Problem Statement
Solve the nuclear physics problem: Solve the nuclear physics problem: Problem 6.212 — Beta Decay: Fermi-Kurie Plot for Neutrino Mass See problem statement for all given quantities. This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equati
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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