Nuclear reactions — Q-value, cross-section: Visual walkthrough

Before we start, three words we will use constantly:

Step 1 — Draw the collision and name the motions

WHAT. Two frames of a movie. Before: the projectile $a$ flies rightward at speed $v_{a}$ ; the target $X$ sits perfectly still. After: two products fly off.

WHY. Every conservation law we use is just "some quantity is the same in the before frame and the after frame." So we must first draw both frames clearly and label what moves.

PICTURE. In the figure, the blue arrow is the projectile's velocity — its length is its speed. The target (orange) has no arrow at all: it is at rest. This "target at rest" setup is called the laboratory frame (the "lab"), because it is what an experimenter actually sees on the bench.

Figure — Nuclear reactions — Q-value, cross-section

Step 2 — Write down what must not change: momentum

WHAT. Total momentum before = total momentum after.

WHY. Momentum is conserved in every collision where no outside push acts — a law we borrow from Conservation laws in collisions. It is the entire reason the threshold formula is not just " $∣ Q ∣$ ".

PICTURE. Before, only the blue projectile carries momentum ( $p_{a} = m_{a} v_{a}$ ); the orange target contributes zero. So the whole system already has a rightward push that cannot vanish. After the collision, the products inherit that push — they cannot all be at rest.

$all the momentum, carried by a m_{a} v_{a} + X is still 0 = products, still moving right p_{Y} + p_{b}$

$m_{a} v_{a}$ — the projectile's momentum, the only momentum in the whole picture at the start.
$0$ — the target's momentum; it is at rest, so this term is genuinely zero.
$p_{Y} + p_{b}$ — the products' total momentum afterward; it must equal $m_{a} v_{a}$ , so it can never be zero.

Step 3 — Write down the energy budget, and meet Q

WHAT. Total energy before = total energy after. Total energy = rest-mass energy $m c^{2}$ plus kinetic energy $T$ .

WHY. Energy is conserved too. But nuclei can convert mass into motion (Einstein's $E=mc^2$), so we must track rest mass and kinetic energy in the same equation.

$rest energy, before (m_{a} + m_{X}) c^{2} + KE we supplied T_{a} = rest energy, after (m_{Y} + m_{b}) c^{2} + KE of products (T_{Y} + T_{b})$

$(m_{a} + m_{X}) c^{2}$ — the energy "frozen" inside the reactant masses. $c$ is the speed of light; $c^{2}$ is the huge conversion factor from mass to energy.
$T_{a}$ — the kinetic energy we gave the projectile (the target is still, so $T_{X} = 0$ ).
$(m_{Y} + m_{b}) c^{2}$ — energy frozen in the product masses.
$(T_{Y} + T_{b})$ — kinetic energy the products fly away with.

Now group the rest-mass terms on one side:

$call this Q [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2} = (T_{Y} + T_{b}) - T_{a}$

PICTURE. The figure stacks two energy bars — a "before" bar and an "after" bar of equal total height. When $Q < 0$ , the rest-energy block grows taller after the reaction, so the kinetic-energy block on top must shrink. The reaction eats motion.

Step 4 — The naive guess (and why it is wrong)

WHAT. A first, tempting answer: "for an endoergic reaction I lack $∣ Q ∣$ of energy, so just give the projectile $T_{a} = ∣ Q ∣$ ."

WHY show a wrong idea? Because it is the single most common mistake, and seeing it fail is what makes the real answer stick.

PICTURE. Imagine we pour exactly $∣ Q ∣$ into the projectile, hoping the products come out frozen still so that all of $∣ Q ∣$ went into re-making mass. But Step 2 forbids frozen products: total momentum was $m_{a} v_{a} \neq = 0$ , so the products must move. Moving products carry kinetic energy $(T_{Y} + T_{b}) > 0$ . That kinetic energy had to come from somewhere — it came from the energy we supplied, leaving less than $∣ Q ∣$ available to actually rebuild the mass. So $∣ Q ∣$ alone falls short.

$Give T_{a} = ∣ Q ∣ \Rightarrow products forced to move \Rightarrow energy left for mass < ∣ Q ∣ \Rightarrow reaction fails.$

Step 5 — Split motion into "drift" and "internal" using the CM frame

WHAT. We separate every particle's motion into two parts:

the drift — the whole system sliding forward together, and
the internal motion — how particles move relative to that common drift.

WHY. Only the internal energy can be spent on rearranging nuclei. The drift energy is "locked": momentum conservation nails it down, so it cannot be touched. To see the split cleanly we jump to a special viewpoint.

PICTURE. Two panels. Left = lab frame: both particles drift rightward (a shared blue drift arrow). Right = CM frame: the same collision but the drift is subtracted out, so $a$ and $X$ rush toward each other with equal and opposite momenta, and their common center stays fixed.

Step 6 — How much energy survives the jump into the CM frame

WHAT. The kinetic energy an experimenter sees in the lab, $E_{l ab} = T_{a}$ , is larger than the useful energy in the CM frame, $E_{c m}$ . The exact discount is:

$E_{c m} = E_{l ab} \cdot \frac{m _{X}}{m _{a} + m _{X}}$

WHY. The factor $\frac{m _{X}}{m _{a} + m _{X}}$ is a number between $0$ and $1$ — it shrinks the lab energy. The chunk that gets removed, $E_{l ab} (1 - \frac{m _{X}}{m _{a} + m _{X}}) = E_{l ab} \frac{m _{a}}{m _{a} + m _{X}}$ , is exactly the drift (CM-motion) energy that Step 5 said we can never touch.

Term by term:

$E_{l ab}$ — the kinetic energy the projectile has in the lab (what we control by accelerating it).
$m_{X}$ (numerator) — the heavier the target, the closer this fraction is to $1$ , so less energy is wasted on drift. (A truck barely budges when a pebble hits it, so almost no energy goes into moving the truck.)
$m_{a} + m_{X}$ (denominator) — the total mass; it sets how the incoming energy divides between "useful internal" and "wasted drift."

PICTURE. A single energy bar sliced into two: a big lower slice = usable $E_{c m}$ , and a thin upper slice = drift energy that is lost to the reaction. As the target grows heavier, the wasted upper slice shrinks toward nothing.

Step 7 — Set the threshold condition and solve

WHAT. The reaction first becomes possible at the exact moment the usable CM energy equals the mass we must rebuild: $E_{c m} = ∣ Q ∣.$

WHY. Below this, the internal energy is too small to pay for the created mass — reaction impossible. At exactly this point the products appear at rest in the CM frame (barely made). We call the lab energy at that moment the threshold energy $E_{t h}$ .

Substitute $E_{c m} = E_{t h} \cdot \frac{m _{X}}{m _{a} + m _{X}}$ from Step 6:

$E_{t h} \cdot \frac{m _{X}}{m _{a} + m _{X}} = ∣ Q ∣$

Solve for $E_{t h}$ by multiplying both sides by $\frac{m _{a} + m _{X}}{m _{X}}$ :

$E_{t h} = ∣ Q ∣ \frac{m _{a} + m _{X}}{m _{X}} = - Q \frac{m _{a} + m _{X}}{m _{X}}$

$∣ Q ∣ = - Q$ — since the reaction is endoergic, $Q$ is negative, so $∣ Q ∣ = - Q$ ; both forms mean "the amount of mass-energy we must supply."
$\frac{m _{a} + m _{X}}{m _{X}}$ — this factor is always bigger than 1 (numerator has an extra $m_{a}$ ), so $E_{t h}$ is always bigger than $∣ Q ∣$ . That is the mathematical shadow of "some energy is wasted on drift."

PICTURE. The threshold snapshot: in the CM frame the products hang motionless at the balance point (zero internal motion), while in the lab frame they all glide forward together carrying the locked drift energy.

Step 8 — Edge cases: check the formula at its extremes

WHAT. A formula is only trustworthy if it behaves sensibly at its limits. Test three.

WHY. The parent contract: the reader must never meet a scenario we did not show. So we push $m_{X}$ , $m_{a}$ , and $Q$ to their extremes.

PICTURE. Three mini-panels, one per case, each drawing the collision in that limit.

Exoergic reaction, $Q > 0$ . Then $∣ Q ∣$ is really "energy released," and no threshold is needed — the formula gives a negative $E_{t h}$ , which we read as " $E_{t h} = 0$ : the reaction can happen even with a barely-moving projectile." No minimum kick required.
Infinitely heavy target, $m_{X} \to \infty$ . Then $\frac{m _{a} + m _{X}}{m _{X}} \to 1$ , so $E_{t h} \to ∣ Q ∣$ . Makes sense: an immovable target absorbs no drift energy, so the naive guess would have worked — momentum barely costs anything.
Equal masses, $m_{a} = m_{X} = m$ . Then $\frac{m _{a} + m _{X}}{m _{X}} = 2$ , so $E_{t h} = 2∣ Q ∣$ . You must supply double — half your energy is doomed to become drift because the two share momentum symmetrically. This is the harshest common case.

Q > 0 : m_{X} \to \infty : m_{a} = m_{X} : E_{t h} \leq 0 E_{t h} \to ∣ Q ∣ E_{t h} = 2∣ Q ∣ no threshold no waste worst waste

Step 9 — Plug in real numbers (Rutherford's reaction)

WHAT. Redo the parent's worked example so the picture attaches to digits: $^{14} N (α, p)^{17} O$ , with $Q = - 1.19 MeV$ , projectile $α$ mass $m_{a} = 4.0026 u$ , target $^{14} N$ mass $m_{X} = 14.0031 u$ .

WHY. To see the "always bigger than 1" factor produce a concrete number.

$E_{t h} = ∣ Q ∣ \frac{m _{a} + m _{X}}{m _{X}} = 1.19 \times \frac{4.0026 + 14.0031}{14.0031} = 1.19 \times 1.286 = 1.53 MeV$

$1.19$ — the $∣ Q ∣$ we must at least rebuild in mass.
$1.286$ — the "drift tax" factor; being $> 1$ pushes the needed energy up.
$1.53 MeV$ — the real minimum: about $0.34 MeV$ more than the naive $1.19$ , all of it lost to forced forward motion.

The one-picture summary

WHAT. The entire argument compressed: lab momentum forces drift → drift steals energy → only CM energy pays for mass → threshold exceeds $∣ Q ∣$ by the mass ratio.

Recall Feynman retelling — say it in plain words

I want to fire a small particle at a still nucleus and make it turn into heavier stuff. Heavier stuff means I have to create mass, and creating mass costs energy — that cost is $∣ Q ∣$ .

Now here's the catch. When my particle is flying rightward and the target is sitting still, the whole pair already has a rightward "push" (momentum). That push cannot vanish, so after the reaction the products are forced to keep gliding forward. Gliding forward is motion, motion is energy, and that energy had to come out of what I supplied. So part of my energy is spent just keeping everything drifting — I can never get it back to pay for the mass.

To see exactly how much is wasted, I hop onto the system's balance point (the center of mass). From there the total push is zero, so the products are allowed to stop dead — meaning every bit of energy in that view is usable. But the energy in that view is smaller than what I put in, shrunk by the factor $\frac{m _{X}}{m _{a} + m _{X}}$ : the heavier the target, the less I waste.

Setting "usable energy = mass cost," I get $E_{t h} = ∣ Q ∣ \frac{m _{a} + m _{X}}{m _{X}}$ . That fraction is always more than one, so I always need more than $∣ Q ∣$ — and for a target as heavy as the projectile I need double. That extra is the drift tax, and momentum conservation is the tax collector.

Recall Retrace the steps

Naive guess $∣ Q ∣$ fails because ::: momentum conservation forces products to move, so some supplied energy becomes unusable drift. The CM frame is special because ::: total momentum is zero there, so products may be at rest and all energy is available for the reaction. Energy discount from lab to CM is ::: $E_{c m} = E_{l ab} m_{X} / (m_{a} + m_{X})$ . Threshold condition is ::: $E_{c m} = ∣ Q ∣$ (products barely made, at rest in CM). Final formula ::: $E_{t h} = - Q (m_{a} + m_{X}) / m_{X}$ , always $> ∣ Q ∣$ . Heavy-target limit gives ::: $E_{t h} \to ∣ Q ∣$ ; equal-mass gives $E_{t h} = 2∣ Q ∣$ .