Radioactive decay — alpha, beta, gamma — mechanisms: Visual walkthrough

This is a companion to Radioactive decay — alpha, beta, gamma — mechanisms. If words like $Q$ -value or nucleon number feel shaky, that parent note builds them first.

Step 1 — Draw the "before": one nucleus, sitting perfectly still

WHAT. We start with a single unstable parent nucleus, alone in space, not moving. Nothing else is around.

WHY start here? Because a nucleus that has been sitting in a sample is, on average, at rest. If we choose our viewpoint to move along with it, then before the decay there is zero total motion. That single fact — "the before-picture is frozen" — is the seed the whole derivation grows from.

PICTURE. The parent is a fat, calm ball. There is one number we track called momentum. Momentum is just "how much oomph of motion something carries": mass times velocity, $p = m v$ . A still object has $v = 0$ , so its momentum is $p = 0$ . The whole system's momentum right now is a big fat zero.

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

Step 2 — Draw the "after": two pieces fly apart

WHAT. The decay happens. Now there are two objects:

the alpha particle, mass $m_{α}$ , flying off with some velocity $v_{α}$ ;
the leftover daughter nucleus, mass $m_{Y}$ , recoiling with velocity $v_{Y}$ .

WHY two, and why opposite? Because the parent broke into exactly these two chunks (the parent note showed $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y + α$ ). And they must fly in opposite directions — here is the reason, coming next.

PICTURE. One small ball (the alpha, coral) shoots right; one big ball (the daughter, lavender) drifts left. Their arrows point opposite ways.

Step 3 — The law that ties them together: momentum is conserved

WHAT. Total momentum after = total momentum before. Before was zero (Step 1). So after must also add to zero.

WHY this law? Momentum conservation is one of the deepest rules in physics: with nothing pushing from outside, the total oomph can't appear or vanish. The nucleus pushed the alpha one way; by Newton's "every push has an equal opposite push," the alpha shoved the nucleus the other way with equal strength.

PICTURE. Two arrows, equal length, pointing opposite ways, cancelling to nothing — like two ice skaters shoving off each other.

$before, Step 1 0 = alpha’s oomph p_{α} - daughter’s oomph p_{Y}$

The minus sign says they point opposite ways. Rearranging:

$p_{α} = p_{Y} \equiv p$

Both pieces carry the same size of momentum $p$ . That is the key that unlocks everything — the small alpha and the big daughter share momentum equally, even though they don't share energy equally.

Step 4 — Energy: writing kinetic energy using momentum

WHAT. We now bring in kinetic energy (KE) — the energy of motion. The usual form is $K E = \frac{1}{2} m v^{2}$ , but we want it in terms of $p$ instead of $v$ .

WHY rewrite it with $p$ ? Because Step 3 handed us a fact about $p$ (both pieces share the same $p$ ), not about $v$ . To use that gift, our energy formula must speak the language of $p$ . So we translate.

HOW (the translation). Start from $K E = \frac{1}{2} m v^{2}$ . From $p = m v$ we get $v = p / m$ . Substitute: $K E = \frac{1}{2} m (\frac{p}{m})^{2} = \frac{1}{2} m \cdot \frac{p ^{2}}{m ^{2}} = \frac{p ^{2}}{2 m}$

PICTURE. Same motion, described two ways — a speedometer view ( $v$ ) versus an oomph view ( $p$ ). We just switched dials.

Step 5 — Where does the released energy go? Enter the Q-value

WHAT. The total kinetic energy of the two flying pieces equals the energy released by the decay, the Q-value: $K E_{α} + K E_{Y} = Q$

WHY? The parent was heavier (in rest-mass-energy) than the daughter + alpha. That missing mass turned into motion energy: $Q = (m_{parent} - m_{products}) c^{2}$ . Before the decay nothing moved, so all of $Q$ shows up as the KE of the two pieces afterward. Energy is conserved.

PICTURE. A tank labelled $Q$ empties into two cups — a small alpha cup and a daughter cup. The cups fill unequally; our job is to find the split.

Write each KE with the Step 4 formula, remembering both pieces share the same $p$ : $K E_{α} \frac{p ^{2}}{2 m _{α}} + K E_{Y} \frac{p ^{2}}{2 m _{Y}} = Q$

Step 6 — Solve for the shared momentum $p$

WHAT. We have one equation with one unknown, $p$ (since $Q$ , $m_{α}$ , $m_{Y}$ are known). Solve it.

WHY? Once we know $p$ , plugging into $K E_{α} = p^{2} /2 m_{α}$ gives the alpha's energy directly — the goal.

HOW. Factor out $p^{2}$ : $p^{2} (\frac{1}{2 m _{α}} + \frac{1}{2 m _{Y}}) = Q$

Combine the fractions over a common bottom $2 m_{α} m_{Y}$ : $\frac{1}{2 m _{α}} + \frac{1}{2 m _{Y}} = \frac{m _{Y} + m _{α}}{2 m _{α} m _{Y}}$

So: $p^{2} \cdot \frac{m _{Y} + m _{α}}{2 m _{α} m _{Y}} = Q ⟹ p^{2} = \frac{2 Q m _{α} m _{Y}}{m _{α} + m _{Y}}$

PICTURE. Two "resistance to motion" terms ( $1/2 m_{α}$ and $1/2 m_{Y}$ ) add up like two springs in a row; the fat daughter contributes a tiny springiness because its $1/ m_{Y}$ is small.

WHAT. Put $p^{2}$ into $K E_{α} = \frac{p ^{2}}{2 m _{α}}$ : $K E_{α} = \frac{1}{2 m _{α}} \cdot \frac{2 Q m _{α} m _{Y}}{m _{α} + m _{Y}}$

The $2$ cancels the $2$ , and one $m_{α}$ cancels the $m_{α}$ : $K E_{α} = Q \cdot \frac{m _{Y}}{m _{α} + m _{Y}}$

WHY it looks right. The fraction $\frac{m _{Y}}{m _{α} + m _{Y}}$ is almost 1 because $m_{Y}$ (the big daughter) hugely outweighs $m_{α}$ . So the alpha grabs nearly all of $Q$ — but not quite, because the daughter still recoils and pockets a sliver.

The handy approximation. Mass is roughly proportional to nucleon number $A$ : the daughter has $A - 4$ nucleons, the alpha has $4$ , the parent had $A$ . So $m_{Y} \approx A - 4$ and $m_{α} + m_{Y} \approx A$ : $K E_{α} \approx Q \cdot \frac{A - 4}{A}$

PICTURE. A see-saw: the big daughter sits close to the pivot (takes little energy), the tiny alpha sits far out and flies up with most of the energy.

Step 8 — Edge and limiting cases (never leave a gap)

WHAT / WHY. A formula is only trustworthy if it behaves sensibly in extreme situations. Let's poke it.

Case A — daughter infinitely heavier ( $m_{Y} \to \infty$ ). Then $\frac{m _{Y}}{m _{α} + m _{Y}} \to 1$ , so $K E_{α} \to Q$ . A truly immovable wall would let the alpha keep all the energy. Reassuring: our formula smoothly approaches the "alpha takes everything" myth as a limit, without ever reaching it for real nuclei.

Case B — equal masses ( $m_{Y} = m_{α}$ ). Then the fraction is $\frac{1}{2}$ : they'd split $Q$ evenly. This never happens in alpha decay (the daughter is always far heavier), but it shows the formula is honest — equal masses share equally.

Case C — $Q \leq 0$ . No decay. If the products weren't lighter than the parent, there's no energy to hand out, and $p^{2}$ in Step 6 would be zero or "negative" (unphysical). This is exactly why alpha decay only occurs for heavy nuclei where $Q > 0$ .

Case D — the daughter left excited. Sometimes the daughter lands in an excited state, so a bit of $Q$ is stored as internal excitation and later leaves as a gamma. Then the available $Q$ for kinetic sharing is smaller, giving a slightly lower $K E_{α}$ line. This is why alpha spectra show several discrete lines (one per daughter state), not just one — each still obeying our split formula with its own reduced $Q$ .

Worked check — Uranium-238

For $_{92}^{238} U \to_{90}^{234} Th + α$ with $Q = 4.27$ MeV: $K E_{α} \approx 4.27 \times \frac{238 - 4}{238} = 4.27 \times \frac{234}{238} = 4.20 MeV$ The daughter thorium quietly keeps the remaining $4.27 - 4.20 = 0.07$ MeV. The alpha grabbed 98% of the budget — exactly what the see-saw of Step 7 predicts.

Reveal-lines:

The alpha of U-238 carries about how much KE?

About

4.20

MeV (≈98% of

Q = 4.27

MeV).

Why doesn't it get the full 4.27 MeV?

The heavier Th-234 daughter must recoil and keeps ≈0.07 MeV.

The one-picture summary

The whole chain in one image: still parent (p=0) → two pieces with equal-and-opposite $p$ → same $p$ , so $K E = p^{2} /2 m$ splits inversely with mass → the light alpha takes the big share, the heavy daughter the sliver → they sum to $Q$ .

Recall Feynman retelling — the whole walk in plain words

Picture a heavy nucleus floating dead still. It suddenly cracks into a tiny alpha and a big leftover chunk. Because nothing was moving before, the two pieces have to fly apart with equal push — like two skaters shoving off each other, one small and zippy, one big and slow. Now, "energy of motion" written in push-language is (push squared) ÷ (twice the mass). Same push for both, but the big chunk has a huge mass on the bottom — so it gets tiny energy, and the little alpha gets almost all of it. All that energy came from the weight the nucleus lost when it decayed — that's the Q-value. Add the two shares back up and you get Q exactly. The alpha's share is $Q$ times (daughter mass)/(total mass), which for a heavy nucleus is nearly $Q$ but never quite — because the daughter always recoils a little. Poke the extremes: an infinitely heavy daughter would give the alpha everything; equal masses would split it evenly; and if the nucleus wouldn't actually get lighter by decaying, there's no energy at all and nothing happens.

Recall Test yourself

State the equal-momentum fact and why. ::: Parent at rest → total $p = 0$ before → after, $p_{α} = p_{Y}$ (equal and opposite), so both share the same magnitude $p$ . Why write $K E = p^{2} /2 m$ instead of $\frac{1}{2} m v^{2}$ ? ::: Because conservation gave us equal $p$ , not equal $v$ ; the $p$ -form lets us use that fact and shows KE falls as mass rises. What does $K E_{α} = Q m_{Y} / (m_{α} + m_{Y})$ become when $m_{Y} \to \infty$ ? ::: It approaches $Q$ — the immovable-daughter limit where the alpha takes all the energy. Why do alpha spectra show several discrete lines? ::: The daughter can land in different excited states, each leaving a different available $Q$ for the split, so a distinct fixed $K E_{α}$ per state.

Related builds: Quantum tunnelling (how the alpha escapes the barrier at all), Nuclear binding energy and mass defect (where $Q$ comes from), Nuclear energy levels and shell model (the discrete daughter states of Case D), Law of radioactive decay (how fast the tunnelling adds up over time).