#nuclear size

Nuclear Size, Density and Structure

What is the approximate radius of a nucleus with mass number 64? (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 = 64 . A¹/³ = 64¹/³ = 4 . R = 1.2 × 10⁻¹⁵ × 4 = 4.8 × 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 ratio of nuclear radii of two nuclei with mass numbers 27 and 125?

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. Radius ratio = (R₁/R₂) = (R₀ A₁¹/³/R₀ A₂¹/³) = ( (A₁/A₂) )¹/³ . A₁ = 27 , A₂ = 125 . (27/125) = 0.216 , (0.216)¹/³ ≈ 0.6 . 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 0.6, consistent with Bohr model and nuclear binding energy systematics.

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Nuclear Size, Density and Structure

What is the radius of a nucleus with mass number 32? (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. R = R₀ A¹/³ . A = 32 , A¹/³ = 32¹/³ ≈ 3.17 . R = 1.2 × 10⁻¹⁵ × 3.17 ≈ 3.8 × 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 3.8

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

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Atomic Models - Rutherford, Thomson and Bohr

In Rutherford’s model, what is the order of magnitude of the ratio of the atomic size to the nuclear size?

**Rutherford's nuclear model** atom has small massive positively charged nucleus with electrons orbiting, size ratio atomic to nuclear ~10⁵, nucleus ~10⁻¹⁵ m atom ~10⁻¹⁰ m, most alpha particles with large impact parameter pass undeflected, small fraction >90° scatter from close approach, centripetal force provided by Coulomb attraction k Z e²/r², fails to explain stability because accelerating charge should radiate and collapse. Atomic size ≈ 10⁻¹⁰ m , nuclear size ≈ 10⁻¹⁵ m . Ratio = (10⁻¹⁰/10⁻¹⁵) = 10⁵ . 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

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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