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#density formula

2 public questions tagged with this topic.

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 .

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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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,

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

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