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#mass number 16

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Radioactive Decay, Nuclear Forces and Stability

A nucleus has a binding energy of \( 127.5 \, \text{MeV} \) and mass number 16. What is its binding energy per nucleon?

**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. Ebₙ = (E_b/A) . E_b = 127.5 MeV , A = 16 . Ebₙ = (127.5/16) ≈ 7.97 MeV ≈ 8 MeV . 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 8.0 MeV, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Radioactive Decay, Nuclear Forces and Stability

What is the radius of a nucleus with mass number 16? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. The radius of a nucleus is given by R = R₀ A¹/³ , where A = 16 . A¹/³ = 16¹/³ = (2⁴)¹/³ = 2⁴/³ ≈ 2.52 . R = 1.2 × 10⁻¹⁵ × 2.52 ≈ 3.0 × 10⁻¹⁵ m . Using E_n = -13.6/n² eV, r_n = n²

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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