together.

Verify: $∣ Q ∣$ equals the deuteron's binding energy from Example 1 to the last digit — because breaking a nucleus is paying back its binding energy. Sign $Q < 0$ ⇒ energy absorbed, as forecast for an unbinding process. ✓

Case G — degenerate inputs (a single nucleon)

Worked example Example 7 — Binding energy of a lone

$_{1}^{1}$ H and a free neutron What is $E_{B}$ of ordinary hydrogen $^{1} H$ ( $Z = 1, N = 0$ ) and of a free neutron?

Forecast: With nothing to bind to, what should the mass defect and binding energy be?

Mass defect of $^{1}$ H. $Z = 1$ , $N = 0$ : $Δ m = [1 (1.007825) + 0] - 1.007825 = 0 u$ Why? The "nucleus" here is a single proton — the sum of parts equals the whole. No mass goes missing.
Binding energy. $E_{B} = 0 \times 931.5 = 0 MeV$ . Why? One nucleon has no partner, so there is no glue to account for.
Free neutron. Same logic: it's already a single free nucleon, $E_{B} = 0$ . Why? Applying the toolkit, its "mass defect" is its own mass minus itself $= 0$ , so $E_{B} = Δ m c^{2} = 0$ — a lone particle can have no binding energy.

Verify: $E_{B} = 0$ for any single nucleon. This is the correct zero/degenerate case and it justifies why in Examples 5–6 the free neutron contributed $0$ . The BE-per-nucleon curve therefore starts at 0 at $A = 1$ . ✓

Case H — limiting behaviour (why the curve flattens then falls)

Worked example Example 8 — Compare the

marginal nucleon: Fe vs U Using Examples 3 and 4, quantify how much less bound (per nucleon) uranium is than iron, and connect it to the curve's downhill slope.

Forecast: We already have $8.79$ (Fe) and $7.40$ (U). Will that ~1.4 MeV gap, spread over 235 nucleons, be a large or small total energy?

Per-nucleon deficit of U vs Fe. $Δ \overline{E_{B}} = 8.79 - 7.40 = 1.39 MeV per nucleon$ Why? This is the vertical drop on the curve from the peak to uranium.
Multiply by nucleon count. If $^{235}$ U's nucleons could each be as tightly bound as iron's, the extra binding would be $1.39 \times 235 = 326.7 MeV$ Why this step? It estimates the total energy "available" by rearranging toward the peak — the origin of fission energy.
Interpret the limit. As $A$ grows past iron, Coulomb repulsion ( $\propto Z^{2}$ , long-range) accumulates while the strong force ( $short-range$ , only nearest neighbours) saturates. So $\overline{E_{B}}$ can only decline — it never recovers. The curve has a single broad maximum and falls gently forever after. Why this step? We must explain why the numbers came out this way — the physics of two competing forces is what forces uranium below iron and shapes the whole curve.

Verify: The real energy released in one $^{235}$ U fission is $\approx 200$ MeV — the same order as our $327$ MeV estimate (the estimate is an upper bound because fission fragments land near, not at, the peak). Order-of-magnitude sanity ✓. Direction: U below Fe ⇒ fission favourable ✓.

Case I — word problem / exam twist

Worked example Example 9 — Reactor bookkeeping

A reactor fissions $^{235}$ U, releasing $\approx 200$ MeV per fission event. (a) How much energy is $200$ MeV in joules? (b) Roughly how many fissions per second give $1 W$ of power? Use $1 MeV = 1.602 \times 1 0^{- 13} J$ (from the toolkit).

Forecast: $1$ watt sounds tiny — but a single fission gives only $\sim 1 0^{- 11}$ J. So do we need thousands of fissions per second, or trillions?

Convert per-event energy to joules. $E_{1} = 200 \times 1.602 \times 1 0^{- 13} = 3.204 \times 1 0^{- 11} J$ Why this step? Power is measured in watts (joules per second), so we must leave MeV and enter SI units before dividing.
Fissions per second for 1 W. Power $=$ energy per event $\times$ events per second, i.e. $P = E_{1} \cdot n$ . Solve for the rate $n$ : $n = \frac{P}{E _{1}} = \frac{1 W}{3.204 \times 1 0 ^{- 11} J} = 3.12 \times 1 0^{10} fissions/s$ Why this step? Rearranging $P = E_{1} \cdot n$ isolates the count $n$ we were asked for.

Verify: $\sim 3 \times 1 0^{10}$ fissions per second for a mere $1$ W — reassuringly huge, confirming nuclei release individually tiny (in joules) but collectively vast energy. Units: $J/s \div J = 1/ s$ ✓. The exam "twist" here was reading " $200$ MeV per event" as a per-nucleon number by mistake — it is per fission (all 235 nucleons together). Watch that trap.

Recall Self-test: which cell is each question?

A gamma ray breaks up $^{4}$ He — energy released or absorbed? ::: Absorbed ( $Q < 0$ , Case F logic): you must supply $^{4}$ He's binding energy back. Which has larger total binding energy, $^{56}$ Fe or $^{235}$ U? ::: $^{235}$ U ( $1740$ vs $492$ MeV) — but iron wins per nucleon (Case D trap). What is $E_{B}$ of a free neutron? ::: Exactly $0$ — no partner to bind (Case G). Why can't any nucleus above iron have higher per-nucleon binding? ::: Coulomb repulsion grows as $Z^{2}$ while strong force saturates (Case H limit).

Connections

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