#nuclear density

Radioactive Decay, Nuclear Forces and Stability

What is the nuclear density of a nucleus with mass \( 3.67 \times 10^{-27} \, \text{kg} \) and radius \( 2.0 \times 10^{

**Stability** belt of stability N≈Z for light, N>Z for heavy due to Coulomb, beyond leads to alpha decay, fission, stability requires balance, nuclear force saturated explains constant density and BE/A. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.0 × 10⁻¹⁵)³ = 8.0 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 8.0 × 10⁻⁴⁵ ≈ 3.35 × 10⁻⁴⁴ m³ . Density = (3.67 × 10⁻²⁷/3.35 × 10⁻⁴⁴) ≈ 1.10 × 10¹⁷ kg/m³ . 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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Radioactive Decay, Nuclear Forces and Stability

What is the nuclear density of a nucleus with mass \( 5.01 \times 10^{-27} \, \text{kg} \) and radius \( 2.1 \times 10^{

**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. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.1 × 10⁻¹⁵)³ = 9.261 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 9.261 × 10⁻⁴⁵ ≈ 3.88 × 10⁻⁴⁴ m³ . Density = (5.01 × 10⁻²⁷/3.88 × 10⁻⁴⁴) ≈ 1.29 × 10¹⁷ kg/m³ .

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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

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

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 nuclear density of a nucleus with mass \( 6.68 \times 10^{-27} \, \text{kg} \) and radius \( 2.4 \times 10^{

**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. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.4 × 10⁻¹⁵)³ = 1.382 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.382 × 10⁻⁴⁴ ≈ 5.79 × 10⁻⁴⁴ m³ . Density = (6.68 × 10⁻²⁷/5.79 × 10⁻⁴⁴) ≈ 1.15 × 10¹⁷ kg/m³ . 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

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Nuclear Size, Density and Structure

What is the nuclear density of a nucleus with mass \( 7.01 \times 10^{-27} \, \text{kg} \) and radius \( 2.5 \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 = (2.5 × 10⁻¹⁵)³ = 1.5625 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.5625 × 10⁻⁴⁴ ≈ 6.54 × 10⁻⁴⁴ m³ . Density = (7.01 × 10⁻²⁷/6.54 × 10⁻⁴⁴) ≈ 1.07 × 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 nuclear density of a nucleus with mass \( 2.33 \times 10^{-27} \, \text{kg} \) and radius \( 1.7 \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 = (1.7 × 10⁻¹⁵)³ = 4.913 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 4.913 × 10⁻⁴⁵ ≈ 2.06 × 10⁻⁴⁴ m³ . Density = (2.33 × 10⁻²⁷/2.06 × 10⁻⁴⁴) ≈ 1.13 × 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

Calculate the nuclear density of a nucleus if its mass is \( 1.67 \times 10^{-27} \, \text{kg} \) and radius is \( 1.2 \

**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. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (1.2 × 10⁻¹⁵)³ = 1.728 × 10⁻⁴⁵ m³ . Volume = (4/3) × 3.14 × 1.728 × 10⁻⁴⁵ = 7.24 × 10⁻⁴⁵ m³ . Density = (1.67 × 10⁻²⁷/7.24 × 10⁻⁴⁵) ≈ 2.31 × 10¹⁷ kg/m³ . 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

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