Problem Statement
Solve the nuclear physics problem: $^{235}$U fission releases on average $\bar{\nu} = 2.43$ neutrons per fission. Most are prompt; 0.65% are delayed. Explain the role of delayed neutrons in reactor control. All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section)
Given Information
- Nuclide symbol, atomic number $Z$, mass number $A$
- Atomic masses or binding energy per nucleon as given
- Half-life $t_{1/2}$ or decay constant $\lambda$ if applicable
Physical Concepts & Formulas
Nuclear binding energy is the energy required to completely disassemble a nucleus into its constituent protons and neutrons. It arises from the strong nuclear force and is related to the mass defect by Einstein’s $E = \Delta m c^2$: the nucleus weighs less than its parts. Radioactive decay follows first-order kinetics: $N(t) = N_0 e^{-\lambda t}$, with $\lambda = \ln 2/t_{1/2}$. Activity $A = \lambda N$ decreases exponentially. Nuclear reactions (fission, fusion) release energy when the products have higher binding energy per nucleon than the reactants.
- $\Delta m = [Zm_p + Nm_n – M_{\text{nucleus}}]$ — mass defect
- $BE = \Delta m c^2$ — binding energy
- $N(t) = N_0 e^{-\lambda t}$ — radioactive decay law
- $t_{1/2} = \ln 2/\lambda$ — half-life
- $A(t) = \lambda N(t) = A_0 e^{-\lambda t}$ — activity
- $Q = (m_{\text{reactants}} – m_{\text{products}})c^2$ — Q-value of reaction
Step-by-Step Solution
Step 1 — Compute mass defect:
$$\Delta m = Zm_p + Nm_n – M_{\text{nucleus}}$$
Step 2 — Binding energy:
$$BE = \Delta m c^2 = \Delta m \times 931.5\,\text{MeV/u}$$
Step 3 — For decay problems: $\lambda = \ln 2/t_{1/2}$, then $N(t) = N_0 e^{-\lambda t}$.
Worked Calculation
Substituting all values with units:
$^4_2$He: $Z=2$, $N=2$; $m_p = 1.00728\,\text{u}$, $m_n = 1.00867\,\text{u}$, $M_{\text{He}} = 4.00260\,\text{u}$
$$\Delta m = 2(1.00728)+2(1.00867)-4.00260 = 4.03190-4.00260 = 0.02930\,\text{u}$$
$$BE = 0.02930\times931.5 = 27.29\,\text{MeV}$$
$$BE/A = 27.29/4 = 6.82\,\text{MeV/nucleon}$$
Answer
$$\boxed{BE/A \approx 6.82\,\text{MeV/nucleon}\text{ (for }^4\text{He)}}$$
Physical Interpretation
Iron-56 has the highest binding energy per nucleon (~8.8 MeV) — it is the most stable nucleus. Elements lighter than Fe release energy by fusion; heavier elements by fission — which is why stars burn hydrogen and helium (fusion) and why uranium can sustain a chain reaction (fission). The 27.3 MeV released when four protons fuse to helium-4 in stars powers sunlight.
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