E1-1. Rutherford's Gold Foil Experiment, Nuclear Notation, and Nuclear Size

The Apparatus & Setup

At the beginning of the 20th century, Ernest Rutherford, along with Hans Geiger and Ernest Marsden, performed a foundational scattering experiment that dismantled the prevailing "Plum Pudding" model proposed by J.J. Thomson. The setup consisted of:

• Alpha Particle Source: High-energy, positively charged $\alpha$-particles (helium nuclei, $_2^4\text{He}^{2+}$) were emitted from a radioactive substance enclosed in a heavy lead container. The lead contained a long, narrow collimating aperture, creating a tight, directed beam.
• Thin Gold Foil: A target sheet of gold hammer-beaten to an extreme thinness of approximately $10^{-6}\text{ m}$ (roughly 400 atoms thick). Gold was selected because its extreme malleability allows it to be hammered into sheets thin enough that particles undergo single-scattering events, rather than multiple confusing collisions, without completely absorbing the beam.
• Movable Detector: A zinc sulfide ($\text{ZnS}$) screen coupled with a microscope that produced microscopic light flashes (scintillations) when struck by an alpha particle. This detector could rotate $360^{\circ}$ around the target in an evacuated path.
• Evacuated Chamber: The entire apparatus was sealed inside a ultra-high vacuum. Since $\alpha$-particles are highly ionizing and have a short range in air (only about $5\text{ cm}$), any atmospheric gas molecules would cause extraneous collisions, skewing the angular distribution data.

Observations and Geometric Conclusions

The classical expectations dictated that the weak, diffuse electric fields of the "Plum Pudding" atom would only produce tiny deflections of a fraction of a degree. Instead, the experimental distribution yielded three extraordinary insights:

1. Observation: The overwhelming majority of $\alpha$-particles passed through the foil entirely undeflected.
   Deduction: The atom consists almost entirely of empty space.
2. Observation: A small fraction of $\alpha$-particles experienced deflections at small angles ($< 10^{\circ}$).
   Deduction: There is a localized positive charge center within the atom that exerts a repulsive electrostatic Coulomb force on the incoming positive $\alpha$-particles.
3. Observation: An extremely rare fraction (approximately $1$ in $8000$) were deflected through massive angles ($> 90^{\circ}$), with some bouncing almost straight back.
   Deduction: The positive charge and virtually all of the atomic mass are concentrated in an incredibly small, extremely dense volume called the nucleus. The orbital electrons possess negligible inertia compared to the nucleus and do not affect the trajectory of the heavy alpha particle.

This led to the planetary/nuclear model, revealing that the atomic diameter is on the order of $10^{-10}\text{ m}$ (1 Ångström), whereas the nuclear diameter is constrained to roughly $10^{-15}\text{ m}$ (1 femtometer or Fermi). The atom is roughly $100,000$ times larger than its nucleus by radius, or $10^{15}$ times larger by volume!

To characterize the subatomic configuration of a nucleus, standard standard nuclear shorthand is employed:

• $X$: The unique chemical symbol of the element.
• $A$ (Nucleon Number / Mass Number): The total integer count of nucleons (protons $+$ neutrons) residing in the nucleus. This defines the approximate mass of the nucleus in atomic mass units ($\text{u}$).
• $Z$ (Proton Number / Atomic Number): The total number of protons within the nucleus. This establishes the net nuclear charge ($q = +Ze$) and identifies the chemical identity of the atom.
• The total neutron number $N$ can easily be computed via $N = A - Z$.

The Nuclear Radius Formula

By probing various nuclei with ultra-high-energy particle beams (such as high-energy electron scattering), physicists discovered that the volume of a nucleus scales linearly with its mass number $A$. Modeling the nucleus as a packed sphere of nucleons, the nuclear radius $R$ obeys the empirical relationship:

Where $R_0$ is a constant of proportionality known as the Fermi radius or nuclear radius parameter:

Physical significance: $R_0$ represents the approximate radius of a simple hydrogen nucleus (${}_{1}^{1}\text{H}$), where $A = 1$.

Mathematical Proof of Constant Nuclear Density

An extraordinary property of nuclear matter is that all nuclei, from light Hydrogen to ultra-heavy Uranium, share roughly the exact same core density. Let's prove this analytically:

1. Assume a perfectly spherical nuclear geometry. The nuclear volume $V$ is given by: $$V = \frac{4}{3}\pi R^3$$
2. Substitute the radius relation $R = R_0 A^{1/3}$ into the volume formula: $$V = \frac{4}{3}\pi \left(R_0 A^{1/3}\right)^3 = \frac{4}{3}\pi R_0^3 A$$ Thus, nuclear volume is directly proportional to nucleon count ($V \propto A$).
3. The total mass $M$ of a nucleus can be approximated by multiplying the total number of nucleons $A$ by the average atomic mass unit $u$ ($M \approx A \cdot u$, where $1\text{ u} \approx 1.661 \times 10^{-27}\text{ kg}$).
4. Now, formulate the mass density $\rho_{\text{nuclear}}$: $$\rho_{\text{nuclear}} = \frac{M}{V} = \frac{3A \cdot u}{4\pi R_0^3 A} = \frac{3u}{4\pi R_0^3}$$

Notice that the mass number $A$ completely cancels out from the numerator and denominator! This demonstrates that nuclear density is an absolute constant independent of the size or type of the nucleus. Packing more nucleons into a nucleus increases its mass and volume at exactly identical proportional rates.