Skip to content

Question

What is the approximate radius of a nucleus with mass number 64? (Given \( R_0 = 1.2 \times 10^{-15} \,
\text{m} \))

Options

Choose one · Correct answer highlighted

Explanation

**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

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.