Worked examples — Nuclear reactions — Q-value, cross-section

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The scenario matrix

Every problem this topic can throw is one of the cells below. Each worked example is tagged with the cell it lands in, so by the end every row is covered.

Cell	What makes it different	Danger / trap	Example
A. Exoergic ($Q>0$)	Mass is lost, energy comes out for free; threshold is zero	Sign slip	Ex 1
B. Endoergic ($Q<0$)	Mass is gained, energy must be paid	Confuse $Q$ and $E_{th}$	Ex 2
C. Threshold	Need the minimum projectile energy	Forgetting the CM factor	Ex 3
D. Degenerate: $Q\approx 0$	Almost no mass change	Rounding kills the answer	Ex 4
E. Binding-energy route	Masses not given, only $B$ values	Wrong sign convention	Ex 5
F. Huge $\sigma$ (thousands of barns)	Beam dies almost instantly	barn↔cm² conversion	Ex 6
G. Tiny $\sigma$ / thick slab	Beam barely attenuates	Approximating $e^{-x}\approx 1-x$	Ex 7
H. Real-world word problem	Reactor flux → reaction rate	Units of $\varphi$, $N$	Ex 8
I. Exam twist	Combine $Q$ + threshold + atomic-mass care	Mixing atomic/nuclear masses	Ex 9

Two constants used everywhere:

The figure above is our map: the horizontal axis is the sign of $Q$ (left = endoergic, right = exoergic), the vertical axis is the size of $\sigma$ (up = huge, down = tiny). Each dot is labelled with its example number and cell letter (e.g. "Ex1 A"), so you can point at any corner and jump straight to the matching worked example below. Keep glancing back to see which corner you are standing in.

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Cell A — Exoergic reaction ($Q>0$)

1. Add the reactant masses. $m_a+m_X=1.008665+6.015123=7.023788\text{u}$. Why this step? The Q-value compares total mass before against total mass after; we need the "before" pile first. Units so far: u (atomic mass units).
2. Add the product masses. $m_Y+m_b=4.002603+3.016049=7.018652\text{u}$. Why this step? This is the "after" pile — mass that survives as rest mass in the products. Still u.
3. Subtract to get the mass change. $\Delta m=7.023788-7.018652=+0.005136\text{u}$. Why this step? $Q=\Delta (\text{rest mass})c^2$. A positive $\Delta m$ means mass disappeared, so energy is released. Units: u.
4. Multiply by 931.5 MeV/u. $Q=0.005136\text{u}\times 931.5\frac{\text{MeV}}{\text{u}}=+4.78\text{MeV}$. Why this step? This is $E=m{c}^2$ in bookkeeping units — the u cancels, leaving MeV of released energy.

Verify: $Q>0$ matches our forecast (thermal neutrons do trigger it — this is a real neutron-detector reaction), and consistently $E_{th}=0$. Final unit: MeV, an energy, as required. ✓

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Cell B — Endoergic reaction ($Q<0$)

1. Reactant sum. $m_a+m_X=15.994915+1.008665=17.003580\text{u}$. Why? The "before" rest mass. Units: u.
2. Product sum. $m_Y+m_b=4.002603+13.003355=17.005958\text{u}$. Why? The "after" rest mass. Units: u.
3. Mass change. $\Delta m=17.003580-17.005958=-0.002378\text{u}$. Why? Products are heavier → mass was gained → energy had to be poured in. Negative $\Delta m$ ⇒ negative $Q$. Units: u.
4. Energy. $Q=-0.002378\text{u}\times 931.5\frac{\text{MeV}}{\text{u}}=-2.22\text{MeV}$. Why? Same $E=m{c}^2$ conversion; u cancels, leaving MeV. The minus sign is the whole physics point.

Verify: $Q<0$ matches the forecast (energy-absorbing); final units MeV ✓. Note: this is not the minimum energy needed to run it — that is the threshold, handled next. ✓

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Cell C — Threshold energy (with the momentum-tax picture)

1. Confirm it is endoergic. $Q=-2.22\text{MeV}<0$, so a threshold exists (an exoergic reaction would give $E_{th}=0$, per Cell A). Why this step? Applying the threshold formula to an exoergic reaction is meaningless — always check the sign first.
2. Apply the derived formula. $E_{th}=|Q|\frac{m_a+m_X}{m_X}$, with $m_a=1.008665$ (neutron), $m_X=15.994915$ (${}^{16}\text{O}$). Why? The bracket $\frac{m_a+m_X}{m_X}$ is exactly the drift-tax factor derived in the figure above — dimensionless (u/u).
3. Plug in. $E_{th}=2.22\text{MeV}\times \frac{1.008665+15.994915}{15.994915}=2.22\text{MeV}\times 1.06306$. Why? The ratio $>1$ always, so the answer must exceed $|Q|$. The bracket is a pure number, so units stay MeV.
4. Compute. $E_{th}=2.22\text{MeV}\times 1.06306=2.36\text{MeV}$.

Verify: $2.36>2.22=|Q|$ ✓ (units MeV throughout) — the tax is small (~6%) because the target is much heavier than the projectile. Limiting check: a very heavy target ($m_X\to \infty$) gives bracket $\to 1$, so $E_{th}\to |Q|$ — the massive target soaks up the recoil for free. ✓

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Cell D — Degenerate case: $Q\approx 0$

1. Full-precision reactant sum. $m_a+m_X=14.003074+4.002603=18.005677\text{u}$. Why? When the answer is a tiny difference of big numbers, every digit matters. Units: u.
2. Full-precision product sum. $m_Y+m_b=16.999132+1.007825=18.006957\text{u}$. Why? Same reason — losing one digit here changes $Q$ by hundreds of keV. Units: u.
3. Mass change. $\Delta m=18.005677-18.006957=-0.001280\text{u}$. Why? Notice it is negative and only ~$10^{-3}\text{u}$ — a whisper of mass. Endoergic, barely. Units: u.
4. Energy. $Q=-0.001280\text{u}\times 931.5\frac{\text{MeV}}{\text{u}}=-1.19\text{MeV}$. Why? Even this whisper of mass is worth over an MeV — that is how enormous $c^2$ is. u cancels ⇒ MeV.

Verify: matches the parent note's Rutherford example ($-1.19$ MeV, units MeV ✓). Trap demo: if you had rounded each mass to 3 decimals ($14.003,4.003,16.999,1.008$) you'd get $\Delta m=18.006-18.007=-0.001\text{u}\Rightarrow -0.93\text{MeV}$ — off by 22%! Lesson: in the $Q\approx 0$ cell, never round until the very end. ✓

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Cell E — Binding-energy route (no masses given)

1. Write $Q$ in binding-energy form. $Q=(B_Y+B_b)-(B_X+B_a)$, justified by the mass-defect sketch above. Why? When masses aren't handed to you, binding energy is the missing-mass information already in energy units — MeV throughout, no 931.5 conversion needed.
2. Sum product binding. $B_Y+B_b=28.296+0=28.296\text{MeV}$. Why? ${}^4\text{He}$ is extremely tightly bound; the free neutron contributes nothing.
3. Sum reactant binding. $B_X+B_a=8.482+2.224=10.706\text{MeV}$. Why? Deuterium and tritium are loosely bound — small $B$.
4. Subtract. $Q=28.296-10.706=+17.59\text{MeV}$. Why? Products far more tightly bound ⇒ big positive $Q$. This is why fusion powers the Sun.

Verify: $Q>0$ (exoergic), matching the forecast; the standard textbook D–T value is $17.6$ MeV, units MeV ✓. Sign check: products' $B$ came first (with $+$), reactants' $B$ subtracted — consistent with "more bound products ⇒ energy out." ✓

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Cell F — Huge cross-section (beam dies fast)

1. Convert barns to cm². $\sigma =755\times 10^{-24}{\text{cm}}^2=7.55\times 10^{-22}{\text{cm}}^2$. Why? $n$ is in ${\text{cm}}^{-3}$; to multiply cleanly, put $\sigma$ in ${\text{cm}}^2$ so units combine to ${\text{cm}}^{-1}$.
2. Macroscopic cross-section. $\Sigma =n\sigma =1.3\times 10^{23}{\text{cm}}^{-3}\times 7.55\times 10^{-22}{\text{cm}}^2=98.2{\text{cm}}^{-1}$. Why? $\Sigma$ bundles "how many targets per cm" into one number — it is the inverse of the mean free path. Units: ${\text{cm}}^{-3}\cdot {\text{cm}}^2={\text{cm}}^{-1}$ ✓.
3. Half-value from the attenuation law. $\frac{1}{2}=e^{-\Sigma x}\Rightarrow \Sigma x=\ln 2$. Why? $I=I_0{e}^{-\Sigma x}$; halving means the exponential equals $\frac{1}{2}$, and $\ln$ undoes the exponential. $\Sigma x$ must be dimensionless — it is (${\text{cm}}^{-1}\cdot \text{cm}$).
4. Solve. $x=\frac{\ln 2}{\Sigma }=\frac{0.693}{98.2{\text{cm}}^{-1}}=7.06\times 10^{-3}\text{cm}=70.6\mu \text{m}$.

Verify: micrometres — matches the "huge $\sigma$ ⇒ tiny half-thickness ⇒ great absorber" logic. Units: $\frac{1}{{\text{cm}}^{-1}}=\text{cm}$ ✓. This is why boron is used in control rods and shielding. ✓

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Cell G — Tiny cross-section (beam barely attenuates)

1. Macroscopic cross-section. $\Sigma =n\sigma =2.7\times 10^{19}{\text{cm}}^{-3}\times 0.33\times 10^{-24}{\text{cm}}^2=8.91\times 10^{-6}{\text{cm}}^{-1}$. Why? Same $\Sigma =n\sigma$ machinery; here it comes out very small. Units: ${\text{cm}}^{-1}$ ✓.
2. Optical depth. $\Sigma x=8.91\times 10^{-6}{\text{cm}}^{-1}\times 10\text{cm}=8.91\times 10^{-5}$ (dimensionless). Why? The dimensionless product $\Sigma x$ is the exponent; if it is $\ll 1$ we are in the thin-target regime.
3. Use the small-$x$ approximation. Since $\Sigma x\ll 1$, $e^{-\Sigma x}\approx 1-\Sigma x$, so the removed fraction is $\frac{I_0-I}{I_0}\approx \Sigma x$ (dimensionless, a pure fraction). Why? When almost nothing is absorbed, the full exponential is overkill; $1-e^{-y}\approx y$ for tiny $y$, saving arithmetic and avoiding calculator round-off.
4. Answer. Fraction reacting $\approx 8.91\times 10^{-5}\approx 0.0089\%$.

Verify: essentially transparent — matches the "tiny $\sigma$, thin gas" forecast. Cross-check with the exact law: $1-e^{-8.91\times 10^{-5}}=8.9096\times 10^{-5}$, agreeing with the approximation to better than 0.01%. Both are dimensionless fractions ✓. ✓

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Cell H — Real-world word problem (reactor flux)

1. Recall the rate formula. $R_{tot}=N\sigma \varphi$. Why? Rate = (number of targets) × (effective area each) × (flux hitting them). Each factor answers "how often a neutron finds a gold nucleus."
2. Convert $\sigma$ to cm². $\sigma =98.7\times 10^{-24}=9.87\times 10^{-23}{\text{cm}}^2$. Why? $\varphi$ is per ${\text{cm}}^2$, so $\sigma$ must be in ${\text{cm}}^2$ for the areas to cancel.
3. Multiply. $R_{tot}=5.0\times 10^{22}[\text{atoms}]\times 9.87\times 10^{-23}[{\text{cm}}^2]\times 2.0\times 10^{13}[{\text{cm}}^{-2}{\text{s}}^{-1}]$. Why? Straight substitution now that units are consistent. Track units: $\text{atoms}\cdot {\text{cm}}^2\cdot {\text{cm}}^{-2}{\text{s}}^{-1}={\text{s}}^{-1}$.
4. Evaluate. $5.0\times 9.87\times 2.0=98.7$; powers: $10^{22-23+13}=10^{12}$. So $R_{tot}=98.7\times 10^{12}=9.87\times 10^{13}\text{nuclei/s}$.

Verify: Units: $[\text{atoms}]\cdot [{\text{cm}}^2]\cdot [{\text{cm}}^{-2}{\text{s}}^{-1}]={\text{s}}^{-1}$ ✓ (a rate, as required). This "$N\sigma \varphi$" is exactly how neutron-activation analysis and isotope production rates are computed. ✓

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Cell I — Exam twist (combine everything, atomic-mass trap)

1. Reactant sum (atomic masses). $m_a+m_X=7.016004+1.007825=8.023829\text{u}$. Why? Use atomic masses consistently. Here the proton is written as ${}^1\text{H}$ (with its electron), and ${}^7\text{Li}$ and ${}^7\text{Be}$ are also atomic — the electron counts balance because total $Z$ before ($3+1=4$) equals total $Z$ after ($4+0=4$). Mixing atomic and nuclear masses is the trap. Units: u.
2. Product sum. $m_Y+m_b=7.016929+1.008665=8.025594\text{u}$. Why? The "after" pile, same atomic-mass convention. Units: u.
3. Mass change and $Q$. $\Delta m=8.023829-8.025594=-0.001765\text{u}$, so $Q=-0.001765\text{u}\times 931.5\frac{\text{MeV}}{\text{u}}=-1.64\text{MeV}$. Why? Negative ⇒ endoergic ⇒ there is a threshold; that flows straight into part (ii). u cancels ⇒ MeV.
4. Threshold. Since $Q<0$, apply $E_{th}=|Q|\frac{m_a+m_X}{m_X}=1.64\text{MeV}\times \frac{1.007825+7.016004}{7.016004}=1.64\text{MeV}\times 1.14364=1.88\text{MeV}$. Why? Endoergic ⇒ apply the CM drift-tax factor derived in Cell C. Here projectile and target have similar mass, so the tax (~14%) is bigger than in Ex 3. Bracket is dimensionless ⇒ MeV stays.

Verify: $Q=-1.64$ MeV is the accepted value for this classic $(p,n)$ neutron-producing reaction; $E_{th}=1.88>|Q|=1.64$ ✓ (both MeV). Electron-count sanity check: $Z$ balanced ⇒ atomic masses were legitimate ⇒ no double-counted electrons. ✓

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Recall — did every cell get covered?

Recall Map each example to its matrix cell

Ex 1 → A (exoergic) ::: ${}^6$Li$(n,\alpha )$, $Q=+4.78$ MeV, threshold $=0$. Ex 2 → B (endoergic) ::: ${}^{16}$O$(n,\alpha )$, $Q=-2.22$ MeV. Ex 3 → C (threshold) ::: $E_{th}=2.36$ MeV $>|Q|$. Ex 4 → D (degenerate $Q\approx 0$) ::: $-1.19$ MeV; keep all decimals! Ex 5 → E (binding-energy route) ::: D–T fusion, $Q=+17.59$ MeV. Ex 6 → F (huge $\sigma$) ::: boron, half-thickness $\approx 71\mu$m. Ex 7 → G (tiny $\sigma$) ::: hydrogen gas, fraction $\approx 8.9\times 10^{-5}$. Ex 8 → H (word problem) ::: gold activation, $9.87\times 10^{13}$ nuclei/s. Ex 9 → I (exam twist) ::: ${}^7$Li$(p,n)$, $Q=-1.64$ MeV, $E_{th}=1.88$ MeV.

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Back to the parent topic · related: Nuclear fission, Nuclear fusion, Conservation laws in collisions, Neutron flux & reactor physics.