Nuclear reactions — Q-value, cross-section: Exercises

Constants you will reuse: $1 u = 931.5 \frac{MeV}{c ^{2}}, 1 barn = 1 0^{- 24} cm^{2}, ln 2 = 0.693.$

Level 1 — Recognition

Recall Solution L1.1

WHAT decides the sign: if mass increases, mass was bought using kinetic energy → energy went in → endoergic ( $Q < 0$ ). $Δ m = m_{reactants} - m_{products} = 18.005677 - 18.006957 = - 0.001280 u$ $Δ m < 0$ means products are heavier → endoergic. $∣ Q ∣ = 0.001280 \times 931.5 = 1.19 MeV .$

Recall Solution L1.2

By definition $1 b = 1 0^{- 24} cm^{2}$ , so it is just a multiplication: $σ = 583 \times 1 0^{- 24} cm^{2} = 5.83 \times 1 0^{- 22} cm^{2} .$ WHY the barn exists: nuclear areas are absurdly tiny in $cm^{2}$ ; the barn keeps numbers human-sized.

Recall Solution L1.3

(i) Energy released → $Q = [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2}$ .
(ii) Likelihood → the cross-section $σ$ (an effective area).
(iii) Beam survival → $I = I_{0} e^{- nσ x}$ , half-thickness $x_{1/2} = ln 2/ (nσ)$ .

Level 2 — Application

Recall Solution L2.1

WHAT: apply $Q = Δ (rest mass) c^{2}$ using atomic masses (electrons balance: $1 + 1$ electrons each side). $Δ m = (2.014102 + 3.016049) - (4.002603 + 1.008665) = 5.030151 - 5.011268 = + 0.018883 u$ Mass lost → exoergic. $Q = 0.018883 \times 931.5 = 17.6 MeV .$ WHY it matters: this is the reaction that powers fusion reactors — see Nuclear fusion.

Recall Solution L2.2

WHY not just $1.19$ MeV: momentum must survive the collision, so the products keep drifting and that drift energy is unavailable. The usable (centre-of-mass) share is the fraction $m_{X} / (m_{a} + m_{X})$ . $E_{t h} = ∣ Q ∣ \frac{m _{a} + m _{X}}{m _{X}} = 1.19 \times \frac{4.0026 + 14.0031}{14.0031} = 1.19 \times 1.286 = 1.53 MeV .$

Recall Solution L2.3

$Σ = nσ = 1.3 \times 1 0^{23} \times 755 \times 1 0^{- 24} = 98.2 cm^{- 1} .$ Half-thickness from $\frac{1}{2} = e^{- Σ x}$ : $x_{1/2} = \frac{l n 2}{Σ} = \frac{0.693}{98.2} = 7.1 \times 1 0^{- 3} cm .$ A giant $σ$ ⇒ tiny half-thickness ⇒ excellent absorber. Look at the survival curve below — boron's curve plunges almost immediately.

Figure — Nuclear reactions — Q-value, cross-section

Level 3 — Analysis

Recall Solution L3.1

$\frac{I}{I _{0}} = e^{- Σ x} = e^{- 98.2 \times 0.02} = e^{- 1.964} = 0.140.$ So about 14% survives — 86% absorbed in just $0.2 mm$ . This is why boron control rods work.

Recall Solution L3.2

WHY $R = σ ϕ$ : rate = (effective area a target presents) × (particles sweeping past per area per second). $R = σ ϕ = 755 \times 1 0^{- 24} \times 2.0 \times 1 0^{12} = 1.51 \times 1 0^{- 9} s^{- 1} .$ Total over $N$ nuclei: $R_{t o t} = N σ ϕ = 6.0 \times 1 0^{20} \times 1.51 \times 1 0^{- 9} = 9.06 \times 1 0^{11} reactions/s .$

Recall Solution L3.3

$λ = \frac{1}{Σ} = \frac{1}{98.2} = 1.02 \times 1 0^{- 2} cm .$ Note $x_{1/2} = λ ln 2 = 0.693 λ$ : the half-thickness is always shorter than the mean free path, because halving needs less penetration than the average single-collision distance. See how the $1/ e$ point (mean free path) sits farther right than the half point on the figure.

Level 4 — Synthesis

Recall Solution L4.1

WHY this works: $Q$ = (energy released as products become more tightly bound) = $\sum B_{products} - \sum B_{reactants}$ — see Binding energy per nucleon curve. $Q = (B_{^{4} He} + B_{n}) - (B_{^{2} H} + B_{^{3} H}) = (28.296 + 0) - (2.224 + 8.482) = 17.59 MeV .$ This matches the $17.6 MeV$ from masses (L2.1) — two roads, one answer. See Mass defect & E=mc².

Recall Solution L4.2

WHAT: even a massless projectile brings momentum, so the same "CM drift" penalty applies. For a photon on a target of rest energy $M c^{2}$ , $E_{t h} = ∣ Q ∣ (1 + \frac{∣ Q ∣}{2 M c ^{2}}) .$ WHY this form: the correction $∣ Q ∣/ (2 M c^{2})$ is the fraction of energy that must stay as recoil. $E_{t h} = 2.224 (1 + \frac{2.224}{2 \times 1876}) = 2.224 (1 + 0.000593) = 2.2253 MeV .$ The recoil penalty is a tiny $+ 0.0013 MeV$ — but it is nonzero, confirming threshold $> ∣ Q ∣$ even for light.

Level 5 — Mastery

Recall Solution L5.1

Plug $Q > 0$ into $E_{t h} = - Q (m_{a} + m_{X}) / m_{X}$ : $E_{t h} = - 17.6 \times \frac{5.03}{3.02} = - 29.3 MeV < 0.$ A negative threshold is physically meaningless as a required energy — it is the formula's way of saying no minimum energy is needed; the reaction can proceed even at (in principle) zero projectile energy. WHY in practice you still need a kick: the Coulomb barrier between two positive nuclei must be overcome — that is a separate effect from the $Q$ -based threshold, and is why fusion needs high temperature (see Nuclear fusion).

Recall Solution L5.2

WHY multiply: each layer attenuates independently, so exponents add. $\frac{I}{I _{0}} = e^{- Σ_{A} x_{A}} e^{- Σ_{B} x_{B}} = e^{- (1.2 \times 0.5 + 4.0 \times 0.3)} = e^{- (0.6 + 1.2)} = e^{- 1.8} = 0.165.$ About 16.5% survives both layers.

Recall Solution L5.3

(i) $σ \to 0 \Rightarrow nσ x \to 0 \Rightarrow e^{0} = 1$ : the beam passes completely undisturbed — the material is transparent. This models an energy below the reaction threshold, where $σ$ vanishes.
(ii) $σ \to \infty \Rightarrow nσ x \to \infty \Rightarrow e^{- \infty} = 0$ : every particle is absorbed in an infinitesimal thickness — a perfect black absorber (a strong resonance). See Neutron flux & reactor physics for how resonances shape reactor design.

Recall wrap-up

Recall Did you internalise the ladder?

Cover the answers; speak them aloud.

Sign convention: heavier products ⇒ ? ::: endoergic, $Q < 0$ . Why $E_{t h} > ∣ Q ∣$ ::: momentum conservation forces centre-of-mass drift, stealing part of the energy. Relation between half-thickness and mean free path ::: $x_{1/2} = λ ln 2 \approx 0.693 λ$ . Two ways to get $Q$ ::: mass difference $\times c^{2}$ , or $\sum B_{products} - \sum B_{reactants}$ . What a negative threshold signals ::: exoergic reaction — no $Q$ -threshold at all. Composite slab survival ::: multiply exponentials; add the $Σ x$ terms.