Nuclear Radius

Nuclear Radius

Nuclei are extraordinarily small — on the order of femtometres (1 fm = 10⁻¹⁵ m). Despite this, a remarkably simple empirical formula describes their size.

The Nuclear Radius Formula

The nuclear radius scales with the cube root of the mass number $A$ (the total number of protons and neutrons):

$R = R_0 \cdot A^{1/3}$

where $R_0 \approx 1.2 \text{ fm} = 1.2 \times 10^{-15}$ m is the nuclear radius constant.

Nuclear Volume

Since $R \propto A^{1/3}$, the nuclear volume $V = \frac{4}{3}\pi R^3$ scales linearly with $A$:

$V = \frac{4}{3}\pi R_0^3 \cdot A$

Nuclear Density

The nuclear density is strikingly constant for all nuclei:

$\rho = \frac{A \cdot m_u}{V} = \frac{m_u}{\frac{4}{3}\pi R_0^3} \approx 2.3 \times 10^{17} \text{ kg/m}^3$

where $m_u = 1.66054 \times 10^{-27}$ kg is the atomic mass unit. This density is roughly $10^{14}$ times denser than water — a teaspoon of nuclear matter would weigh about 500 million tonnes.

Your Task

Implement three functions. All constants must be defined inside each function.

• nuclear_radius(A) — returns $R = R_0 \cdot A^{1/3}$ in metres
• nuclear_volume(A) — returns $V = \frac{4}{3}\pi R^3$ in m³
• nuclear_density(A) — returns $\rho = A \cdot m_u / V$ in kg/m³

Use $R_0 = 1.2 \times 10^{-15}$ m and $m_u = 1.66054 \times 10^{-27}$ kg.