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#size estimation

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Nuclear Size, Density and Structure

The radius of a nucleus is \( 6.0 \times 10^{-15} \, \text{m} \). What is its approximate mass number? (Given \( R_0 = 1

**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. R = R₀ A¹/³ . 6.0 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (6.0/1.2) = 5 . A = 5³ = 125 . 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 125, consistent

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

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Nuclear Size, Density and Structure

What is the approximate radius of a nucleus with mass number 8? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**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. The radius of a nucleus is given by R = R₀ A¹/³ , where A = 8 . A¹/³ = 8¹/³ = 2 . R = 1.2 × 10⁻¹⁵ × 2 = 2.4 × 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

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

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Nuclear Size, Density and Structure

What is the approximate mass number of a nucleus with radius \( 3.3 \times 10^{-15} \, \text{m} \)? (Given \( R_0 = 1.2

**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. R = R₀ A¹/³ . 3.3 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (3.3/1.2) = 2.75 . A = (2.75)³ ≈ 20.8 ≈ 21 . 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 21, consistent with

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

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