Question 1 of 13intermediate🔧 ApplyNumerical2 marks

Given the mass of iron nucleus as 55.85u and A=56, find the nuclear density?

Correct Answer

The nuclear density is 2.29×102.29 \times 10¹⁷ kg m⁻³.

Exercise: Example 13.1 | Q: Example 13.1 | (Page 5)
For More Understanding

Explanation

This example from the textbook demonstrates that nuclear density is extremely large and independent of mass number A. The calculation uses the formula for nuclear radius R = R₀A^(1/3) where R₀ = 1.2×101.2 \times 10⁻¹⁵ m, and assumes a spherical nucleus. The result shows that nuclear matter has enormous density compared to ordinary matter like water (10³ kg m⁻³).

Solution Steps

  1. Step 1: Given mass of iron nucleus = 55.85u = 9.27×109.27 \times 10⁻²⁶ kg (since 1u ≈\approx 1.66×101.66 \times 10⁻²⁷ kg, and 55.85u ≈\approx 9.27×109.27 \times 10⁻²⁶ kg)

  2. Step 2: Calculate radius of nucleus using R = R₀A^(1/3), where R₀ = 1.2×101.2 \times 10⁻¹⁵ m and A = 56

  3. Step 3: Calculate volume of nucleus = (4π/3)R³ = (4π/3)(R₀A^(1/3))³ = (4π/3)R₀³A = (4π/3)(1.2×101.2 \times 10⁻¹⁵)³ × 56

  4. Step 4: Calculate nuclear density = mass/volume = (9.27×109.27 \times 10⁻²⁶)/((4π/3)(1.2×101.2 \times 10⁻¹⁵)³ × 56) = 2.29×102.29 \times 10¹⁷ kg m⁻³

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