Radioactive decay — alpha, beta, gamma — mechanisms: Worked examples

The scenario matrix

Every decay problem is one of these cells. Think of the columns as which decay and the rows as what makes it tricky.

Cell	Case class	What is nasty about it	Example
A1	Alpha — energy split	daughter recoils, sharing $Q$	Ex 1
A2	Alpha — limiting mass	what happens as $A \to \infty$ ?	Ex 2
B1	Beta-minus — endpoint	continuous spectrum, max energy	Ex 3
B2	Beta-plus vs capture — threshold	$Q$ can be zero or negative (degenerate)	Ex 4
C1	Gamma — recoil correction	photon has momentum too	Ex 5
D1	Chain — track $Z, A$	multiple decays, bookkeeping	Ex 6
E1	Real-world word problem	smoke detector ( $^{241}$ Am)	Ex 7
E2	Exam twist — spot the impossible	a decay that cannot happen ( $Q < 0$ )	Ex 8

We use these constants throughout. $1 u = 931.5 MeV / c^{2}$ (this is just the energy hidden in one atomic mass unit, from $E = m c^{2}$ ). The electron rest energy is $m_{e} c^{2} = 0.511 MeV$ .

Definition The symbols, from zero

$A$ = mass number = total count of protons + neutrons (the "chunk count").
$Z$ = atomic number = number of protons (fixes which element it is).
$Q$ = energy released, in MeV. Positive $Q$ ⇒ decay is allowed.
$K E$ = kinetic energy = energy of motion, $K E = \frac{1}{2} m v^{2}$ at low speed.
$p$ = momentum = $m v$ (a measure of "quantity of motion"). Conserved: total before = total after.
$c$ = speed of light ( $3 \times 1 0^{8} m/s$ ). It appears only through $E = m c^{2}$ (mass ↔ energy) and, for massless photons, $p = E / c$ (a photon's momentum is its energy divided by $c$ ). Whenever you see $c$ , read "the exchange rate between mass and energy".
$m_{α}$ = mass of the alpha particle; $m_{Y}$ = mass of the daughter nucleus (we call any daughter $Y$ , as in $X \to Y + α$ ); $M$ = a nucleus's total mass in general. All three are the same kind of thing — a rest mass — just for different objects.

Ex 1 · Cell A1 — Alpha: who gets the energy?

Step 1 — Draw the split (why: to see momentum conservation). The uranium starts at rest, so the total momentum is zero. After the decay the alpha flies one way and the thorium recoils the opposite way with equal and opposite momentum. In the figure below, the orange arrow is the alpha's momentum $p_{α}$ and the teal arrow is the thorium's momentum $p_{T h}$ . Both arrows have the same length (equal size of momentum) but point in opposite directions — that is exactly what "total momentum stays zero" looks like.

Figure — Radioactive decay — alpha, beta, gamma — mechanisms

Figure: U-238 sits at rest (black dot). It splits into a light alpha (orange, right) and a heavy thorium daughter (teal, left). The two momentum arrows are equal in length and opposite in direction, so they cancel to zero — momentum is conserved. Because $K E = p^{2} /2 m$ and the alpha's mass is far smaller, the alpha carries almost all the kinetic energy.

Step 2 — Write momentum conservation. Here $p_{T h}$ means the momentum of the thorium daughter (our general daughter $Y$ is thorium in this case). $0 = p_{α} - p_{T h} \Rightarrow p_{α} = p_{T h} \equiv p$ Why this step? Because "total momentum stays zero" is the only handle that links the two speeds we don't yet know — it forces both particles to share one common momentum size $p$ .

Step 3 — Write kinetic energy using $K E = p^{2} /2 m$ . We use $K E = p^{2} /2 m$ (not $\frac{1}{2} m v^{2}$ ) because we know the shared momentum $p$ , not the velocities. (This form comes straight from $p = m v$ : substitute $v = p / m$ into $\frac{1}{2} m v^{2}$ to get $\frac{1}{2} m (p / m)^{2} = p^{2} /2 m$ .) Same $p$ for both, so: $K E_{α} = \frac{p ^{2}}{2 m _{α}}, K E_{T h} = \frac{p ^{2}}{2 m _{T h}}$ Why this step? This form puts the mass in the denominator, so the lighter particle (smaller $m$ ) gets the bigger $K E$ . The alpha wins.

Step 4 — Derive the share formula (no formula pulled from thin air). Add the two kinetic energies; they must equal all the released energy $Q$ : $Q = K E_{α} + K E_{T h} = \frac{p ^{2}}{2 m _{α}} + \frac{p ^{2}}{2 m _{T h}} = \frac{p ^{2}}{2} (\frac{1}{m _{α}} + \frac{1}{m _{T h}}) = \frac{p ^{2}}{2} \cdot \frac{m _{α} + m _{T h}}{m _{α} m _{T h}}$ Why this step? We want $K E_{α}$ in terms of $Q$ alone, so first express $p^{2}$ using $Q$ . Rearranging: $p^{2} = \frac{2 Q m _{α} m _{T h}}{m _{α} + m _{T h}}$ Now feed this into $K E_{α} = p^{2} /2 m_{α}$ ; the $m_{α}$ cancels once: $K E_{α} = \frac{p ^{2}}{2 m _{α}} = Q \cdot \frac{m _{T h}}{m _{α} + m _{T h}}$ Why this step? This is the formula the parent note quoted — but now you have seen it fall out of momentum + energy conservation, not been handed it. Since $m \approx A u$ (see the ratio-safe box above), $m_{T h} / (m_{α} + m_{T h}) \approx 234/ (4 + 234) = 234/238 = (A - 4) / A$ . Plug in: $K E_{α} = 4.27 \times \frac{234}{238} = 4.198 MeV$ Then the recoil takes the rest: $K E_{T h} = Q - K E_{α} = 4.27 - 4.198 = 0.072 MeV$

Verify: $K E_{α} + K E_{T h} = 4.198 + 0.072 = 4.27 = Q$ . ✓ Energy is fully accounted for. The alpha got ~98% — the light particle wins, as forecast by Step 3.

Ex 2 · Cell A2 — Alpha: the limiting case

Step 1 — Take the limit. $lim_{A \to \infty} \frac{A - 4}{A} = lim_{A \to \infty} (1 - \frac{4}{A}) = 1$ Why this step? A limit answers the question "what value does this expression approach as $A$ grows without bound?" — the cleanest way to read off long-run behaviour. As $A \to \infty$ the term $4/ A$ shrinks to zero, so the fraction climbs toward $1$ . A super-heavy nucleus barely recoils, so the alpha takes essentially all of $Q$ .

Step 2 — Check the small end (degenerate input). The formula needs a real daughter, so $A - 4 \geq 1$ , i.e. $A \geq 5$ . At the smallest sensible case $A = 5$ : fraction $= (5 - 4) /5 = 0.2$ . The alpha gets only 20% — the tiny daughter recoils hard. Why this step? Testing the extreme small input is how you find where a formula breaks or behaves oddly; here it shows the alpha's share can drop as low as one-fifth, the opposite extreme from Step 1.

Verify: monotonic and bounded — $(A - 4) / A$ rises from $0.2$ (at $A = 5$ ) toward $1$ (at $A = \infty$ ) and never exceeds $1$ . Physically sane: the alpha can never carry more than 100% of $Q$ . ✓

Ex 3 · Cell B1 — Beta-minus: the endpoint energy

Step 1 — Compute the mass difference. $Δ m = 14.003242 - 14.003074 = 0.000168 u$ Why this step? Decay energy comes from vanished mass, so the first move is always to find how much mass disappeared. For $β^{-}$ , using atomic masses automatically bundles in the electron masses correctly, so we don't add $m_{e}$ separately.

Step 2 — Convert mass to energy. $Q = Δ m \times 931.5 MeV/u = 0.000168 \times 931.5 = 0.1565 MeV$ Why this step? $E = m c^{2}$ (the $c$ is the mass–energy exchange rate): the mass that vanished reappears as released energy $Q$ . The factor $931.5 MeV/u$ is exactly $c^{2}$ written in these units.

Step 3 — Interpret the spectrum. This is a three-body decay ( $N + e^{-} + \overset{ν}{ˉ}_{e}$ ). The electron and the antineutrino share $Q$ continuously. So the electron's energy runs from $0$ up to a maximum (the endpoint), reached when the antineutrino takes ~zero energy: $K E_{e, m a x} \approx Q = 0.156 MeV$ Why this step? Answering the Forecast: unlike a two-body split (one fixed energy), three bodies can divide $Q$ infinitely many ways, giving a smooth band — the historical fingerprint of the neutrino.

Verify: matches the parent note's stated $Q \approx 0.156$ MeV. ✓ The continuous spread (Forecast) is why the neutrino had to exist — see Neutrino and lepton number conservation.

Ex 4 · Cell B2 — Beta-plus vs electron capture: a threshold and a degenerate case

Step 1 — Write the $β^{+}$ Q-value with atomic masses. $Q_{β^{+}} = (m_{parent}^{atom} - m_{daughter}^{atom} - 2 m_{e}) c^{2}$ Why this step? Atomic masses count the orbital electrons, and $β^{+}$ both loses an orbital electron (daughter has one fewer proton) and creates a positron; writing the $Q$ this way exposes that double cost cleanly. The daughter has one fewer proton, so a neutral daughter atom carries one fewer orbital electron; on top of that we created a positron. Two extra electron-masses' worth of energy must be paid — a real threshold, not bookkeeping fluff.

Step 2 — Set the threshold. $Q_{β^{+}} > 0$ requires $(m_{parent}^{atom} - m_{daughter}^{atom}) c^{2} > 2 m_{e} c^{2} = 1.022 MeV$ Why this step? "Decay allowed" means $Q > 0$ ; rearranging that inequality isolates the exact mass gap the parent must beat, which is the number an examiner wants.

Step 3 — Electron capture Q-value. $Q_{E C} = (m_{parent}^{atom} - m_{daughter}^{atom}) c^{2} > 0$ Why this step? Comparing the two channels side by side shows the whole difference is the missing $2 m_{e}$ : capture absorbs an existing electron instead of making one, so it pays no positron-creation penalty.

Step 4 — The degenerate window. If $0 < (m_{parent} - m_{daughter}) c^{2} < 1.022 MeV$ , then $Q_{β^{+}} < 0$ (forbidden) but $Q_{E C} > 0$ (allowed). At exactly a $0.5$ MeV gap only electron capture occurs — that's the Forecast answer. Why this step? This is the "zero/degenerate input" cell: it shows a whole band of masses where one channel is dead and the other alive, the subtlety the matrix promised.

Verify: threshold $2 m_{e} c^{2} = 2 \times 0.511 = 1.022 MeV$ . ✓ And $0.5 < 1.022$ , so $Q_{β^{+}} = 0.5 - 1.022 = - 0.522 < 0$ (forbidden) while $Q_{E C} = 0.5 > 0$ (allowed). ✓

Ex 5 · Cell C1 — Gamma: the recoil correction

Step 1 — Photons carry momentum. A photon of energy $E_{γ}$ carries momentum $p_{γ} = E_{γ} / c$ (recall $c$ = speed of light; for a massless particle this replaces $p = m v$ ). Momentum conservation — the nucleus started at rest — gives the nucleus an equal, opposite recoil momentum $p = E_{γ} / c$ . Why this step? A photon has no rest mass, so $p = m v$ is useless; $p = E / c$ is the right momentum tool, and momentum conservation is what couples the photon to the recoiling nucleus.

Step 2 — Recoil kinetic energy. $K E_{recoil} = \frac{p ^{2}}{2 M} = \frac{E _{γ}^{2}}{2 M c ^{2}}$ Energy conservation: $E_{0} = E_{γ} + K E_{recoil}$ . Since recoil is tiny, set $E_{γ} \approx E_{0}$ inside it: $\frac{K E _{recoil}}{E _{0}} \approx \frac{E _{0}}{2 M c ^{2}}$ Why this step? We reuse $K E = p^{2} /2 M$ (same tool as Ex 1) because we know the recoil momentum, not the recoil speed; forming a ratio of energies makes the answer dimensionless and lets the crude $M \approx A u$ mass cancel out to 3 figures.

Step 3 — Plug numbers. $M c^{2} = 57 \times 931.5 MeV = 53, 095 MeV = 5.31 \times 1 0^{10} eV$ , and $E_{0} = 1.44 \times 1 0^{4} eV$ : $\frac{K E _{recoil}}{E _{0}} = \frac{1.44 \times 1 0 ^{4}}{2 \times 5.31 \times 1 0 ^{10}} \approx 1.36 \times 1 0^{- 7}$ Why this step? Putting in real numbers turns the symbolic ratio into the concrete "how tiny is tiny" the Forecast asked for.

Verify: the fraction is $\sim 1 0^{- 7}$ — utterly tiny (Forecast: tiny). ✓ This is why gamma lines look sharp; the famous Mössbauer effect exploits exactly this near-zero recoil.

Ex 6 · Cell D1 — A decay chain: bookkeeping $Z$ and $A$

Step 1 — Recall the per-decay changes. $α$ : $A - 4, Z - 2$ . $β^{-}$ : $A$ same, $Z + 1$ . (From the parent's comparison table.) Why this step? A chain is just repeated arithmetic on two numbers, so we must nail the single-step rule before iterating it.

Step 2 — Track $A$ . Two alphas remove $4$ each; betas leave $A$ alone. $A = 238 - 4 - 4 = 230$ Why this step? $A$ and $Z$ change independently, so it's cleaner to run each ledger separately and avoid mixing them up.

Step 3 — Track $Z$ . Two alphas remove $2$ each ( $- 4$ ); two betas add $1$ each ( $+ 2$ ). $Z = 92 - 2 + 1 + 1 - 2 = 90$ Why this step? Now the second ledger; summing the signed contributions gives the final proton count.

Step 4 — Name it. $Z = 90$ is thorium: $_{90}^{230} Th$ . Why this step? $Z$ alone fixes the element, so reading off the periodic-table name is the final translation from numbers to a real isotope.

Verify: total $Δ A = - 8$ (four missing chunks = two alphas ✓), total $Δ Z = - 2$ . Order didn't matter for the totals — good sanity check. ✓

Ex 7 · Cell E1 — Real-world: the smoke-detector alpha

Step 1 — Balance the decay. $α$ removes $A - 4$ , $Z - 2$ : from $A = 241, Z = 95$ we get $A = 237, Z = 93$ = neptunium. $_{95}^{241} Am \to_{93}^{237} Np + α$ Why this step? We need the daughter's mass number before we can split the energy, and balancing $A, Z$ also names the daughter.

Step 2 — Split the energy. $K E_{α} = Q \cdot \frac{m _{N p}}{m _{α} + m _{N p}} \approx Q \cdot \frac{A - 4}{A} = 5.64 \times \frac{237}{241} = 5.546 MeV$ Why this step? We reuse the derived share formula from Ex 1 (with the daughter now neptunium); the mass ratio $\approx (A - 4) / A$ is valid to 3 figures as justified in the ratio box.

Step 3 — Physics of "safe outside, useful inside". Alphas have low penetration (stopped by paper or skin). Outside: harmless — the skin's dead layer blocks them. Inside: the alpha ionises air in a small chamber, letting a tiny current flow; smoke particles disrupt this current, triggering the alarm. Ionising power is high, range is short — perfect for a sealed sensor. Why this step? Part (c) asks for the physical consequence, not a number; linking penetration to real behaviour is what turns kinematics into an engineering answer.

Verify: $K E_{α} = 5.546$ MeV is $98.3%$ of $Q$ (Forecast: closer to $Q$ ). ✓ Recoil $N p$ takes $5.64 - 5.546 = 0.094$ MeV, and $5.546 + 0.094 = 5.64 = Q$ . ✓

Ex 8 · Cell E2 — Exam twist: spot the impossible decay

Step 1 — Recall the spontaneity rule. A decay happens spontaneously only if $Q > 0$ (products lighter than parent, energy released). This is the master conservation rule turned into a yes/no test. Why this step? One number ( $Q$ ) decides everything, so checking its sign first saves you from any further calculation.

Step 2 — Read the sign. $Q_{α} = - 7.6 MeV < 0$ . Negative $Q$ means the products are heavier than the parent — you'd have to add $7.6$ MeV to force it. So it cannot occur on its own. Why this step? Interpreting the sign physically (negative ⇒ energy must be supplied) is what converts a bare number into the verdict.

Step 3 — Why iron especially. $^{56} Fe$ sits near the peak of the binding-energy-per-nucleon curve (see Nuclear binding energy and mass defect) — it is one of the most tightly bound nuclei. Nothing to gain by breaking apart; every path out costs energy. Why this step? Explaining why the sign came out negative for iron (maximal binding) shows the result isn't a fluke but a deep feature of the mass curve.

Verify: $Q = - 7.6 < 0 \Rightarrow$ forbidden. Student is wrong. ✓ Sanity: iron being maximally stable is exactly why stars can't get energy by fusing past it — consistent physics.

Recall Self-test — cover the answers

Which particle gets more KE in alpha decay, and why? ::: The alpha (lighter); with shared momentum $p$ , $K E = p^{2} /2 m$ is bigger for smaller $m$ . As $A \to \infty$ , the alpha's share of $Q$ approaches what? ::: $1$ (100%) — the huge daughter barely recoils. Why is the beta electron energy a continuous band? ::: Three-body decay; the antineutrino shares $Q$ continuously with the electron. What extra energy must $β^{+}$ pay that electron capture does not? ::: $2 m_{e} c^{2} = 1.022$ MeV (create a positron + lose one orbital electron). How big is the gamma recoil energy fraction for a heavy nucleus? ::: Order $1 0^{- 7}$ — negligible; lines stay sharp. One number decides if any decay is allowed — what is it? ::: The Q-value; decay is spontaneous only if $Q > 0$ .

Back to parent: Radioactive decay mechanisms.