Nuclear reactions — Q-value calculation: Exercises

Constants used throughout: $1 u = 931.5 MeV / c^{2}$ . Atomic/particle masses (u): $^{1}$ H $= 1.007825$ , n $= 1.008665$ , $^{2}$ H $= 2.014102$ , $^{3}$ H $= 3.016049$ , $^{3}$ He $= 3.016029$ , $^{4}$ He $= 4.002603$ , $^{6}$ Li $= 6.015123$ , $^{7}$ Li $= 7.016004$ , $^{9}$ Be $= 9.012182$ , $^{12}$ C $= 12.000000$ , $^{14}$ N $= 14.003074$ , $^{17}$ O $= 16.999132$ , $^{235}$ U $= 235.043930$ , $^{236}$ U $= 236.045568$ , $m_{e} = 0.000549$ .

Level 1 — Recognition

L1.1 — Sign of Q

For each reaction below, state whether it is exothermic ( $Q > 0$ ) or endothermic ( $Q < 0$ ), given only the mass change $Δ m = m_{before} - m_{after}$ : (a) $Δ m = + 0.0189$ u (b) $Δ m = - 0.0013$ u (c) $Δ m = 0$ .

Recall Solution

The rule (mnemonic BAM): $Q = (Before - After) \times 931.5$ . Positive mass loss ⇒ energy pops out.

(a) $Δ m > 0 \Rightarrow Q > 0 \Rightarrow$ exothermic. Products are lighter; the missing mass left as kinetic energy.
(b) $Δ m < 0 \Rightarrow Q < 0 \Rightarrow$ endothermic. Products are heavier; you had to push energy in.
(c) $Δ m = 0 \Rightarrow Q = 0 \Rightarrow$ elastic — no rest mass changed hands.

L1.2 — One-line conversion

A reaction has $Δ m = 0.0189$ u. Convert to $Q$ in MeV.

Recall Solution

$Q = 0.0189 \times 931.5 = 17.6 MeV .$ Why multiply by 931.5? That is how many MeV live inside one atomic mass unit of vanished mass — the same energy written in nuclear-physics clothes.

Level 2 — Application

L2.1 — Direct Q from a mass table

Compute $Q$ for the D–D fusion branch: $_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n .$

Recall Solution

Before: $2.014102 + 2.014102 = 4.028204$ u. After: $3.016029 + 1.008665 = 4.024694$ u. $Δ m$ : $4.028204 - 4.024694 = 0.003510$ u. Convert: $Q = 0.003510 \times 931.5 = 3.27 MeV$ . Exothermic (fusion releases energy).

L2.2 — Neutron-induced, atomic masses cancel

Compute $Q$ for $_{3}^{6} Li +_{0}^{1} n \to_{1}^{3} H +_{2}^{4} He .$

Recall Solution

Check electrons first: left has $Z = 3$ (Li) $+ 0$ (n) $= 3$ ; right has $Z = 1 + 2 = 3$ . Balanced ⇒ atomic masses cancel electrons safely. Before: $6.015123 + 1.008665 = 7.023788$ u. After: $3.016049 + 4.002603 = 7.018652$ u. $Δ m$ : $7.023788 - 7.018652 = 0.005136$ u. $Q = 0.005136 \times 931.5 = 4.78 MeV$ (exothermic; this reaction breeds tritium).

L2.3 — Q straight from kinetic energies

$^{210}$ Po decays by $α$ emission at rest. The $α$ leaves with $K_{α} = 5.30$ MeV, the $^{206}$ Pb daughter recoils with $K_{D} = 0.10$ MeV. Find $Q$ without a mass table.

Recall Solution

$Q = K_{final} - K_{initial} = (5.30 + 0.10) - 0 = 5.40 MeV .$ Why this works: the disappeared rest mass reappears as the total kinetic energy of ALL fragments — the daughter's recoil counts too.

Figure — Nuclear reactions — Q-value calculation

Level 3 — Analysis

L3.1 — Endothermic + threshold

For Rutherford's transmutation $_{7}^{14} N +_{2}^{4} He \to_{8}^{17} O +_{1}^{1} H,$ find $Q$ and the threshold $α$ kinetic energy (N target at rest).

Recall Solution

Before: $14.003074 + 4.002603 = 18.005677$ u. After: $16.999132 + 1.007825 = 18.006957$ u. $Δ m = - 0.001280$ u ⇒ heavier products ⇒ endothermic. $Q = - 0.001280 \times 931.5 = - 1.19 MeV$ . Threshold (projectile $a = α$ , target $X =$ N): $K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}}) = 1.19 (1 + \frac{4.0026}{14.0031}) = 1.19 \times 1.286 = 1.53 MeV .$ Why bigger than $∣ Q ∣$ ? Products cannot stop dead — the center-of-mass keeps drifting forward to conserve momentum, and that bulk KE is unavailable for the reaction.

L3.2 — Split of Q in $α$ -decay

$^{210}$ Po $\to^{206}$ Pb $+ α$ has $Q = 5.40$ MeV. Using momentum conservation, find $K_{α}$ and $K_{D}$ separately. Take $m_{α} \approx 4$ u, $m_{D} \approx 206$ u.

Recall Solution

Parent at rest ⇒ fragments fly apart with equal, opposite momentum $p$ : $p_{α} = p_{D} = p$ . Kinetic energy $K = p^{2} / (2 m)$ , so $K \propto 1/ m$ : the lighter $α$ gets the bigger slice. $K_{α} = Q \frac{m _{D}}{m _{D} + m _{α}} = 5.40 \times \frac{206}{210} = 5.30 MeV,$ $K_{D} = Q \frac{m _{α}}{m _{D} + m _{α}} = 5.40 \times \frac{4}{210} = 0.103 MeV .$ Check: $5.30 + 0.103 = 5.40 = Q$ . ✓ Consistent with L2.3.

Level 4 — Synthesis

L4.1 — Positron-emission correction

$^{?}$ ... consider a $β^{+}$ decay $X \to Y + e^{+} + ν$ where the atomic masses satisfy $m_{X} - m_{Y} = 0.00220$ u. Find the true $Q$ , remembering the electron-count mismatch. (Take $m_{e} = 0.000549$ u; neutrino massless.)

Recall Solution

In $β^{+}$ decay the nucleus loses a proton, gaining an electron in the daughter's atom count mismatch — using atomic masses you must subtract $2 m_{e} c^{2}$ : $Q = [(m_{X} - m_{Y}) - 2 m_{e}] c^{2} .$ $Δ m = 0.00220 - 2 (0.000549) = 0.00220 - 0.001098 = 0.001102 u .$ $Q = 0.001102 \times 931.5 = 1.03 MeV .$ Why $2 m_{e}$ ? Atomic-mass tables include $Z$ electrons per atom. In $β^{+}$ the daughter has one fewer proton (one fewer atomic electron), AND a positron is created — two electron masses are unaccounted for, so they must be removed by hand. This is the caveat flagged in Beta decay.

L4.2 — Binding-energy route to Q

Compute $Q$ for D–T fusion $^{2}$ H $+^{3}$ H $\to^{4}$ He $+$ n using binding energies instead of masses. Given total binding energies: $B (^{2} H) = 2.224$ MeV, $B (^{3} H) = 8.482$ MeV, $B (^{4} He) = 28.296$ MeV, $B (n) = 0$ .

Recall Solution

$Q = (final total B) - (initial total B)$ — a reaction releases energy when products are more tightly bound (Binding energy and mass defect). $Q = B (^{4} He) + B (n) - B (^{2} H) - B (^{3} H)$ $= 28.296 + 0 - 2.224 - 8.482 = 17.59 MeV .$ This matches the mass-table answer ( $\approx 17.6$ MeV) — two doors to the same room.

Level 5 — Mastery

L5.1 — Fission energy budget

Estimate $Q$ for the neutron-induced fission step forming the compound nucleus $^{236}$ U: $^{235} U + n \to^{236} U^{*} .$ Then comment on why full fission releases $\sim 200$ MeV even though this capture step is small.

Recall Solution

Capture step: Before: $235.043930 + 1.008665 = 236.052595$ u. After: $236.045568$ u. $Δ m = 236.052595 - 236.045568 = 0.007027$ u. $Q = 0.007027 \times 931.5 = 6.55 MeV$ — the excitation energy that lets $^{236}$ U $^{*}$ deform and split. Why $\sim 200$ MeV overall: the huge release comes from the actual splitting into two mid-mass fragments, which sit much higher on the binding-energy-per-nucleon curve than $^{236}$ U (Nuclear fission). Rearranging $\sim 236$ nucleons to gain $\sim 0.85$ MeV/nucleon in binding gives $\sim 200$ MeV. The capture $Q$ here is only the trigger, not the payoff.

L5.2 — Full synthesis: reaction, sign, threshold, split

Consider $^{9}$ Be $(α, n)^{12}$ C, i.e. $_{2}^{4} He +_{4}^{9} Be \to_{6}^{12} C +_{0}^{1} n .$ (a) Find $Q$ . (b) Is it exo- or endothermic? (c) If endothermic, give threshold energy; if exothermic, explain why no threshold. (d) State which conservation laws you invoked.

Recall Solution

(a) Electron check: $Z$ left $= 2 + 4 = 6$ ; right $= 6 + 0 = 6$ . Balanced. Before: $4.002603 + 9.012182 = 13.014785$ u. After: $12.000000 + 1.008665 = 13.008665$ u. $Δ m = 13.014785 - 13.008665 = 0.006120$ u. $Q = 0.006120 \times 931.5 = 5.70 MeV$ . (b) $Q > 0$ ⇒ exothermic (this is the classic lab neutron source). (c) No threshold: energy is released, so even a slow $α$ that surmounts the Coulomb barrier can react. Threshold energy is a purely endothermic concept ( $K_{th} = ∣ Q ∣ (1 + m_{a} / m_{X})$ only makes sense when $Q < 0$ ). (d) We used: conservation of charge and nucleon number (to balance the equation), conservation of mass-energy (to get $Q$ ), and implicitly momentum (needed only if we split the KE, as in L3.2).