Problem 6.71 — Characteristic X-ray K-alpha for Tungsten

Problem Statement

Solve the nuclear physics problem: Estimate the $K_\alpha$ energy and wavelength for tungsten ($Z=74$) via Moseley’s law. $$E_{K_\alpha} = 10.2(Z-1)^2 = 10.2\times(73)^2 = 10.2\times5329 = 54356 \text{ eV} \approx 54.4 \text{ keV}$$ $$\lambda = 1240/54400 = 0.0228 \text{ nm}$$ (Actual: 59.3 keV — Moseley’s law is approximate)

Given Information

  • All quantities, constants, and constraints stated in the problem above
  • Physical constants used as needed (see Concepts section)

Physical Concepts & Formulas

This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.

  • Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
  • State the mathematical form of that law as it applies here
  • Check dimensions at every step: both sides of an equation must have the same units

Step-by-Step Solution

Problem Statement

Solve the nuclear physics problem: Estimate the $K_\alpha$ energy and wavelength for tungsten ($Z=74$) via Moseley’s law. $$E_{K_\alpha} = 10.2(Z-1)^2 = 10.2\times(73)^2 = 10.2\times5329 = 54356 \text{ eV} \approx 54.4 \text{ keV}$$ $$\lambda = 1240/54400 = 0.0228 \text{ nm}$$ (Actual: 59.3 keV — Moseley’s law is approximate)

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.

Worked Calculation

Substituting all given numerical values with their units into the derived formula:

$$\text{Numerical result} = \text{given expression substituted with values}$$

Answer

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

Physical Interpretation

The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.


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