1. Calculate the Volume of the Nucleus
* The nucleus is assumed to be spherical.
* Volume of a sphere = (4/3)πr³
* Where r is the radius of the nucleus (3.555 fm = 3.555 x 10⁻¹⁵ m)
Volume = (4/3) * π * (3.555 x 10⁻¹⁵ m)³
Volume ≈ 1.88 x 10⁻⁴³ m³
2. Calculate the Mass of the Nucleus
* The atomic mass of iron is 56, which means it has 56 nucleons (protons and neutrons).
* The mass of one nucleon is approximately 1.67 x 10⁻²⁷ kg.
Mass of the nucleus = 56 nucleons * 1.67 x 10⁻²⁷ kg/nucleon
Mass of the nucleus ≈ 9.35 x 10⁻²⁶ kg
3. Calculate the Density
* Density = Mass / Volume
Density = (9.35 x 10⁻²⁶ kg) / (1.88 x 10⁻⁴³ m³)
Density ≈ 4.97 x 10¹⁷ kg/m³
Therefore, the density of an iron nucleus is approximately 4.97 x 10¹⁷ kg/m³.
Important Note: This calculation assumes a uniform distribution of mass within the nucleus. In reality, the distribution of protons and neutrons within the nucleus is more complex.
