Nuclear reactions — Q-value calculation: Question bank

Before we start, one word we lean on constantly: the Q-value is the leftover energy of a reaction $a + X \to Y + b$ , defined as $Q = (m_{before} - m_{after}) c^{2}$ — mass that vanishes ( $Q > 0$ ) becomes kinetic energy, mass that appears ( $Q < 0$ ) must be paid for with kinetic energy. Keep the mnemonic BAM (Before Minus After) in your head as you go.

True or false — justify

True or false: A reaction with

Q > 0

can happen even if the projectile has almost zero kinetic energy.

True — an exothermic reaction has no threshold; the reaction supplies its own energy, so any tiny nudge (or none, for spontaneous decay) is enough.

True or false: If the products of a reaction are heavier than the reactants, energy was released.

False — heavier products mean rest mass increased, so

Q < 0

; energy had to be absorbed (put in) to create that extra mass.

True or false: The Q-value depends on the reference frame you measure it in.

False —

Q

is a difference of rest masses times

c^{2}

, and rest masses are frame-independent, so

Q

is the same in every frame.

True or false: In spontaneous alpha decay the Q-value must be positive.

True — nothing supplies external energy, so the process can only run "downhill";

Q > 0

means the released rest mass appears as the kinetic energy of the alpha plus the recoiling daughter.

True or false: You may always substitute atomic masses for nuclear masses when computing

Q

False — it works only when the electron count balances (

Z

conserved); for positron emission or electron capture the electrons are mismatched and you need correction terms.

True or false: A larger mass defect

Δ m

always means a larger energy release.

True — since

Q = Δ m c^{2}

and

c^{2}

is a fixed constant,

Q

scales directly with the mass that disappeared.

True or false: The threshold kinetic energy equals the magnitude of the Q-value for an endothermic reaction.

False — threshold energy is

∣ Q ∣ (1 + m_{a} / m_{X})

, larger than

∣ Q ∣

because momentum forces the products to keep moving, wasting some energy as center-of-mass motion.

True or false: In alpha decay the emitted alpha particle carries the entire Q-value.

False — momentum conservation makes the daughter recoil, so

Q = K_{α} + K_{daughter}

; the alpha gets the larger share

Q \cdot m_{D} / (m_{D} + m_{α})

but not all of it.

True or false: A reaction can have

Q = 0

True — this is an elastic collision where no rest mass changes; total kinetic energy before equals total kinetic energy after.

Spot the error

Find the flaw: "For

^{14}

+ α \to^{17}

+ p

with

Q = - 1.19

MeV, an alpha of exactly

1.19

MeV will trigger the reaction."

Wrong — you need the threshold

∣ Q ∣ (1 + m_{a} / m_{X}) \approx 1.53

MeV; part of the

1.19

MeV must remain as forward CM motion and cannot be used to rearrange the nucleus.

Find the flaw: "

Q = m_{after} - m_{before}

times

c^{2}

Sign is flipped — it's Before minus After (BAM); as written it turns every exothermic reaction into an endothermic one.

Find the flaw: "Since the daughter nucleus is heavy and slow, its kinetic energy is negligible, so I ignore it in

Q = K_{final}

Its speed is small but its momentum equals the alpha's;

K = p^{2} /2 m

, so the recoil energy is real and small but not zero — dropping it makes

Q

slightly too low.

Find the flaw: "

1 u = 931.5

MeV, so

Q

in MeV is just

Δ m

in kilograms times

931.5

The

931.5

factor converts atomic mass units, not kilograms; you must have

Δ m

in u first (or use

E = Δ m c^{2}

directly in SI).

Find the flaw: "This reaction releases

17.6

MeV, so each product particle gets

8.8

MeV."

The energy is not split equally — momentum conservation gives the lighter fragment more kinetic energy, so the neutron and helium share

17.6

MeV unequally by their inverse mass ratio.

Find the flaw: "The reaction absorbs energy, so its binding energy went up."

Backwards — absorbing energy means products are less tightly bound (more massive); energy is released only when products become more tightly bound, per binding energy.

Why questions

Why does a tiny mass change produce such a huge energy?

Because

Q = Δ m c^{2}

and

c^{2} \approx 9 \times 1 0^{16}

is enormous; even

0.02

u (

\sim 3 \times 1 0^{- 29}

kg) becomes tens of MeV via [[Mass-energy equivalence E=mc^2|

E = m c^{2}

]].

Why is

Q

equal to

K_{final} - K_{initial}

and not just

K_{final}

Total energy (rest + kinetic) is conserved; the rest-mass change

Q

shows up as the change in total kinetic energy, so any initial KE must be subtracted out.

Why do we say

Q

equals "final binding energy minus initial binding energy"?

A more tightly bound nucleus weighs less, so gaining binding energy loses mass; that lost mass

\times c^{2}

is exactly the released

Q

Why does fusion of light nuclei and fission of heavy nuclei both release energy?

Both move toward the iron-peak of maximum binding energy per nucleon, so products end up more tightly bound and lighter, giving

Q > 0

in each direction.

Why is threshold energy always greater than

∣ Q ∣

The projectile carries momentum that must be conserved, so the products cannot all be at rest; the unavoidable forward CM kinetic energy is unavailable for rearrangement, so you must supply extra.

Why do the electron masses cancel when using atomic masses (in the usual case)?

Charge conservation means the same total number of protons — hence electrons — appears before and after, so their masses subtract to zero in

m_{before} - m_{after}

Why does positron emission need a

2 m_{e} c^{2}

correction with atomic masses?

The daughter atom has one fewer proton but the same electron count as written, so an electron mass is unmatched on each side; the imbalance amounts to two electron masses that must be subtracted.

Edge cases

Edge case: What is

Q

if the reactants and products have identical total rest mass?

Q = 0

; the collision is perfectly elastic and total kinetic energy is unchanged, just redistributed.

Edge case: For alpha decay where the parent is at rest, how do you find

Q

from measured energies?

Add the alpha's kinetic energy and the daughter's recoil energy:

Q = K_{α} + K_{daughter}

; no mass table needed.

Edge case: Can an endothermic reaction (

Q < 0

) ever happen with zero projectile kinetic energy?

No — with nothing to supply

∣ Q ∣

plus the CM-motion tax, there is no source for the mass that must be created, so it cannot proceed.

Edge case: If the target is much heavier than the projectile (

m_{X} ≫ m_{a}

), what does the threshold energy approach?

K_{th} = ∣ Q ∣ (1 + m_{a} / m_{X}) \to ∣ Q ∣

; a heavy stationary target barely recoils, so almost none of the energy is wasted as CM motion.

Edge case: If the projectile is much heavier than the target (

m_{a} ≫ m_{X}

), what happens to the threshold?

The factor

(1 + m_{a} / m_{X})

becomes very large, so

K_{th} ≫ ∣ Q ∣

; most of the incoming energy is locked in bulk CM motion and little is usable.

Edge case: A gamma photon strikes a nucleus (photodisintegration). What plays the role of the projectile "rest mass" in the threshold logic?

A photon is massless but carries momentum

E / c

; that momentum must still be shared with the products, so the photon energy needed still exceeds

∣ Q ∣

Recall Fast recap of the traps you just cleared

Sign ::: BAM — Before minus After; lighter products give $Q > 0$ . Threshold ::: always $∣ Q ∣ (1 + m_{a} / m_{X})$ , never just $∣ Q ∣$ . Energy split ::: never equal — momentum favours the lighter fragment. Atomic masses ::: fine when $Z$ balances, corrected for $β^{+}$ /capture. Frame ::: $Q$ is frame-independent; kinetic-energy splits are not.