Nuclear radius (AQA A-Level Physics): Revision Notes

Nuclear radius

Understanding the size of a nucleus

When we model the nucleus as a collection of tightly packed protons and neutrons (collectively called nucleons), we can explore the relationship between nuclear size and the number of nucleons present. The volume of a nucleus with nucleon number A (mass number) is directly proportional to the number of nucleons it contains:

$V = A \times \text{volume of one nucleon}$

Deriving the nuclear radius formula

If we treat the nucleus as spherical with radius $R$, its volume can be expressed using the standard formula for the volume of a sphere:

$V = \frac{4}{3}\pi R^3$

Since the total nuclear volume equals the nucleon number multiplied by the volume of a single nucleon, we can write:

$\frac{4}{3}\pi R^3 = A \times \text{volume of one nucleon}$

Rearranging this expression gives:

$R^3 = \frac{A \times \text{volume of one nucleon}}{\frac{4}{3}\pi}$

This can be simplified to:

$R^3 = \frac{A}{\frac{4\pi}{3}} \times \text{volume of one nucleon}$

If we assume that the volume occupied by each nucleon remains constant across all nuclei, then $R^3$ is proportional to $A$. This leads to the relationship:

$R \propto A^{1/3}$

This proportionality is typically written as:

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

where $R_0$ is the constant of proportionality. Experimental measurements have determined that $R_0 = 1.05$ fm (where fm stands for femtometre, equal to $10^{-15}$ m).

Experimental methods to measure nuclear radius

Historical approach - Rutherford's alpha scattering

Rutherford's famous alpha particle scattering experiment, which involved firing alpha particles at thin gold foil, provided an early estimate for nuclear size. By analyzing the electrostatic repulsion between alpha particles and gold nuclei, Rutherford established an upper limit for the gold nucleus radius of approximately $10^{-14}$ m. However, this method could not provide precise measurements of nuclear radii for different elements.

Modern approach - electron diffraction

A more accurate and versatile method uses high-energy electron beams directed at thin samples of various elements. The electrons scatter through different angles, and detectors count the number of electrons scattered at each angle within a specified time interval. This produces a diffraction pattern with characteristic maxima (peaks) and minima (troughs).

The scattering occurs because electrons carry charge and experience an attractive electromagnetic force when passing near the positively charged protons in the nucleus. The resulting diffraction pattern depends on the size of the nucleus, and by analyzing the angles at which minima occur, we can determine nuclear radii for different elements.

Why electrons are superior to alpha particles

Electrons offer several advantages over alpha particles as probing particles:

Interaction mechanism: Electrons, being leptons with no internal structure, interact with nuclei solely through the electromagnetic force arising from their charge. This interaction is well understood theoretically. Alpha particles, being composite particles made of protons and neutrons (hadrons), also interact via the strong nuclear force, which is less well understood and complicates the analysis.

Particle mass: Electrons have a much smaller mass than alpha particles. When accelerated to very high speeds in a particle accelerator, electrons can achieve de Broglie wavelengths comparable to nuclear dimensions.

De Broglie wavelength matching: The de Broglie wavelength of electrons accelerated through a potential difference of 400 MV is approximately $10^{-15}$ m, which matches the typical size of nuclear radii. This similarity is essential for producing clear diffraction patterns. When a wave encounters an obstacle of similar size to its wavelength, it produces pronounced diffraction effects that can be analyzed to determine the size of the obstacle.

Experimental results

When electron diffraction data from many different elements are collected and plotted as a graph of nuclear radius $R$ against $A^{1/3}$, the result is a straight line passing through the origin. The gradient of this line is 1.05 fm, confirming the relationship $R = R_0 A^{1/3}$ where $R_0 = 1.05$ fm. This experimental verification supports the theoretical prediction that nuclear radius depends on the cube root of the nucleon number.

Nuclear density

Calculating nuclear density

The density $\rho$ of any material is defined as its mass divided by its volume:

$\rho = \frac{M}{V}$

For a nucleus containing $A$ nucleons, where each nucleon has an atomic mass unit (u) equal to $1.661 \times 10^{-27}$ kg, the total mass is:

$M = A \times 1.661 \times 10^{-27} \text{ kg}$

The volume of the nucleus, using $R = R_0 A^{1/3}$, is:

$V = \frac{4}{3}\pi(R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$

Therefore, the density becomes:

$\rho = \frac{A \times 1.661 \times 10^{-27}}{\frac{4}{3}\pi R_0^3 A}$

Notice that the nucleon number $A$ appears in both the numerator and denominator, so it cancels out:

$\rho = \frac{1.661 \times 10^{-27}}{\frac{4}{3}\pi R_0^3}$

Substituting $R_0 = 1.05 \times 10^{-15}$ m:

$\rho = 3.4 \times 10^{17} \text{ kg} \cdot \text{m}^{-3}$

The constant density result

This remarkable result shows that the density of a nucleus does not depend on the nucleon number. Whether we examine a light nucleus like carbon or a heavy nucleus like uranium, the density remains the same at approximately $3.4 \times 10^{17}$ kg m$^{-3}$. This confirms our initial assumption that nucleons are packed together with constant density throughout the nucleus.