Nuclear reactions — Q-value calculation

WHAT is a Q-value?

The masses can be rest masses of nuclei, OR (more conveniently) atomic masses — because the electrons cancel out (we check this below).

WHY does mass turn into energy?

HOW to derive Q from first principles

Start with conservation of total energy (rest + kinetic), since nucleon number and charge are conserved but rest mass is not:

$\underbrace{(m_a + m_X)c^2 + (K_a + K_X)}_{\text{before}} = \underbrace{(m_Y + m_b)c^2 + (K_Y + K_b)}_{\text{after}}$

Rearrange to group rest masses on one side and kinetic energies on the other:

$\underbrace{\left[(m_a+m_X)-(m_Y+m_b)\right]c^2}_{\equiv\,Q} = (K_Y+K_b) - (K_a+K_X)$

Handy unit fact (derive once, use forever)

$1\ \text{u} = 931.494\ \text{MeV}/c^2$. So if $\Delta m$ is in atomic mass units u: $Q\,(\text{MeV}) = \Delta m\,(\text{u}) \times 931.5$

Why atomic masses are safe to use

In $a+X\to Y+b$ the total proton number (electrons) is conserved: $Z_a+Z_X = Z_Y+Z_b$. Each atomic mass carries its $Z$ electrons. When you take the difference, the same number of electron masses appear on both sides and cancel. (Caveat: this fails for positron emission / electron capture, where electron count is not matched — there you must add correction terms.)

Endothermic reactions: the THRESHOLD energy

For $Q<0$ you must supply kinetic energy. But you can't just supply $|Q|$ — momentum must also be conserved, so the products can't be left at rest. The threshold kinetic energy (projectile $a$ hits stationary target $X$) is:

$K_{\text{th}} = |Q|\left(1 + \frac{m_a}{m_X}\right)$

Worked Example 1 — Exothermic (the classic D-T fusion)

Reaction: $\ {}^2_1\text{H} + {}^3_1\text{H} \to {}^4_2\text{He} + {}^1_0\text{n}$ Atomic/particle masses (u): $^2$H = 2.014102, $^3$H = 3.016049, $^4$He = 4.002603, n = 1.008665.

Step 1 — mass before. $2.014102 + 3.016049 = 5.030151$ u. Why? Sum reactant rest masses.

Step 2 — mass after. $4.002603 + 1.008665 = 5.011268$ u. Why? Sum product rest masses.

Step 3 — $\Delta m$. $5.030151 - 5.011268 = 0.018883$ u. Why? Mass that disappeared.

Step 4 — convert. $Q = 0.018883 \times 931.5 \approx \mathbf{17.6\ MeV}$. Why? $\Delta m>0 \Rightarrow Q>0$, exothermic — this is why fusion powers stars.

Worked Example 2 — Endothermic + threshold

Reaction: $\ {}^{14}_7\text{N} + {}^4_2\text{He} \to {}^{17}_8\text{O} + {}^1_1\text{H}$ (Rutherford's first transmutation). Masses (u): $^{14}$N = 14.003074, $^4$He = 4.002603, $^{17}$O = 16.999132, $^1$H = 1.007825.

Step 1 — before: $14.003074 + 4.002603 = 18.005677$ u. Step 2 — after: $16.999132 + 1.007825 = 18.006957$ u. Step 3 — $\Delta m = 18.005677 - 18.006957 = -0.001280$ u. Why negative? Products are heavier → energy absorbed. Step 4 — $Q = -0.001280\times931.5 = -1.19\ \text{MeV}$. Endothermic. Step 5 — threshold (α projectile, N target): $K_{\text{th}} = 1.19\left(1+\frac{4.0026}{14.0031}\right) = 1.19\times1.286 \approx \mathbf{1.53\ MeV}$. Why this step? You need at least this much α kinetic energy or the reaction can't proceed.

Worked Example 3 — Q from kinetic energies (no mass table!)

α-decay $^{210}$Po $\to$ $^{206}$Pb $+\ \alpha$. Suppose the α emerges with $K_\alpha = 5.30$ MeV and the daughter recoils with $K_{\text{Pb}}=0.10$ MeV; parent at rest.

$Q = K_{\text{final}} - K_{\text{initial}} = (5.30+0.10) - 0 = \mathbf{5.40\ MeV}$ Why? The KE form of $Q$ — released rest mass shows up as the total kinetic energy of fragments. No need to look up masses.

Connections

• Mass-energy equivalence E=mc^2
• Binding energy and mass defect
• Nuclear fission and Nuclear fusion
• Conservation of momentum in decays
• Alpha decay / Beta decay energetics
• Threshold energy in particle reactions

Concept Map

Hinglish (regional understanding)