advanced34 min studyLesson 581 in Physics

Reaction Q-values and threshold energies

Track mass conventions and derive the difference between reaction energetics and a fixed-target beam threshold.

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What you’ll understand

  • Calculate a Q-value from consistently defined masses.
  • Derive a fixed-target threshold from invariant energy.
  • Distinguish a kinematic threshold from a Coulomb suppression scale.

A reaction can conserve energy and still require more beam energy than its negative Q-value suggests. The target is initially at rest, so momentum conservation forces some energy into the translation of the products. Nuclear bookkeeping also demands a consistent choice between bare-nucleus and neutral-atom masses. A missing electron rest mass can overwhelm the small energy difference being sought, especially in a beta-decay calculation.

Define the masses before subtracting them

For a two-body reaction, define Q as initial rest energy minus final rest energy for the specified internal states. Positive Q releases rest energy into kinetic energy; negative Q requires kinetic input. Exciting a product lowers the available Q by that excitation energy. Rest masses already contain nuclear binding energy, so adding a binding-energy release again would double count it. State whether a quoted mass corresponds to a ground state or an isomer before using the difference.

Neutral-atom masses contain electron rest masses and electron binding energies. They can be used directly for many reactions when the summed nuclear charge, and hence the number of neutral-atom electrons, balances on both sides, provided electronic binding corrections are negligible at the desired precision. Bare projectiles, ionized products, and beta processes need explicit care. For neutral-atom beta-minus masses the electron rest contribution cancels in the usual Q formula; positron emission instead requires subtracting two electron rest energies, up to the relevant atomic corrections.

Q=(ma+mA−mb−mB)c2,Matom=Mnucleus+Zme−Be/c2Q=(m_a+m_A-m_b-m_B)c^2,\qquad M_{\rm atom}=M_{\rm nucleus}+Zm_e-B_e/c^2
The masses are in kilograms or a declared common mass unit, Q and total electronic binding Be are energies, and c is the speed of light. All masses in the first expression must refer to the declared particles and internal states; electron bookkeeping must be consistent.↔ Scroll wide equations horizontally.

The centre of mass cannot stop in a fixed-target experiment

A projectile and stationary target carry nonzero total momentum. At an endothermic two-body threshold, the final particles have zero relative momentum in their centre-of-mass frame, but move together in the laboratory. The invariant total energy reaches the sum of their rest energies at this point. Solving that condition for the laboratory projectile kinetic energy gives an exact relativistic threshold, assuming the channel is otherwise allowed and no extra particles are present.

At nuclear beam energies small compared with rest energies, the leading expression is the magnitude of Q multiplied by one plus the projectile-to-target mass ratio. The extra term is the energy tied up in centre-of-mass translation. It becomes small for a light projectile on a very heavy target, but is substantial for comparable masses. The nonrelativistic approximation discards a calculable correction quadratic in Q; quoting an exact threshold and then using inconsistent approximate masses would defeat its precision.

Ta,th=[(mb+mB)2−(ma+mA)2]c22mA=−Q(1+mamA)+Q22mAc2(Q<0)T_{a,\rm th}=\frac{[(m_b+m_B)^2-(m_a+m_A)^2]c^2}{2m_A}=-Q\left(1+\frac{m_a}{m_A}\right)+\frac{Q^2}{2m_Ac^2}\quad(Q<0)
Ta,th and Q are energies; mA is the initially stationary target mass. This exact two-body invariant threshold assumes positive physical masses and includes final excitation in mb and mB. Its leading low-energy form drops the final term when |Q| is small compared with rest energies.↔ Scroll wide equations horizontally.

Allowed energetics does not guarantee a large yield

A positive Q-value removes an endothermic kinematic threshold, but it does not remove charged-particle Coulomb repulsion, centrifugal barriers, or restrictive quantum numbers. Tunneling permits sub-barrier reactions, so a Coulomb barrier is a suppression scale rather than an absolute energy-conservation threshold. Similarly, a channel with the correct energy may have a tiny matrix element or negligible phase space near opening. Rates require dynamics in addition to the mass balance.

A useful calculation separates three questions: whether the listed particles conserve charge and nucleon number, whether total energy and momentum can realize the channel, and how likely the transition is. These checks also clarify inverse reactions, whose Q-values have opposite signs for the same specified states. Experimental thresholds can appear broadened by target thermal motion, beam spread, energy loss, and detector sensitivity; an observed onset is not automatically the ideal threshold of a target at rest.

A deliberately simple endothermic channel

Use toy nuclear rest energies mac2=1000 MeVm_ac^2=1000\,\mathrm{MeV} and mAc2=4000 MeVm_Ac^2=4000\,\mathrm{MeV}. The final rest-energy sum is 5002 MeV5002\,\mathrm{MeV}. Find Q and both the leading and exact fixed-target threshold.

  1. The initial rest-energy sum is 5000 MeV, so Q=5000−5002=−2 MeVQ=5000-5002=-2\,\mathrm{MeV}. This channel needs energy rather than releasing it; the numbers describe a toy channel, not an identified isotope reaction.

  2. The leading threshold includes the recoil factor: −Q(1+ma/mA)=2(1+1/4)=2.5 MeV-Q(1+m_a/m_A)=2(1+1/4)=2.5\,\mathrm{MeV}. A 2 MeV beam would fail because some of its energy must sustain the total momentum.

  3. The exact correction is Q2/(2mAc2)=4/8000=0.0005 MeVQ^2/(2m_Ac^2)=4/8000=0.0005\,\mathrm{MeV}. The invariant expression therefore gives 2.5005 MeV2.5005\,\mathrm{MeV}, consistent with a small nonrelativistic correction.

The result

The Q-value is −2 MeV. The leading laboratory threshold is 2.5 MeV and the exact two-body threshold is 2.5005 MeV; the magnitude of Q alone is insufficient.

Explore the model

Interactive lab

Interactive lab · function

The unavoidable energy cost of total momentum

Represent a normalized translational energy as y = a x². Reset and set both controls to a = 0.5 and x = 1; predict y = 0.5. Reset and set both to a = 0.5 and x = 2; predict y = 2, four times larger. This illustrates the quadratic momentum cost of centre-of-mass motion, with x a normalized momentum. The lab does not compute masses, Q-values, or reaction thresholds.

The unavoidable energy cost of total momentumThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNonzero total momentum reserves kinetic energy even at a channel’s relative-motion threshold.

Check your understanding

Why does a fixed-target endothermic threshold generally exceed −Q?

Try it yourself

Neutral-atom parent and daughter masses in a toy positron-decay branch differ by 1.500 MeV/c21.500\,\mathrm{MeV}/c^2. Neglect electronic binding differences and use mec2=0.511 MeVm_ec^2=0.511\,\mathrm{MeV}. Find the available positron-decay Q and its value if the daughter is excited by 0.200 MeV.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A neutral-atom mass difference for positron emission needs two electron rest energies subtracted.

Compare with the solution

The ground-state branch has Qβ+=1.500−2(0.511)=0.478 MeVQ_{\beta^+}=1.500-2(0.511)=0.478\,\mathrm{MeV}. Excitation consumes another 0.200 MeV, leaving 0.278 MeV. That energy is shared by positron kinetic energy, neutrino energy, and recoil; it is not all the positron’s kinetic energy in every event. Using only one electron rest energy would confuse atomic and bare-nuclear mass conventions.

Take it with you

  • Use consistent particle and electronic mass conventions.
  • A fixed-target threshold contains a centre-of-mass recoil cost.
  • Kinematic accessibility and a measurable reaction rate are separate questions.

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