Problem Statement
Solve the nuclear physics problem: Solve the nuclear physics problem: Problem 6.201 — Neutron Star: Nuclear Density $M = 1.4M$ $\rho_{NS}/\rho_N = 6.65/$ This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving
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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