Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF: Exercises

Level 1 — Recognition

Exercise 1.1

On the Binding Energy per Nucleon Curve, the value $B / A$ rises for light nuclei, peaks near $A \approx 56$ , then falls. Which direction along the curve — left-to-peak or right-to-peak — corresponds to fusion, and why does that direction release energy?

Recall Solution

Left-to-peak. Fusion combines light nuclei (small $A$ , low on the left of the curve) into a heavier product nearer the peak. Because $B / A$ is larger at the product, each nucleon in the product is more tightly bound than it was in the reactants. That "extra" binding energy — the deepening of the well — is released as kinetic energy. Right-to-peak (splitting heavy nuclei down toward the peak) is Nuclear Fission. Both routes move toward the peak, which is exactly where the tightest binding lives.

Exercise 1.2

State the D–T reaction and identify which product is neutral and which is charged. Which one can a magnetic bottle hold?

Recall Solution

$_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n$ The $_{2}^{4} He$ (an $α$ particle) carries charge $+ 2 e$ → charged. The neutron $_{0}^{1} n$ has charge $0$ → neutral. A magnetic bottle steers particles by the force $q v \times B$ , which is zero for $q = 0$ . So it holds the $α$ but not the neutron — the neutron flies straight out to the blanket. (See Plasma Physics & Magnetic Confinement.)

Level 2 — Application

Exercise 2.1

Compute the Q-value of the D–D reaction $_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n$ using $m (^{2} H) = 2.014102 u$ , $m (^{3} He) = 3.016029 u$ , $m (n) = 1.008665 u$ .

Recall Solution

Step 1 — sum reactants. $2.014102 + 2.014102 = 4.028204 u$ . Step 2 — sum products. $3.016029 + 1.008665 = 4.024694 u$ . Step 3 — mass defect. $Δ m = 4.028204 - 4.024694 = 0.003510 u$ . Step 4 — convert. $Q = 0.003510 \times 931.5 \approx 3.27 MeV$ . Interpretation: far smaller than D–T's 17.6 MeV — this is one reason D–T is our lab fuel of choice, not D–D.

Exercise 2.2

A D–T reaction releases $17.6$ MeV per event. How many such reactions per second give $1$ watt of fusion power? ( $1 MeV = 1.602 \times 1 0^{- 13} J$ .)

Recall Solution

Step 1 — energy per event in joules. $17.6 MeV \times 1.602 \times 1 0^{- 13} \frac{J}{MeV} = 2.819 \times 1 0^{- 12} J .$ Step 2 — rate = power ÷ energy-per-event. $\frac{1 J/s}{2.819 \times 1 0 ^{- 12} J} \approx 3.55 \times 1 0^{11} reactions/s .$ Why this step? Power is energy per second; dividing the demanded joules-per-second by joules-per-reaction leaves reactions-per-second.

Level 3 — Analysis

Exercise 3.1

Estimate the height of the Coulomb barrier for two deuterons ( $Z_{1} = Z_{2} = 1$ ) meeting at a nuclear separation $r_{0} = 5 \times 1 0^{- 15} m$ . Give the answer in keV. (See figure below.)

Figure — Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF

Recall Solution

Tool choice — why Coulomb's law? Before the short-range strong force can act, the only force acting is the electrostatic repulsion between the two positive charges. Its potential energy is the barrier they must climb. Step 1 — write the barrier. $U_{C} = \frac{1}{4 π ε _{0}} \frac{Z _{1} Z _{2} e ^{2}}{r _{0}} .$ Step 2 — plug numbers (joules). $U_{C} = 8.988 \times 1 0^{9} \cdot \frac{( 1.602 \times 1 0 ^{- 19} ) ^{2}}{5 \times 1 0 ^{- 15}} = 4.61 \times 1 0^{- 14} J .$ Step 3 — convert to keV. Divide by $1.602 \times 1 0^{- 16} J/keV$ : $U_{C} \approx 288 keV .$ Look at the red barrier in the figure: classically a particle with less than $288$ keV of energy stops at the wall and rolls back. The green dashed line is where Quantum Tunnelling lets it leak through the wall anyway.

Exercise 3.2

In the D–T reaction the products start essentially at rest (D and T fuse slowly). Show from momentum conservation that the kinetic energies split as $E_{n} : E_{α} = m_{α} : m_{n}$ , and hence recover $E_{n} \approx 14.1$ MeV, $E_{α} \approx 3.5$ MeV from $Q = 17.6$ MeV.

Recall Solution

Step 1 — momentum conservation. Total momentum before $\approx 0$ , so after the split the two products carry equal and opposite momenta: $p_{n} = p_{α} \equiv p$ . Step 2 — write KE in terms of momentum. For a non-relativistic particle, $E = \frac{p ^{2}}{2 m}$ . With the same $p$ : $\frac{E _{n}}{E _{α}} = \frac{p ^{2} /2 m _{n}}{p ^{2} /2 m _{α}} = \frac{m _{α}}{m _{n}} .$ Step 3 — put in mass numbers. $m_{α} : m_{n} \approx 4 : 1$ , so $E_{n} : E_{α} = 4 : 1$ . Step 4 — share out $Q$ . The lighter neutron gets $\frac{4}{5}$ of the total: $E_{n} = \frac{4}{5} \times 17.6 = 14.1 MeV, E_{α} = \frac{1}{5} \times 17.6 = 3.5 MeV .$ Why the neutron is faster: with equal momenta, KE $\propto 1/ m$ , so the lighter partner carries the larger kinetic energy.

Level 4 — Synthesis

Exercise 4.1

The Lawson triple product for D–T ignition is $n T τ_{E} ≳ 3 \times 1 0^{21} keV\cdots\cdotm^{- 3} .$ A tokamak operates at $T = 15 keV$ (about $1.7 \times 1 0^{8}$ K) and density $n = 1 \times 1 0^{20} m^{- 3}$ . What minimum energy-confinement time $τ_{E}$ does it need? Then contrast with an ICF pellet reaching $n = 1 0^{31} m^{- 3}$ at the same $T$ .

Recall Solution

Step 1 — solve for $τ_{E}$ . $τ_{E} \geq \frac{3 \times 1 0 ^{21}}{n T} = \frac{3 \times 1 0 ^{21}}{( 1 \times 1 0 ^{20} ) ( 15 )} = \frac{3 \times 1 0 ^{21}}{1.5 \times 1 0 ^{21}} = 2 s .$ Step 2 — repeat for ICF. $τ_{E} \geq \frac{3 \times 1 0 ^{21}}{( 1 0 ^{31} ) ( 15 )} = \frac{3 \times 1 0 ^{21}}{1.5 \times 1 0 ^{32}} = 2 \times 1 0^{- 11} s = 20 ps .$ Interpretation. Same target product, opposite strategy: the tokamak wins with a long confinement (seconds) at low density; ICF wins with enormous density and needs only tens of picoseconds — "burn before it blows apart." (See Plasma Physics & Magnetic Confinement.)

Exercise 4.2

The net solar p–p chain is $4^{1} H \to^{4} He + 2 e^{+} + 2 ν_{e}$ , releasing $\approx 26.7$ MeV per helium (as $E=mc^2$ applied to the mass defect). The Sun's luminosity is $L = 3.8 \times 1 0^{26}$ W. Estimate how many helium nuclei the Sun makes each second, and the hydrogen mass consumed per second. ( $m (^{1} H) = 1.6726 \times 1 0^{- 27}$ kg.)

Recall Solution

Step 1 — energy per He event in joules. $26.7 MeV \times 1.602 \times 1 0^{- 13} \frac{J}{MeV} = 4.277 \times 1 0^{- 12} J .$ Step 2 — He nuclei per second. $\frac{L}{E _{He}} = \frac{3.8 \times 1 0 ^{26}}{4.277 \times 1 0 ^{- 12}} \approx 8.9 \times 1 0^{37} He/s .$ Step 3 — protons consumed. Each He uses 4 protons: $4 \times 8.9 \times 1 0^{37} \approx 3.55 \times 1 0^{38} protons/s .$ Step 4 — hydrogen mass per second. $3.55 \times 1 0^{38} \times 1.6726 \times 1 0^{- 27} kg \approx 5.9 \times 1 0^{11} kg/s .$ That is about $600$ million tonnes of hydrogen fused every second — yet with $\sim 1 0^{57}$ protons in store, the Sun runs for billions of years (the slow weak step from Beta Decay & the Weak Interaction throttles the rate).

Level 5 — Mastery

Exercise 5.1

The mean thermal energy of a plasma at temperature $T$ is $⟨ E ⟩ \sim k_{B} T$ . For a tokamak at $T = 1.5 \times 1 0^{8}$ K, compute $k_{B} T$ in keV and compare it to the $288$ keV Coulomb barrier of Exercise 3.1. If the average particle falls so far short, how does fusion still occur? Reference the Maxwell–Boltzmann Distribution and Quantum Tunnelling.

Recall Solution

Step 1 — thermal energy. $k_{B} = 1.381 \times 1 0^{- 23}$ J/K, so $k_{B} T = 1.381 \times 1 0^{- 23} \times 1.5 \times 1 0^{8} = 2.07 \times 1 0^{- 15} J .$ Step 2 — convert to keV (divide by $1.602 \times 1 0^{- 16}$ J/keV): $k_{B} T \approx 12.9 keV .$ Step 3 — compare. $12.9$ keV vs a $288$ keV barrier: the average particle has only $\sim 1/22$ of the needed energy. Step 4 — why fusion still happens (two effects stacking).

The high-energy tail (Maxwell–Boltzmann): the fraction of particles with energy $E$ falls off like $e^{- E / k_{B} T}$ . A tiny minority in the far tail carries tens of keV — enough to climb most of the barrier.
Quantum tunnelling: even those tail particles don't fully clear $288$ keV. Quantum mechanics lets them tunnel through the remaining wall (the green dashed line in the figure). The overlap of "tail particles" $\times$ "tunnelling probability" is the Gamow window — a narrow energy band where nearly all fusion actually occurs. So fusion runs not on the average particle but on a rare, fast, tunnelling few.

Exercise 5.2

Order these fuels by Q-value per reaction and explain what the ordering reveals about why the Sun and an Earth reactor make different choices: D–T ( $17.6$ MeV), D–D→ $^{3}$ He+n ( $3.27$ MeV, from Ex 2.1), p–p first step $^{1} H +^{1} H \to^{2} H + e^{+} + ν_{e}$ ( $Q \approx 0.42$ MeV).

Recall Solution

Ordering by Q: p–p first step ( $0.42$ MeV) $<$ D–D ( $3.27$ MeV) $<$ D–T ( $17.6$ MeV). But Q is not the deciding factor — the rate is:

The Sun cannot pick D–T (no free tritium) and cannot pick a fast reaction; it is stuck starting from bare protons via the weak first step. That step is astronomically slow (needs a proton to $β^{+}$ -decay mid-collision, per Beta Decay & the Weak Interaction), which is why the Sun burns for $\sim 1 0^{10}$ years despite its low per-event Q. Its huge fuel tank ( $\sim 1 0^{57}$ protons) makes the total power vast. (Longer chains build heavier elements — Stellar Nucleosynthesis.)
Earth reactors need a fast reaction at reachable temperatures. D–T wins on both the largest Q and the largest cross-section at $\sim 10$ – $20$ keV, so it lights first — even though tritium must be bred (from the $14.1$ MeV neutron hitting lithium). Lesson: fuel choice is set by rate × availability, with Q a bonus, not the referee.

Recall One-line self-check

Which single number appears in both the tokamak and ICF Lawson estimates and forces their opposite strategies? ::: The target triple product $3 \times 1 0^{21}$ keV·s·m $^{- 3}$ — fixed target, so large $τ_{E}$ (tokamak) trades against large $n$ (ICF).