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

The bottom number is $Z$ (protons), the top number is $A$ (total nucleons).

• $Z=8$
• $A=16$
• $N=A-Z=16-8=8$ What this looks like: 8 positive protons and 8 neutral neutrons packed in one clump — a nice even-even, doubly magic nucleus that sits on an early peak of the curve.

Using atomic masses so the 8 electron masses cancel: $\Delta m=[8m_{{}^1\text{H}}+8m_n]-M{(}^{16}\text{O})$ $\Delta m=[8(1.007825)+8(1.008665)]-15.994915$ Why atomic masses? Tables list atoms (nucleus + electrons). Using $m_{{}^1\text{H}}$ (which carries one electron) for each proton means the total $Z=8$ electrons on the left exactly match the $Z=8$ electrons inside $M{(}^{16}\text{O})$ on the right — they cancel, and we never need $m_e$ at all.

Step 1 — mass defect (why: it is the source of $E_B$): $8(1.007825)+8(1.008665)=8.062600+8.069320=16.131920u$ $\Delta m=16.131920-15.994915=0.137005u$ Step 2 — energy (why: $E=m{c}^2$ turns missing mass into glue energy): $E_B=0.137005\times 931.5=127.62\text{MeV}$ Step 3 — per nucleon (why: fair stability comparison): $\overline{E_B}=\frac{127.62}{16}=7.98\text{MeV}$ This sits just below iron's $\approx 8.8$ MeV peak — ¹⁶O is very tightly bound, consistent with its early sharp peak on the curve.

$Z=26$, $N=56-26=30$. Step 1: $26(1.007825)+30(1.008665)=26.203450+30.259950=56.463400u$ $\Delta m=56.463400-55.934939=0.528461u$ Step 2: $E_B=0.528461\times 931.5=492.26\text{MeV}$ Step 3: $\overline{E_B}=\frac{492.26}{56}=8.79\text{MeV}$ What it means: This is the top of the curve — the most tightly bound region. No fusion or fission of iron releases energy, because you can't climb higher.

Compute B's per-nucleon value: $\overline{E_B}{(}^{238}\text{U})=\frac{1802}{238}=7.57\text{MeV}$ Compare: $8.79>7.57$, so iron-56 is more stable. Why totals mislead: Total $E_B$ always rises with $A$ (more nucleons → more attractive bonds), so ²³⁸U's larger total (1802 vs 492) just reflects that it has more particles, not tighter bonding. Stability is about how hard it is to remove ONE nucleon, which is exactly $\overline{E_B}$. Uranium's lower per-nucleon value is precisely why it can fission and release energy — it climbs toward the peak by splitting.

Why this works: Every nucleon's glue improves by $\Delta \overline{E_B}=8.5-7.6=0.9$ MeV. Binding energy that increases is energy released (the products sit deeper in the energy well). $Q\approx A\times \Delta \overline{E_B}=236\times 0.9=212.4\text{MeV}$ What it looks like: On the curve, the reaction moves up the right slope toward the peak. The vertical rise, times the number of nucleons carried up, is the energy budget. This ~200 MeV per fission is the textbook figure for uranium.

Step 1 — mass balance (why: $Q=(\text{reactant mass}-\text{product mass})c^2$): $M_{\text{react}}=2.014102+3.016049=5.030151u$ $M_{\text{prod}}=4.002602+1.008665=5.011267u$ $\Delta m=5.030151-5.011267=0.018884u$ Step 2 — energy: $Q=0.018884\times 931.5=17.59\text{MeV}$ Electron check: left side has $1+1=2$ electrons (from the two hydrogen atoms), right side has $2$ (from helium). They balance, so atomic masses are safe here. See Q-value of nuclear reactions. Why positive: ⁴He sits high on the curve (≈7.07 MeV/nucleon vs the fuel's low values) — glue improved, energy came out. This powers fusion in the Sun.

Why BE bookkeeping works: The free neutron has zero binding energy, so $Q=E_B(\text{products})-E_B(\text{reactants})=E_B{(}^4\text{He})-[E_B{(}^2\text{H})+E_B{(}^3\text{H})]$ $Q=28.30-(2.224+8.482)=28.30-10.706=17.59\text{MeV}$ Match: identical to L4.1 (17.59 MeV) up to rounding. Two independent routes — raw masses and binding energies — agree, because both encode the same $E=m{c}^2$ accounting.

(a) Energy per fission (why: nucleons climb toward the peak): $Q\approx 236\times (8.45-7.59)=236\times 0.86=202.96\text{MeV}$ In joules: $Q=202.96\times 1.602\times 10^{-13}=3.251\times 10^{-11}\text{J}$.

(b) Fission rate (why: electrical power / efficiency = thermal power, then divide by energy each fission gives): Thermal power needed $=\frac{1000\text{MW}}{0.33}=3.030\times 10^9\text{W}$. $\text{fissions/s}=\frac{3.030\times 10^9}{3.251\times 10^{-11}}=9.32\times 10^{19}{\text{s}}^{-1}$

(c) Daily mass (why: count nuclei in a day, convert to moles then grams): Fissions/day $=9.32\times 10^{19}\times 86400=8.05\times 10^{24}$. Moles $=\frac{8.05\times 10^{24}}{6.022\times 10^{23}}=13.37\text{mol}$. Mass $=13.37\times 235=3142\text{g}\approx 3.14\text{kg/day}$.

What it means: A few kilograms a day of fuel powers a gigawatt city — the payoff of moving nucleons up the curve. Contrast with chemical fuel (coal: thousands of tonnes/day), the numeric signature of nuclear vs chemical energy scales.

Argument: Fusion releases energy only when the product sits higher on the $\overline{E_B}$ curve than the reactants. Iron (8.79 MeV) is the maximum. Gold (7.91 MeV) lies on the descending right slope — it is lower than iron. So fusing lighter nuclei up to gold would mean climbing past the peak and back down, which would cost energy, not release it. A star can only power itself by moving up toward the peak; once it reaches iron there is nowhere higher to go, energy generation halts, and the core collapses. Gold and heavier elements are forged not by ordinary fusion but by rapid neutron capture in supernovae. See Stability of nuclei & N-Z curve and Strong nuclear force.

Energy released when nucleons move to higher $\overline{E_B}$ equals ==$A\times \Delta \overline{E_B}$== (nucleons times per-nucleon gain). Fission of a heavy nucleus releases about ::: ~200 MeV D–T fusion releases about ::: ~17.6 MeV