Nucleon number and nuclear radius

The nucleon number (A) represents the total number of protons and neutrons within a nucleus. Although protons were known during Rutherford's era, neutrons were not discovered until 1932 because their lack of charge made them challenging to detect using early 20th-century research methods that depended on ionisation effects.

The radius (R) of a nucleus is proportional to the cube root of its total nucleon number (A), assuming the nucleus is spherical. This relationship can be expressed mathematically as:

$R=R_0{A}^{\frac{1}{3}}$

Where:

• R = Radius (m)
• R_0_ = Fermi radius, an experimentally determined value equal to $1.2\times {10}^{-15}\text{ m}$.
• A = nucleon number

Worked example - Calculating nuclear radius

To calculate the nuclear radius for a nucleus with a nucleon number (A) of 64, using the Fermi radius $R_0=1.2\times {10}^{-15}\text{ m}$, follow these steps:

Step 1: Formula

$R=R_0{A}^{\frac{1}{3}}$

Step 2: Substitution and correct evaluation

$R=1.2\times {10}^{-15}\times {64}^{\frac{1}{3}}{\text{ = 4.8 x 10}}^{-15}\text{ m}$

Calculating nuclear density

Nuclear density (${\rho }_n$) can be estimated by dividing the mass of the nucleus by its volume.

total mass of nucleus = A u

radius = $R_0{\text{ A}}^{\frac{1}{3}}$

volume = $\frac{4}{3}\pi {\left(R_0{A}^{\frac{1}{3}}\right)}^3=\frac{4}{3}\pi R_0^{3}A$

This leads to the equation:

${\rho }_n=\frac{mass}{volume}=\frac{A\times u}{\frac{4}{3}\pi R_0^{3}A}=\frac{3u}{4\pi R_0^3}$

Where:

• $\rho$n = nuclear density (kg m^-3^)
• A = nucleon number
• u = unified atomic mass unit ($1.661\times {10}^{-27}\text{ kg}$)
• R_0_ = Fermi radius (1.2 x 10^-15^ m)

Notice the nucleon number cancels out due to appearing in the numerator and denominator.

This means nuclear density is not dependent on nucleon number and all nuclei have the same density.

Worked example - Calculating nuclear density

To calculate the nuclear density, given the unified atomic mass unit $u=1.661\times {10}^{-27}\text{ kg}$ and the Fermi radius $R_0=1.2\times {10}^{-15}\text{ m}$, follow these steps:

Step 1: Formula

${\rho }_n=\frac{3u}{4\pi R_0^3}$

Step 2: Substitution and correct evaluation

${\rho }_n\text{ = }\frac{3u}{4\pi R_0^3}\text{ = }\frac{3\times 1.661\times {10}^{-27}}{4\pi \times {(1.2\times {10}^{-15})}^3}\text{ = }2.29\times {10}^{17}{\text{ kg m}}^{-3}$