Practice question
Question
Why is the binding energy per nucleon lower in very heavy nuclei?
Explanation
**Radioactive decay** occurs when nucleus unstable, alpha decay emits He-4, beta decay neutron→proton+electron+antineutrino, gamma decay photon emission, decay law N=N₀ e^{-λt}, half-life T½=ln2/λ, nuclear density ~10¹⁷ kg/m³, nuclear force saturated means BE/A constant for A>20. In very heavy nuclei (A > 170), the increased Coulomb repulsion between protons reduces the net binding energy per nucleon, as the nuclear force cannot fully counteract this effect. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Increased Coulomb repulsion, consistent with Bohr model and nuclear binding energy systematics.
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