Exercises — Fusion — solar fusion, tokamak (concept)

WHAT the rule says: joining two nuclei releases energy only if the product sits higher on the Binding Energy per Nucleon Curve than the reactants — i.e. only for nuclei below the iron peak.

(a) Deuterium and tritium are extremely light (mass numbers 2 and 3), far below iron. Fusing them climbs the curve → energy released. Yes.

(b) Uranium-235 is above iron (mass number 235). Adding a neutron and combining does not climb the curve; instead ${}^{235}\text{U}$ splits (fission) to release energy. Fusion here does not release energy — No.

Answer: (a) Yes, (b) No.

(a) The Coulomb (electrostatic) repulsion — both protons are positive and like charges repel. (b) The strong nuclear force — attractive, but only over a tiny range (~femtometres), so the protons must get very close first. (c) Quantum tunnelling — see Quantum Tunnelling; particles have a small probability of passing through the Coulomb barrier even without enough classical energy to go over it.

Step 1 — reactant mass. $2\times m_D=2\times 2.014102=4.028204$ u. Why sum? Rest mass holds the energy of both inputs. Step 2 — product mass. $m_{{}^{3}He}+m_n=3.016029+1.008665=4.024694$ u. Why sum the products? The rest mass locked in the two output particles is what we must compare against the inputs to see how much vanished. Step 3 — mass defect. $\Delta m=4.028204-4.024694=0.003510$ u. Why subtract products from reactants? The mass that disappeared became energy. Step 4 — convert. $Q=\Delta m\times 931.5=0.003510\times 931.5=3.27\text{MeV}$. Why $\times 931.5$? That is $1\text{u}\cdot c^2$ in MeV, so it turns mass-in-u into energy-in-MeV.

Answer: $Q\approx 3.27\mathbf{MeV}$.

Step 1 — reactant mass. $2\times 2.014102=4.028204$ u. Why sum? Both deuterons bring their rest mass into the reaction, so we add them. Step 2 — product mass. $3.016049+1.007825=4.023874$ u. Why sum the products? The tritium and proton carry away their own rest masses; adding them lets us compare inputs to outputs. Step 3 — mass defect. $\Delta m=4.028204-4.023874=0.004330$ u. Why subtract? The decrease in rest mass is exactly the mass that got converted into energy. Step 4 — convert. $Q=0.004330\times 931.5=4.03\text{MeV}$. Why $\times 931.5$? Multiplying the mass defect in u by $1\text{u}\cdot c^2=931.5$ MeV turns the vanished mass into its energy equivalent.

Answer: $Q\approx 4.03\mathbf{MeV}$. Notice both D–D branches give only ~3–4 MeV — much less than D–T's 17.6 MeV, which is why D–T is the preferred reactor fuel.

Step 1 — how many D nuclei in 1 g? Mass of one D atom $=2.014\times 1.6605\times 10^{-27}=3.344\times 10^{-27}$ kg. Why? Convert u to kg with $1\text{u}=1.6605\times 10^{-27}$ kg. Number of nuclei $N=\frac{10^{-3}\text{kg}}{3.344\times 10^{-27}\text{kg}}=2.99\times 10^{23}$. Step 2 — energy per reaction in joules. $17.6\text{MeV}\times 1.602\times 10^{-13}\text{J/MeV}=2.82\times 10^{-12}$ J. Why convert to joules first? We are asked for the answer in joules, and MeV are not joules; multiplying an unconverted MeV value by $N$ would give a number in the wrong unit family, off by a factor of $1.602\times 10^{-13}$. Step 3 — total. $E=N\times (2.82\times 10^{-12})=2.99\times 10^{23}\times 2.82\times 10^{-12}=8.4\times 10^{11}$ J. Why multiply by $N$? Every one of the $N$ reactions releases the same packet of energy, so the total is the per-reaction energy stacked $N$ times over.

Answer: $E\approx 8.4\times 10^{11}\mathbf{J}$ — about the energy of 200 tonnes of TNT from one gram of deuterium. Why the point: this staggering energy density is the whole promise of fusion power.

Step 1 — energy per He in joules. $26.7\text{MeV}\times 1.602\times 10^{-13}=4.28\times 10^{-12}$ J. Why convert? $L$ is in watts (joules per second), so $Q$ must also be in joules for the division to make sense. Step 2 — rate. Number per second $=\frac{L}{Q}=\frac{3.8\times 10^{26}}{4.28\times 10^{-12}}=8.9\times 10^{37}$ per second. Why divide? Each helium event delivers a fixed packet $Q$; total power $÷$ packet size $=$ events per second.

Answer: $\approx 8.9\times 10^{37}$ helium nuclei per second. Since each uses 4 protons, the Sun burns $\sim 3.6\times 10^{38}$ protons every second — and still has enough for billions of years.

Step 1 — multiply the three. $nT{\tau }_E=(1.0\times 10^{20})(10)(3)=3.0\times 10^{21}{\text{keV⋅s⋅m}}^{-3}$. Why multiply all three? Losses scale down when you are denser, hotter, or leak slower; the product measures whether their combined effect beats the losses. That's why it's a product, not a sum — improving any one factor multiplies the whole. Step 2 — compare. $3.0\times 10^{21}\ge 3\times 10^{21}$ → just meets the threshold. (b) Yes, marginally at ignition. (c) Since it exactly reaches the bar, no change is needed; but any single factor could be dropped by up to $\times$ its own margin. For robust margin (say $2\times$ over threshold) you would, e.g., double ${\tau }_E$ to 6 s, or double $n$, or double $T$ — each multiplies the triple product by 2.

Answer: (a) $3.0\times 10^{21}$; (b) reaches ignition (just); (c) doubling any one of $n$, $T$, or ${\tau }_E$ doubles the product.

Step 1 — what work means. The rate of work (power) done by a force is $P=\vec{F}\cdot \vec{v}$ — force along the direction of motion. Why the dot product? It picks out only the component of force parallel to velocity; a sideways force transfers no energy. Step 2 — plug in Lorentz. $P=(q\vec{v}\times \vec{B})\cdot \vec{v}$. Step 3 — the key geometric fact. The cross product $\vec{v}\times \vec{B}$ is perpendicular to $\vec{v}$ by definition. A vector perpendicular to $\vec{v}$ dotted with $\vec{v}$ gives $0$. So $P=q(\vec{v}\times \vec{B})\cdot \vec{v}=0$ always, for any $\vec{v}$, $\vec{B}$. Step 4 — meaning. Zero work → the field never changes the particle's speed (kinetic energy), only its direction. So the magnetic cage confines (bends paths into circles/helices around field lines) but cannot heat. Heating must come from other sources: currents (ohmic), radio-frequency waves, neutral-beam injection.

Answer: $P=0$ identically; magnetic field confines but does no heating.

Fusion, per kg. Step 1: energy per reaction $=17.6\text{MeV}\times 1.602\times 10^{-13}=2.82\times 10^{-12}$ J. Step 2: fuel mass per reaction $=5.03\text{u}\times 1.6605\times 10^{-27}=8.35\times 10^{-27}$ kg. Step 3: reactions per kg $=1/8.35\times 10^{-27}=1.198\times 10^{26}$. Step 4: energy per kg $=1.198\times 10^{26}\times 2.82\times 10^{-12}=3.38\times 10^{14}$ J/kg.

Chemical (carbon), per kg. Step 1: energy per atom $=4\text{eV}=4\times 1.602\times 10^{-19}=6.41\times 10^{-19}$ J. Step 2: mass per atom $=12\times 1.6605\times 10^{-27}=1.993\times 10^{-26}$ kg. Step 3: atoms per kg $=1/1.993\times 10^{-26}=5.02\times 10^{25}$. Step 4: energy per kg $=5.02\times 10^{25}\times 6.41\times 10^{-19}=3.22\times 10^7$ J/kg.

Ratio. $\frac{3.38\times 10^{14}}{3.22\times 10^7}\approx 1.0\times 10^7$.

Answer: fusion delivers about ten million times more energy per kilogram than chemical burning. Why: nuclear binding energies (MeV) dwarf chemical bond energies (eV) by roughly a factor of a million per particle, amplified further by the lighter fuel mass.

(a) Mass-to-energy rate. From Mass-Energy Equivalence (E=mc^2), $E=m{c}^2\Rightarrow \dot{m}=L/c^2$. $\dot{m}=\frac{3.8\times 10^{26}}{(3.0\times 10^8{)}^2}=\frac{3.8\times 10^{26}}{9.0\times 10^{16}}=4.2\times 10^9\text{kg/s}$. Why divide by $c^2$? $c^2$ is the "exchange rate" from mass to energy; dividing energy-per-second by it gives mass-per-second. (b) Total over its life. Seconds in $10^{10}$ years $=10^{10}\times 3.156\times 10^7=3.156\times 10^{17}$ s. Total mass converted $=4.2\times 10^9\times 3.156\times 10^{17}=1.33\times 10^{27}$ kg. Fraction $=\frac{1.33\times 10^{27}}{2.0\times 10^{30}}=6.6\times 10^{-4}\approx 0.07\%$.

Answer: (a) $\approx 4.2\times 10^9$ kg/s ($\sim$4 million tonnes/s); (b) $\approx 1.3\times 10^{27}$ kg total, only about 0.07% of the Sun's mass. That tiny fraction powers 10 billion years of sunlight — the signature of how much energy hides in $c^2$.

Chain of reasoning (each link earns the next):

1. Classically impossible. $1.3\text{keV}\ll$ several hundred keV barrier, so almost no proton has enough energy to climb over the repulsive Coulomb barrier. (Cross-check: even the fast tail of the thermal distribution rarely reaches barrier-top.)
2. Tunnelling opens a door. By Quantum Tunnelling, a proton has a small but nonzero probability of passing through the barrier. This probability rises steeply with energy, so fusion is dominated by the rare, above-average-energy protons (the "Gamow peak").
3. The weak-force bottleneck. Even after tunnelling, the very first step $p+p\to {}^2\text{D}+e^{+}+{\nu }_e$ requires a proton to convert to a neutron via the weak interaction — an intrinsically slow process.
4. Result: each individual proton waits, on average, billions of years to fuse. The Sun shines steadily only because it contains an astronomical number ($\sim 10^{57}$) of protons, so the total rate is enormous even though the per-proton rate is tiny.

Answer: fusion proceeds by quantum tunnelling in the plasma's hot ionized core, gated by the slow weak interaction; slowness per proton + huge proton count = steady multi-billion-year burn.

Fusion Q comes from ==mass defect $\times c^2$; D–T gives 17.6 MeV; magnetic field does zero work (confines, never heats); ignition needs the triple product $nT{\tau }_E$ above threshold; the Sun fuses only via quantum tunnelling==.