Nuclear reactions — Q-value calculation: Visual walkthrough

Step 1 — Weigh the pieces before and after

WHAT. A nuclear reaction is written $a + X \to Y + b$ . Read it like a recipe: two things come in on the left (a small flying particle $a$ and a target nucleus $X$ ), two things come out on the right ( $Y$ , the big leftover nucleus, and $b$ , a small ejected particle). We put all the reactants on a scale, then put all the products on a scale, and compare.

WHY. Everything about $Q$ starts from a shocking experimental fact: the two scale readings are not equal. In everyday chemistry the "before" and "after" weigh the same. In nuclear reactions they do not — and the tiny difference is the whole story.

PICTURE. Two pans. Left pan = $a + X$ . Right pan = $Y + b$ . The left pan sits slightly lower — it is heavier. That missing mass on the right has to go somewhere.

Step 2 — Name the missing mass: $Δ m$

WHAT. Define the mass that vanished: $"how much lighter the after-pile is" Δ m = before, left pan (m_{a} + m_{X}) - after, right pan (m_{Y} + m_{b})$

The symbol $Δ$ (Greek "delta") always means "the change in" — here, before minus after.

WHY. We need one number that captures "how unbalanced the scale is." That single number is $Δ m$ . If $Δ m > 0$ the before-pile was heavier (mass disappeared). If $Δ m < 0$ the after-pile is heavier (mass was created — which, we'll see, costs energy).

PICTURE. The same two pans, now with the height difference labelled as a little ruler marked $Δ m$ . Positive ruler pointing down on the left = mass lost.

Step 3 — Where did the mass go? Einstein's exchange rate

WHAT. The vanished mass reappears as energy. Einstein's mass–energy relation gives the exchange rate: $energy E = mass m a fixed number c^{2}$ Here $c = 3.0 \times 1 0^{8} m/s$ is the speed of light, and $c^{2}$ is just a conversion factor turning kilograms into joules.

WHY. We asked "where does the missing mass go?" Nothing else in the reaction changes — same number of protons, same number of neutrons overall. The only place the extra weight can hide is as motion energy of the flying-apart pieces. $E = m c^{2}$ is the rule for how much energy each unit of missing mass is worth.

PICTURE. The little slice of mass $Δ m$ from Step 2 gets fed into a "converter box" and comes out the other side as a burst of yellow energy $Δ m c^{2}$ . Because $c^{2}$ is a gigantic number, a speck of mass makes a firework of energy.

Step 4 — The same story told with kinetic energy

WHAT. Now watch energy, not mass. Total energy is conserved: the total before equals the total after. Each side has two ingredients — the rest-mass energy ( $m c^{2}$ ) and the kinetic energy $K$ (energy of motion, from being flung around).

$total energy before (m_{a} + m_{X}) c^{2} + (K_{a} + K_{X}) = total energy after (m_{Y} + m_{b}) c^{2} + (K_{Y} + K_{b})$

Every symbol: $m c^{2}$ terms are "locked-up" rest energy; each $K$ is the energy of a moving particle.

WHY. We already have $Q$ from a mass table (Step 3). But in a real lab you don't weigh a single flying nucleus — you measure how fast it moves, i.e. its $K$ . So we need a second face of $Q$ written in things you can measure. Energy conservation is the bridge.

PICTURE. A horizontal energy bar. Before: a tall stack of rest energy plus a small sliver of kinetic energy. After: a shorter rest-energy stack (mass was lost) plus a taller kinetic sliver. The bar total is the same height — energy just moved from the "rest" compartment into the "motion" compartment.

Rearrange — move all rest masses to one side, all kinetic energies to the other: $= Q (Step 3!) [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2} = motion after (K_{Y} + K_{b}) - motion before (K_{a} + K_{X})$

Step 5 — The clean-up trick: 931.5 MeV per u

WHAT. Masses live in atomic mass units u; energies live in MeV (mega-electron-volts). One conversion swallows all the physics constants: $1 u = 931.5 MeV / c^{2} ⟹ Q (MeV) = Δ m (u) \times 931.5$

WHY. Carrying $c^{2} = (3 \times 1 0^{8})^{2}$ around by hand is error-prone. We do that multiply once, bake it into the number 931.5, and never touch $c$ again.

PICTURE. A currency exchange counter: hand over "1 u", receive "931.5 MeV". The teller is $c^{2}$ .

Step 6 — Read the sign: three worlds of $Q$

WHAT. $Q$ can be positive, negative, or zero. Each is a different physical world.

Sign of $Q$	Mass	Meaning
$Q > 0$	after-pile lighter	exothermic — energy pops out
$Q < 0$	after-pile heavier	endothermic — energy must be pushed in
$Q = 0$	equal	elastic — nothing released

WHY. Step 4 said $Q = K_{final} - K_{initial}$ . If $Q > 0$ the pieces speed up — the reaction gives kinetic energy away for free. If $Q < 0$ the pieces would have to slow down below zero, which is impossible unless you first supplied energy from outside. That impossibility is exactly why endothermic reactions need a push (next step).

PICTURE. Three scale-and-bar panels side by side: left pan lower ( $Q > 0$ , green), right pan lower ( $Q < 0$ , red), balanced ( $Q = 0$ , grey).

Step 7 — The degenerate case: why endothermic needs more than $∣ Q ∣$

WHAT. For $Q < 0$ you must supply kinetic energy — but supplying exactly $∣ Q ∣$ is not enough. If the projectile $a$ hits a stationary target $X$ , the least kinetic energy that lets the reaction happen is: $K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}})$ Every symbol: $K_{th}$ is the threshold (minimum) kinetic energy of $a$ ; the factor $(1 + m_{a} / m_{X})$ is always bigger than 1.

WHY. Momentum is conserved. The incoming $a$ carries forward momentum, so the products cannot all sit still afterward — something must keep moving forward. That forced forward-motion energy (the "center-of-mass" energy) is unavailable for breaking the nucleus. Only the energy left over after paying that momentum tax can do the reaction — hence you must supply extra.

PICTURE. Top: $a$ flies in, $X$ sits still. Bottom: even at threshold the products drift forward together (they can't stop). A yellow bracket marks the "wasted" forward energy; a green bracket marks the part that actually did the reaction.

Step 8 — Splitting the payout: not all to the light particle

WHAT. When $Q > 0$ is released as motion, how is it shared? For a decay at rest ( $X$ at rest, splitting into $Y$ and $b$ ), momentum conservation forces: $K_{b} = Q \cdot \frac{m _{Y}}{m _{Y} + m _{b}}, K_{Y} = Q \cdot \frac{m _{b}}{m _{Y} + m _{b}}$ The light particle $b$ gets the lion's share (because $m_{Y}$ on top is big).

WHY. The two fragments fly apart with equal and opposite momentum. Equal momentum but unequal mass ⇒ the light one moves fast (much $K$ ), the heavy one barely recoils. So $Q$ is $K_{b} + K_{Y}$ , both terms — never just the fast particle's energy.

PICTURE. α-decay cartoon: heavy daughter recoils slowly left (short arrow), light α zips right (long arrow), momenta equal and opposite. Energy pie split mostly to the α.

The one-picture summary

Everything at once: a scale (mass lost, Steps 1–2) feeding a converter box ( $E = m c^{2}$ , Step 3) whose output splits into the kinetic energies of the products (Step 4), with the 931.5 exchange rate (Step 5) and the sign key (Step 6) in the corners.

Recall Feynman retelling — the whole walkthrough in plain words

Put the ingredients on a scale and write the number down. Do the reaction. Put the leftovers on the scale — the number is a hair smaller. That missing hair of weight didn't leak away; Einstein's rule $E = m c^{2}$ turned it into a burst of motion, and because $c^{2}$ is huge, a speck of weight becomes a firework. That firework of energy is the Q-value. You can find it two ways: weigh the mass that vanished, or clock how fast the pieces fly apart — both give the same $Q$ . If the leftovers came out heavier instead, you had to shove energy in to make them — and thanks to the rule that momentum keeps things moving, you have to shove in even more than the mass gap, because the products can't be left standing still. Finally, when energy pops out, the little fast piece grabs most of it while the big piece barely nudges — but you must count both nudges to get $Q$ right.