#atomic nucleus

Nuclear Reactions - Fission, Fusion and Energy Release

What is the mass number of a nucleus with radius \( 5.4 \times 10^{-15} \, \text{m} \)? (Given \( R_0 = 1.2 \times 10^{-

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. R = R₀ A¹/³ . 5.4 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (5.4/1.2) = 4.5 . A = (4.5)³ = 91.125 ≈ 91 . 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

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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 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

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Nuclear Size, Density and Structure

What is the radius of a nucleus with mass number 4? (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. The radius of a nucleus is given by R = R₀ A¹/³ , where A = 4 . A¹/³ = 4¹/³ ≈ 1.59 . R = 1.2 × 10⁻¹⁵ × 1.59 ≈ 1.9 × 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 1.9 × 10⁻¹⁵ m, consistent with

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De Broglie Hypothesis and Quantization in Bohr Model

What was the primary evidence from Rutherford’s alpha-particle scattering experiment that led to the nuclear model of th

**De Broglie hypothesis** λ = h/p, p=mv momentum, suggests electron as wave, in Bohr model circumference 2πr = n λ, standing wave condition, n wavelengths fit into orbit, for n=6, 6 wavelengths, for n=4, 4 wavelengths, explains quantization of angular momentum L = r p = r h/λ = r h n/(2πr)= n h/2π = n ħ, physical basis for Bohr quantization. The observation that a small fraction of alpha-particles were deflected by large angles (e.g., more than 90°) indicated a small, dense, positively charged nucleus at the atom’s center. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =

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