Q-Value of Nuclear Reaction

Equivalent forms

Q = (\sum m_{reactants} - \sum m_{products}) c^2

A single scalar that predicts whether a reaction ignites stars or absorbs energy from them.

Symbol	Name	SI unit	Dimension	Range
$Q$	Reaction energyoutput Net energy released (Q > 0) or absorbed (Q < 0)	J	M·L²·T⁻²	-1e-11 – 1e-11
$mᵢ$	Initial mass Total rest mass of reactants	kg	M	0 – 1e-25
$m_f$	Final mass Total rest mass of products	kg	M	0 – 1e-25
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Nuclear
$Q$	J	erg	MeV
$mᵢ$	kg	g	u
$m_f$	kg	g	u
$c$	m/s	cm/s	1 (with 1 u·c² = 931.494 MeV)

Where it holds

Valid for any nuclear reaction where rest masses of all reactants and products are known. Assumes the calculation is done in the center-of-mass frame or that kinetic energies are properly accounted for. Does not include energy carried away by neutrinos unless explicitly included in the mass/energy balance.

Dimensional analysis

Q \to [M\cdot L^{2}\cdot T^{-2}]

(mᵢ

- m_f)\cdot c^{2} \to [M]\cdot [L\cdot T^{-1}]^{2} = [M\cdot L^{2}\cdot T^{-2}]

Both sides are

[M\cdot L^{2}\cdot T^{-2}]

= Joules.

\checkmark

Discovery

Nuclear physics community · 1932

Q-values became standard after Cockcroft and Walton's first artificial nuclear reaction confirmed Einstein's mass-energy equivalence experimentally.

Try this

How do we know a reaction will release energy before running it?

In the reaction D + T → ⁴He + n, the mass difference is 0.01889 u. Find Q in MeV.

Research status: stable

Real-world applications

Fusion reactor design (D-T fusion: $Q = 17.6\,\mathrm{MeV}$ , the most energetically favorable fusion reaction)
Nuclear fission energy budgets $(^{235}U$ fission: $Q \approx 200\,\mathrm{MeV}$ per event)
Astrophysical nucleosynthesis models (predicting which reactions power different stellar burning stages)
Medical isotope production planning (determining energy requirements for transmutation reactions)

Common misconceptions

Q-value is NOT the kinetic energy of any single product — it is the total kinetic energy released (or absorbed) across ALL products minus all reactant kinetic energies.
A positive Q does not mean the reaction happens spontaneously — there may be a Coulomb barrier or other activation energy to overcome.
Q-value calculations using atomic masses automatically include electron masses, which cancel for most nuclear reactions but must be handled carefully for beta decay.

Experimental verification

Cockcroft and Walton (1932) verified Q-values in the first artificial nuclear reaction:

^{7}Li + p \to 2 ^{4}He

, measuring

Q = 17.3\,\mathrm{MeV}

in agreement with mass tables. Modern accelerator experiments routinely confirm Q-values to keV precision using magnetic spectrometers.

Derivation

By conservation of total relativistic energy in a nuclear reaction:

\Sigma (m

_reactants

\cdot c^{2} + KE

_reactants

) = \Sigma (m

_products

\cdot c^{2} + KE

_products).

Rearranging:

Q = \Sigma KE

_products

- \Sigma KE

_reactants

= (\Sigma m

_reactants

- \Sigma m

_products

)\cdot c^{2} = (m

ᵢ

- m_f)\cdot c^{2}

.

Positive Q means kinetic energy increased (exothermic); negative Q means kinetic energy decreased (endothermic).

Limiting cases

mᵢ

= m_f

⟶

Q = 0

No mass difference means no energy is released or absorbed — the reaction is at the threshold of being possible.

mᵢ > m_f⟶ Q > 0 (exothermic)Products are lighter; the missing mass appears as kinetic energy of products. Fusion of light nuclei and fission of heavy nuclei are exothermic.

mᵢ < m_f⟶ Q < 0 (endothermic)Products are heavier; kinetic energy must be supplied to create the extra mass. The reaction has a threshold energy.

What if…

What if Q is exactly zero?

The reaction is at threshold — products have zero kinetic energy in the center-of-mass frame. In the lab frame, a minimum projectile kinetic energy (threshold energy) is still needed to conserve momentum.

What if Q is negative and we supply insufficient kinetic energy?

The reaction cannot proceed. The minimum lab-frame kinetic energy required is KE_threshold $= |Q| \times (1 + m$ _projectile/m_target) to conserve both energy and momentum.

What if we use atomic masses instead of nuclear masses?

Electron masses cancel in most reactions since the same number of electrons appear on each side. Exception: beta decay, where one electron is created or absorbed, requiring explicit correction.

1

D-T fusion Q-value

Given ·

mᵢ u:: 5.03016
m f u:: 5.01127
c:: 299792458

Find · Q

Steps

Step 1: Reactant masses: $m_D = 2.01410\,\mathrm{u}$ , $m_T = 3.01605\,\mathrm{u} \to m$ ᵢ $= 5.03016\,\mathrm{u}$ .
Step 2: Product masses: m_He $4 = 4.00260\,\mathrm{u}$ , $m_n = 1.00867\,\mathrm{u} \to m_f = 5.01127\,\mathrm{u}$ .
Step 3: Mass difference: $\Delta m = 5.03016 - 5.01127 = 0.01889\,\mathrm{u}$ .
Step 4: $Q = 0.01889 \times 931.494\,\mathrm{MeV/u} = 17.6\,\mathrm{MeV}$ (exothermic).

Result ·

Q = (5.03016 - 5.01127) \times 931.494 = 17.6\,\mathrm{MeV}

2

Lithium-7 proton capture

Given ·

mᵢ u:: 8.02383
m f u:: 8.00521
c:: 299792458

Find · Q

Steps

Step 1: Reactant masses: $m_Li7 = 7.01600\,\mathrm{u}$ , $m_p = 1.00783\,\mathrm{u} \to m$ ᵢ $= 8.02383\,\mathrm{u}$ .
Step 2: Product masses: $2 \times m$ _He $4 = 2 \times 4.00260 = 8.00521\,\mathrm{u}$ .
Step 3: $\Delta m = 8.02383 - 8.00521 = 0.01862\,\mathrm{u}$ .
Step 4: $Q = 0.01862 \times 931.494 = 17.3\,\mathrm{MeV}$ — this was the first artificially verified nuclear reaction (Cockcroft–Walton, 1932).

Result ·

Q = (8.02383 - 8.00521) \times 931.494 = 17.3\,\mathrm{MeV}

Mass–Energy Equivalence→Nuclear Binding Energy→Semi-Empirical Mass Formula→Relativistic Energy–Momentum Relation→