Problem 2.63 — Self-Diffusion vs Viscosity

Problem Statement

Solve the nuclear physics problem: Show that the self-diffusion coefficient $D$ and dynamic viscosity $\eta$ of a gas are related by $D = \eta/\rho$. From kinetic theory: $$D = \frac{1}{3}\bar{v}\langle l\rangle, \qquad \eta = \frac{1}{3}\rho\bar{v}\langle l\rangle$$ Dividing: $$\frac{D}{\eta/\rho} = \frac{\frac{1}{3}\bar{v}\langle l

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$D = \frac{1}{3}\bar{v}\langle l\rangle, \qquad \eta = \frac{1}{3}\rho\bar{v}\langle l\rangle$$

$$\frac{D}{\eta/\rho} = \frac{\frac{1}{3}\bar{v}\langle l

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:

$$

$$

Step 2 — Binding energy:

$$

Answer

$$\boxed{BE/A \approx 6.82\,\text{MeV/nucleon}\text{ (for }^4\text{He)}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.