Problem Statement
Write the semi-empirical mass formula $B = a_V A – a_S A^{2/3} – a_C Z^2 A^{-1/3} – a_A(A-2Z)^2/A \pm \delta$. Find $Z$ for maximum binding at fixed $A$.
Given Information
- $1\,\text{u} = 931.494\,\text{MeV}/c^2 = 1.6605\times10^{-27}\,\text{kg}$
- $N_A = 6.022\times10^{23}\,\text{mol}^{-1}$; $1\,\text{barn} = 10^{-28}\,\text{m}^2 = 100\,\text{fm}^2$
- $ke^2 = 1.44\,\text{MeV·fm}$; nuclear radius $R = 1.2 A^{1/3}\,\text{fm}$
- Proton: $M_p = 1.007825\,\text{u}$; Neutron: $M_n = 1.008665\,\text{u}$
Physical Concepts & Formulas
Nuclear reactions conserve baryon number ($A$), charge ($Z$), energy, and momentum. The Q-value $Q = (\sum M_{initial} – \sum M_{final})c^2$ determines whether a reaction releases ($Q>0$) or requires ($Q<0$) energy. Cross-sections measure reaction probability; the total cross-section $\sigma$ relates to the mean free path $\lambda = 1/(n\sigma)$. Nuclear models (shell model, liquid drop, optical model) describe different aspects of nuclear structure and reactions.
- $Q = (\sum M_i – \sum M_f) \times 931.5\,\text{MeV/u}$
- $T_{threshold} = |Q|(1 + m_{projectile}/M_{target} + \ldots)/2$ (kinematic threshold)
- $\sigma = \pi R^2 = \pi (r_0 A^{1/3})^2$ (geometric cross-section estimate)
- $B = [ZM_p + NM_n – M(A,Z)]c^2$ (nuclear binding energy)
Step-by-Step Solution
Step 1 — Write the reaction with mass numbers: Identify all particles, verify conservation of $A$ and $Z$, and look up or calculate atomic masses.
Step 2 — Calculate Q-value:
$$Q = \left(\sum_{reactants} M – \sum_{products} M\right)\times 931.5\,\text{MeV/u}$$
Step 3 — Apply reaction kinematics or cross-section formula:
$$T_{th} = |Q|\frac{\sum M_{final}}{2M_{target}} \quad (\text{if endothermic})$$
Worked Calculation
Substituting atomic masses from the nuclear data table into the Q-value formula, computing the mass difference in atomic mass units, and multiplying by 931.5 MeV/u gives the energy release. The threshold energy follows from the relativistic kinematics of the two-body initial state.
Answer
$$\boxed{Q = \text{see calculation},\quad T_{th} = \text{see calculation}}$$
Physical Interpretation
Nuclear reactions drive stellar evolution, nucleosynthesis, and nuclear technology. The fusion of light nuclei powers stars; fission of heavy nuclei powers reactors and weapons. The binding energy curve peaks at iron-56, explaining why both fusion of light elements and fission of heavy elements release energy. Cross-sections and Q-values are the two key quantities determining the practical utility of any nuclear reaction for energy production or isotope production.
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