Nuclear Size, Density and Structure

Nuclear Size, Density and Structure

What is the energy equivalent of \( 0.001 \, \text{kg} \) of matter in Joules? (Given \( c = 3 \times 10^8 \, \text{m/s}

**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. E = m c² . m = 0.001 kg , c² = (3 × 10⁸)² = 9 × 10¹⁶ m²/s² . E = 0.001 × 9 × 10¹⁶ = 9 × 10¹³ J . 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 9 × 10¹³ J, consistent with Bohr model and nuclear binding energy systematics.

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Nuclear Size, Density and Structure

What is the binding energy of a nucleus with mass defect \( 0.12 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \, \tex

**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. E_b = Δ M · c² . Δ M = 0.12 u . E_b = 0.12 × 931.5 = 111.78 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 111.78 MeV, consistent with Bohr model and nuclear binding energy systematics.

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 8.35 \times 10^{-27} \, \text{kg} \) and radius \( 2.7 \times 10^{

**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. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.7 × 10⁻¹⁵)³ = 1.9683 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.9683 × 10⁻⁴⁴ ≈ 8.25 × 10⁻⁴⁴ m³ . Density = (8.35 × 10⁻²⁷/8.25 × 10⁻⁴⁴) ≈ 1.01 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n

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Nuclear Size, Density and Structure

What is the binding energy of a nucleus with a mass defect of \( 0.09 \, \text{u} \)? (Given \( 1 \, \text{u} = 931.5 \,

**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. Binding energy = Δ M · c² . Δ M = 0.09 u . E_b = 0.09 × 931.5 = 83.835 MeV ≈ 83.84 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 83.84 MeV, consistent

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Nuclear Size, Density and Structure

Why is the nuclear density nearly constant across all nuclei?

**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. The volume of a nucleus is proportional to A (since R ∝ A¹/³ and V ∝ R³ ), and the mass is also proportional to A , making the density (mass/volume) independent of A . 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 Volume proportional to mass number, consistent with Bohr

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Nuclear Size, Density and Structure

A nucleus with mass number 50 has a binding energy of \( 425 \, \text{MeV} \). What is its binding energy per nucleon?

**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. Ebₙ = (E_b/A) . E_b = 425 MeV , A = 50 . Ebₙ = (425/50) = 8.5 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 8.5 MeV, consistent with Bohr model and nuclear binding energy systematics.

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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

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 1.66 \times 10^{-27} \, \text{kg} \) and radius \( 1.5 \times 10^{

**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. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (1.5 × 10⁻¹⁵)³ = 3.375 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 3.375 × 10⁻⁴⁵ ≈ 1.41 × 10⁻⁴⁴ m³ . Density = (1.66 × 10⁻²⁷/1.41 × 10⁻⁴⁴) ≈ 1.18 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀,

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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

Which factor primarily determines the nuclear radius?

**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. The nuclear radius is given by R = R₀ A¹/³ , where A (mass number) is the key factor determining the size, as the radius scales with the cube root of the number of nucleons. 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

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Nuclear Size, Density and Structure

What is the radius of a nucleus with mass number 200? (Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

**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. R = R₀ A¹/³ . A = 200 , A¹/³ = 200¹/³ ≈ 5.85 . R = 1.2 × 10⁻¹⁵ × 5.85 ≈ 7.0 × 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 7.0 × 10⁻¹⁵ m, consistent with Bohr model and nuclear binding energy systematics.

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Nuclear Size, Density and Structure

A nucleus has a radius of \( 3.6 \times 10^{-15} \, \text{m} \). What is its approximate mass number? (Given \( R_0 = 1.

**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. R = R₀ A¹/³ . 3.6 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (3.6/1.2) = 3 . A = 3³ = 27 . 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 27, consistent with Bohr model and nuclear binding energy systematics.

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