Skip to content

Question

What is the nuclear density of a nucleus with mass \( 5.01 \times 10^{-27} \, \text{kg} \) and radius
\( 2.1 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

Options

Choose one · Correct answer highlighted

Explanation

**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 .

Discussion

Comments

0 comments

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