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#mass number 216

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 216 has a radius of \( 7.2 \times 10^{-15} \, \text{m} \). What is the value of \( R_0 \)?

**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. R = R₀ A¹/³ . A = 216 , A¹/³ = 6 . R₀ = (R/A¹/³) = (7.2 × 10⁻¹⁵/6) = 1.2 × 10⁻¹⁵ m . 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 1.2 × 10⁻¹⁵ m, 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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Nuclear Size, Density and Structure

What is the ratio of nuclear radii of nuclei with mass numbers 216 and 8?

**Nuclear radius** R = R₀ A^{1/3}, R₀=1.2×10⁻¹⁵ m, A mass number, for A=16 R=1.2×10⁻¹⁵×2.52=3.02×10⁻¹⁵ m, for A=36 R=1.2×10⁻¹⁵×3.30=3.96×10⁻¹⁵ m, nuclear density ρ = mass/volume = (A×1.66×10⁻²⁷ kg)/(4/3 π R³) ≈2.3×10¹⁷ kg/m³ independent of A, extremely high, mass 3.67×10⁻²⁷ kg radius 2.0×10⁻¹⁵ m gives density =3.67×10⁻²⁷/(4/3 π×8×10⁻⁴⁵)=1.09×10¹⁷ kg/m³. Radius ratio = (R₁/R₂) = ( (A₁/A₂) )¹/³ . A₁ = 216 , A₂ = 8 . (216/8) = 27 , (27)¹/³ = 3 . 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 3.0, consistent with Bohr

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Size, Density and Structure

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