3.8.1.5
Nuclear density is constant and independent of the size of the nucleus
Nuclear Structure & Radiation — AQA A-Level Physics
- Assuming the nucleus is spherical: $V = (4/3)\pi R^{3}$.
- Substituting R = R₀\(A^{1/3}\): $V = (4/3)\pi R_{0}^{3} A$.
- Volume is proportional to A (the number of nucleons).
- Mass $m = Au$, where u = 1.661 × 10⁻²⁷ kg is the atomic mass unit.
$$\rho = \frac{3u}{4\pi R_0^3}$$
- $ρ$: nuclear densityMass per unit volume of a material. Measured in kg m⁻³. (kg m⁻³)
- $u$: atomic mass unit = 1.661 × 10⁻²⁷ kg
- $R₀$: constant ≈ 1.05 × 10⁻¹⁵ m
- Since A cancels out, nuclear densityMass per unit volume of a material. Measured in kg m⁻³. is constant — it does not depend on which element you have.
- ρ ≈ 3.4 × 10¹⁷ kg m⁻³. This is enormously larger than atomic densityMass per unit volume of a material. Measured in kg m⁻³. (~10³ kg m⁻³).
- This confirms that almost all the mass of an atom is concentrated in its tiny nucleus, and atoms are predominantly empty space.
- Nucleons are evenly separated throughout the nucleus regardless of its size.
Examiner Tips and Tricks
- Nuclear density should always be of the order 10¹⁷ kg m⁻³.
- If your answer is significantly different, check your value of R₀ and the unit conversions.