Worked examples — Nuclear reactions — Q-value calculation

Before anything else, three quick reminders that EVERY example below leans on:

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The scenario matrix

Every Q-value problem you can be handed falls into one of these cells. The examples that follow each carry a [Cell N] tag so you can see the whole grid gets covered.

Cell	Scenario class	What is tricky about it	Example
1	$Q>0$, exothermic, mass data	straightforward sign check	Ex 1 (fusion)
2	$Q<0$, endothermic + threshold	products can't be at rest	Ex 2
3	$Q$ from kinetic energies only (no masses)	use the kinetic face	Ex 3
4	Splitting $Q$ between fragments (momentum)	daughter recoils, share by mass	Ex 4
5	$Q=0$ / degenerate (elastic, no rest-mass change)	limiting case, $\Delta m=0$	Ex 5
6	Positron (${\beta }^{+}$) decay — atomic-mass trap	extra $2m_e{c}^2$ correction	Ex 6
7	${\beta }^{-}$ decay — atomic masses cancel cleanly	contrast with cell 6	Ex 7
8	Limiting behaviour of threshold ($m_X\to \infty$, $m_X\to 0$)	sanity-check the formula	Ex 8
9	Exam twist / real-world: reactor fission energy per event → per kg	scale-up word problem	Ex 9

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Ex 1 — Exothermic, mass data [Cell 1]

Forecast: two very tightly-bound helium nuclei come out — guess: strongly positive $Q$, several MeV. Write your guess down.

1. Before: $7.016004+1.007825=8.023829$ u. Why this step? Sum the reactant rest masses — that is the total mass we start with.
2. After: $4.002603+4.002603=8.005206$ u. Why this step? Sum the product rest masses.
3. $\Delta m$: $8.023829-8.005206=0.018623$ u. Why this step? The vanished mass. Positive ⇒ mass disappeared ⇒ energy comes out.
4. Convert: $Q=0.018623\times 931.5\approx 17.35\mathbf{MeV}$. Why this step? $\Delta m$ in u times the unit-bridge $931.5$ MeV/u gives energy.

Verify: Sign is $+$ (products lighter) → exothermic, matching the forecast. Units: $\text{u}\times (\text{MeV}/\text{u})=\text{MeV}$. ✓ Two alphas are among the most bound nuclei, so a large release is physically sensible.

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Ex 2 — Endothermic + threshold [Cell 2]

Forecast: knocking a lighter fragment loose usually costs energy → guess $Q<0$, and threshold slightly bigger than $|Q|$.

Matching the template $a+X\to Y+b$: here the neutron is the projectile $a$ (so $m_a=m_n=1.008665$ u), oxygen is the target $X$ (so $m_X=15.994915$ u), ${}^{13}$C is $Y$ and ${}^4$He is $b$.

1. Before: $15.994915+1.008665=17.003580$ u. Why? Reactant masses $m_a+m_X$.
2. After: $13.003355+4.002603=17.005958$ u. Why? Product masses $m_Y+m_b$.
3. $\Delta m$: $17.003580-17.005958=-0.002378$ u. Why negative? Products are heavier — mass had to be created, so energy is absorbed.
4. $Q$: $-0.002378\times 931.5=-2.215\mathbf{MeV}$. Endothermic.
5. Threshold — full algebraic derivation. Recall the symbols: $m_a$ = projectile mass, $m_X$ = (stationary) target mass, $K_a$ = projectile kinetic energy in the lab, $K_{\text{th}}$ = the minimum value of $K_a$ that lets the reaction happen. We build the formula in four honest steps.
   • (i) Momentum in. The projectile carries lab momentum $p_a$; the target is at rest, so total momentum is $p_a$. Non-relativistically $K_a=\frac{p_a^2}{2m_a}$, so $p_a=\sqrt{2m_a{K}_a}$.
   • (ii) The products must keep moving. Total momentum can't become zero, so the products cannot all be at rest. At the threshold (bare minimum energy) the products are created just barely moving together — a single lump of mass $(m_a+m_X)$ carrying the whole momentum $p_a$. Its kinetic energy is the unavoidable centre-of-mass energy $K_{\text{cm}}=\frac{p_a^2}{2(m_a+m_X)}=\frac{2m_a{K}_a}{2(m_a+m_X)}=\frac{m_a}{m_a+m_X}{K}_a.$
   • (iii) Energy that is actually available. Only the kinetic energy above this bulk-motion floor can be spent on the reaction's mass deficit $|Q|$. So the available energy is $K_a-K_{\text{cm}}=K_a\left(1-\frac{m_a}{m_a+m_X}\right)=K_a\cdot \frac{m_X}{m_a+m_X}.$
   • (iv) Set available energy $=|Q|$ and solve for $K_a$. At threshold $K_a=K_{\text{th}}$: $K_{\text{th}}\cdot \frac{m_X}{m_a+m_X}=|Q|⟹K_{\text{th}}=|Q|\frac{m_a+m_X}{m_X}=|Q|\left(1+\frac{m_a}{m_X}\right).$ Now plug numbers ($m_a=1.008665$, $m_X=15.994915$, $|Q|=2.215$ MeV): $K_{\text{th}}=2.215\left(1+\frac{1.008665}{15.994915}\right)=2.215\times 1.06306=2.355\mathbf{MeV}.$ Why this step? The bracket is exactly the extra energy stuck in unavoidable bulk motion; you must supply it on top of $|Q|$.

Verify: $K_{\text{th}}>|Q|$ ✓ (as it must be). The correction factor $1.063$ is close to $1$ because the projectile is light compared to the target — see Ex 8 for the limits.

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Ex 3 — Q straight from kinetic energies [Cell 3]

Forecast: parent is at rest so nothing goes in; $Q$ is just the total that comes out. Guess $\approx 4.87$ MeV.

1. $K_{\text{initial}}$: parent at rest $\Rightarrow K_{\text{initial}}=0$. Why? Nothing was moving before.
2. $K_{\text{final}}$: $4.78+0.086=4.866$ MeV. Why? Total kinetic energy of ALL fragments — the daughter recoil counts too.
3. $Q$: $Q=K_{\text{final}}-K_{\text{initial}}=4.866-0=4.866\mathbf{MeV}$. Why this step? The kinetic face of $Q$: vanished rest mass reappears as fragment kinetic energy. We never needed the masses.

Verify: $Q>0$ (spontaneous decay is always exothermic — otherwise it wouldn't happen). Forgetting the $0.086$ MeV recoil would under-count $Q$ — that is exactly the common mistake this example is built to prevent.

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Ex 4 — Splitting Q between fragments [Cell 4]

Figure s01 (referenced below): it shows the whole event on a single horizontal axis (position along the recoil line). The parent sits at rest at the centre (black dot, momentum $=0$); it then splits into the fast light alpha (red, long arrow, pointing right) and the slow heavy radon daughter (black, short arrow, pointing left). The two labelled arrows are the momenta $p_{\alpha }$ and $p_{\text{Rn}}$ — drawn equal in length but opposite in direction, the key visual of equal and opposite momenta. The alpha ends far to the right because it moves faster.

Forecast: the light alpha should sprint off fast; the heavy daughter barely creeps. Guess: alpha gets almost all of $Q$.

1. Momentum conservation. Parent at rest ⇒ total momentum $=0$, so the two fragments fly apart with equal and opposite momenta (look at the red alpha arrow and the black recoil arrow in Figure s01 — same length, opposite way): $p_{\alpha }=p_{\text{Rn}}\equiv p.$ Why this step? Nothing was moving before; momentum can't appear from nowhere, so the momenta must cancel.
2. Write kinetic energy via momentum. For a non-relativistic fragment (speed $\ll c$, which holds here since alpha/daughter energies of a few MeV give speeds of only a few percent of $c$), kinetic energy is $K=\frac{p^2}{2m}$. Why this step? Both fragments share the SAME $p$, so writing $K$ in terms of $p$ lets us compare them directly. (For MeV-scale nuclear fragments this classical formula is accurate to better than $1\%$; only at hundreds of MeV would we need the relativistic version.)
3. Share ratio. $\frac{K_{\alpha }}{K_{\text{Rn}}}=\frac{p^2/2m_{\alpha }}{p^2/2m_{\text{Rn}}}=\frac{m_{\text{Rn}}}{m_{\alpha }}.$ Why this step? The lighter fragment gets the bigger share — energy is split inversely to mass.
4. Solve for $K_{\alpha }$. Since $Q=K_{\alpha }+K_{\text{Rn}}$: $K_{\alpha }=Q\cdot \frac{m_{\text{Rn}}}{m_{\text{Rn}}+m_{\alpha }}=4.87\times \frac{222}{226}=4.784\mathbf{MeV}.$ Why this step? This is the exact split rule quoted in the parent note's mistake box, now derived.

Verify: $K_{\text{Rn}}=Q-K_{\alpha }=4.87-4.784=0.086$ MeV — the very recoil energy given in Ex 3! ✓ The two examples are consistent. Alpha keeps $98.2\%$ of $Q$, matching the forecast.

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Ex 5 — Degenerate case: $Q=0$ (elastic) [Cell 5]

Forecast: the "before pile" and the "after pile" are the same particles — nothing was created or destroyed. Guess $Q=0$.

1. Before: $m_n+m_{\text{C}}$. After: $m_{\text{C}}+m_n$. Why? List them — they are literally identical.
2. $\Delta m$: $(m_n+m_{\text{C}})-(m_{\text{C}}+m_n)=0$. Why? Same terms cancel term-by-term.
3. $Q=0\times 931.5=0$. Why this step? No rest-mass change ⇒ no rest-mass energy released or absorbed. Total kinetic energy is merely redistributed between the two, not changed.

Verify: This is the limiting middle of the sign axis: $Q>0$ (Ex 1), $Q<0$ (Ex 2), $Q=0$ here. It matches the parent's third bullet: "$Q=0$ → elastic." Kinetic energy is conserved but repackaged — the whole point of using carbon to moderate (slow down) reactor neutrons.

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Ex 6 — Positron decay: the atomic-mass trap [Cell 6]

Forecast: the correction $2m_e{c}^2=2\times 0.511=1.022$ MeV is a real chunk. If you forget it your $Q$ comes out about $1$ MeV too big.

1. Nuclear-mass truth. In terms of nuclear masses $M$, $Q=[M_{\text{C}}-M_{\text{B}}-m_e]c^2.$ Why this step? A positron ($+m_e$) is genuinely created, and the neutrino is massless — so we lose one electron-mass of rest energy explicitly.
2. Convert nuclear to atomic masses. Atomic mass $=$ nuclear mass $+Z{m}_e$ (ignoring electron binding), where $Z$ (the atomic number) is the proton/electron count. So $M_{\text{C}}=m{(}^{11}\text{C})-6m_e$ (carbon has $Z=6$) and $M_{\text{B}}=m{(}^{11}\text{B})-5m_e$ (boron has $Z=5$). Why this step? We were handed atomic masses; substitute to use them.
3. Substitute and simplify. $Q=[(m_{\text{C}}-6m_e)-(m_{\text{B}}-5m_e)-m_e]c^2=[m_{\text{C}}-m_{\text{B}}-2m_e]c^2.$ Why this step? The electron bookkeeping collapses to a single $-2m_e$ correction — the extra term the parent warned about.
4. Numbers. $m_{\text{C}}-m_{\text{B}}=11.011434-11.009305=0.002129$ u. Subtract $2m_e=0.001098$ u: $\Delta m=0.001031$ u. $Q=0.001031\times 931.5=0.960\mathbf{MeV}.$

Verify: Known experimental $Q$ for ${}^{11}$C ${\beta }^{+}$ decay is $\approx 0.96$ MeV ✓. If you'd (wrongly) skipped the $-2m_e$: $0.002129\times 931.5=1.983$ MeV — nearly double, and off by exactly $1.022$ MeV, the $2m_e{c}^2$ you dropped. That is the trap.

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Ex 7 — ${\beta }^{-}$ decay: atomic masses cancel cleanly [Cell 7]

Forecast: contrast with Ex 6 — here $Z$ goes up by one, and the emitted electron's mass is already accounted for inside the daughter's atomic mass. Guess: no correction term.

1. Nuclear truth: $Q=[M_{\text{C}}-M_{\text{N}}-m_e]c^2$ ($-m_e$ for the created electron). Why? One electron is genuinely emitted.
2. Atomic substitution: $M_{\text{C}}=m_{\text{C}}-6m_e$ (carbon $Z=6$), $M_{\text{N}}=m_{\text{N}}-7m_e$ (nitrogen $Z=7$). Why? Convert to the atomic masses we were given.
3. Simplify: $Q=[(m_{\text{C}}-6m_e)-(m_{\text{N}}-7m_e)-m_e]c^2=[m_{\text{C}}-m_{\text{N}}]c^2.$ Why this step? The $-6m_e$, $+7m_e$, and $-m_e$ exactly cancel — the emitted electron slots neatly into the daughter's electron cloud. No correction survives, unlike Ex 6.
4. Numbers: $\Delta m=14.003242-14.003074=0.000168$ u. $Q=0.000168\times 931.5=0.156\mathbf{MeV}.$ Why this step? $\Delta m$ in u times $931.5$ MeV/u gives the released energy.

Verify: Textbook endpoint energy of the ${}^{14}$C beta spectrum is $0.156$ MeV ✓. Notice the symmetry: ${\beta }^{-}$ needs no correction (Ex 7), ${\beta }^{+}$ needs $-2m_e{c}^2$ (Ex 6). Same idea, opposite bookkeeping.

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Ex 8 — Limiting behaviour of the threshold [Cell 8]

Figure s02 (referenced below) plots the threshold formula's key ratio $K_{\text{th}}/|Q|$ (vertical axis) against the mass ratio $m_a/m_X$ = projectile-over-target (horizontal axis). The red curve is the formula $K_{\text{th}}/|Q|=1+m_a/m_X$; the dashed black line at height $1$ is the floor it approaches for heavy targets; the black dot marks the equal-mass case $m_a=m_X$ where the ratio is exactly $2$.

Forecast: hitting a brick wall (heavy target) should waste almost no energy on recoil; hitting a feather (light target) should waste almost everything.

1. Heavy target, $m_X\to \infty$: $\frac{m_a}{m_X}\to 0$, so $K_{\text{th}}\to |Q|$. Why this step? A massive target barely recoils, so almost none of the incoming energy is lost to bulk motion — you need just $|Q|$. This is why Ex 2's factor was only $1.063$: oxygen is $16\times$ heavier than the neutron. On the graph the red curve flattens onto the dashed floor.
2. Light target, $m_X\to 0$: $\frac{m_a}{m_X}\to \infty$, so $K_{\text{th}}\to \infty$. Why this step? A tiny target flies off carrying enormous momentum; almost all incoming energy becomes wasted bulk motion, so the threshold blows up. On the graph the red curve shoots upward on the right.
3. Middle check, $m_X=m_a$: $K_{\text{th}}=2|Q|$. Why this step? Equal masses share the recoil symmetrically — exactly half the energy is unavailable, so you need twice $|Q|$. The curve passes cleanly through the marked black dot.

Verify: All three limits are monotonic and sensible: heavier target ⇒ cheaper threshold. The formula never dips below $|Q|$ (the bracket is always $\ge 1$ for positive masses) ✓, guaranteeing $K_{\text{th}}\ge |Q|$ always — the parent note's promise.

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Ex 9 — Exam twist / real-world: fission energy per kilogram [Cell 9]

Forecast: nuclear energy densities are famously "a million times" chemical. Guess the answer lands around $\sim 10^{13}$–$10^{14}$ J from 1 kg, and the ratio to coal near a million.

1. Energy per fission in joules: $200\text{MeV}\times 1.602\times 10^{-13}\text{J/MeV}=3.204\times 10^{-11}$ J. Why this step? Convert the Q-value to SI joules so we can scale it up. (The MeV itself already came from $\Delta m{c}^2$, so $c^2$ is baked in.)
2. Number of nuclei in 1 kg: $N=\frac{1000\text{g}}{235\text{g/mol}}\times N_A=4.2553\text{mol}\times 6.022\times 10^{23}=2.5626\times 10^{24}$ nuclei. Why this step? Total energy $=$ (energy per fission) $\times$ (how many nuclei there are), so we first count the nuclei.
3. Total energy: $E=N\times (3.204\times 10^{-11}\text{J})=2.5626\times 10^{24}\times 3.204\times 10^{-11}=8.21\times 10^{13}\mathbf{J}$. Why this step? Straight multiplication — the payoff.
4. Compare to coal: $\frac{8.21\times 10^{13}\text{J}}{3\times 10^7\text{J}}\approx 2.7\times 10^6$. Why this step? The whole point: 1 kg of nuclear fuel releases about 2.7 million times more energy than 1 kg of coal.

Verify: Units: $\text{nuclei}\times \text{J/nucleus}=\text{J}$ ✓. Order of magnitude $10^{13}$ J matches the forecast, and the ratio $\sim 10^6$ matches "a million times." $8.2\times 10^{13}$ J $\approx 23$ GWh — roughly a small power plant's daily output from a single kilogram, the famous headline number. ✓

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Prerequisites & neighbours: Mass-energy equivalence E=mc^2 · Binding energy and mass defect · Conservation of momentum in decays · Alpha decay · Beta decay · Threshold energy in particle reactions · Nuclear fission · Nuclear fusion