Deuteron problem (introductory)
Nuclear physics · Physics
Study notes
Worked example: Estimate the deuteron's size from its binding energy using the uncertainty principle. Step 1: Binding E_b = 2.225 MeV = 3.56×10⁻¹³ J. Reduced mass μ = m_p/2 ≈ 4.7×10⁻²⁸ kg. Step 2: Rough size from Δx·Δp ~ ℏ with Δp ~ √(2μE_b): Δp = √(2×4.7×10⁻²⁸×3.56×10⁻¹³) = 1.83×10⁻²⁰ kg·m/s. Step 3: Δx ~ ℏ/Δp = 1.055×10⁻³⁴/1.83×10⁻²⁰ ≈ 5.8×10⁻¹⁵ m = 5.8 fm. Step 4: Measured rms radius ≈ 2.1 fm — same order; the estimate overshoots because the true wave function has a long exponential tail (the hallmark of weak binding). Step 5: Compare the force range (~1–2 fm from pion exchange): the deuteron is bigger than the force's reach — it lives mostly outside the potential well! Conclusion: weak binding → extended halo-like wave function — the deuteron is less a tight ball than two nucleons holding hands at arm's length.