42-1 Discovering the Nucleus

42-1 Discovering the Nucleus#

Prompts

  • Before 1911, what was the prevailing model of the atom (plum pudding)? Why would that model predict only small deflections for alpha particles passing through a foil?

  • Describe the Rutherford scattering experiment: What are alpha particles? What was the setup? What was the surprising result?

  • Why did large-angle scattering (including backscattering) imply that the positive charge must be concentrated in a small nucleus rather than spread through the atom?

  • For an alpha particle headed directly toward a gold nucleus, use energy conservation to find the distance of closest approach \(d\). How does \(d\) depend on the initial kinetic energy \(K_i\)?

  • Rutherford concluded the nuclear radius is smaller than the atomic radius by a factor of about \(10^4\). What does that say about the atom?

Lecture Notes#

Overview#

  • In 1911, Rutherford (with Geiger and Marsden) discovered the nucleus by scattering alpha particles from thin metal foils.

  • The plum pudding model (positive charge spread through the atom) predicted only tiny deflections; the experiment showed large-angle and even backscattering — implying a small, dense, positively charged nucleus.

  • Energy conservation relates the alpha particle’s initial kinetic energy to its distance of closest approach to the nucleus.

  • The nucleus is much smaller than the atom (radius ratio \(\sim 10^{-4}\)); the atom is mostly empty space.

Before Rutherford: The Plum Pudding Model#

Around 1900, atoms were known to contain electrons (discovered by J. J. Thomson in 1897). Atoms are neutral, so positive charge must balance the electrons. Thomson’s plum pudding model proposed:

  • Positive charge spread uniformly through the volume of the atom.

  • Electrons (the “plums”) embedded in this sphere of positive charge (the “pudding”).

In this model, the maximum deflecting force on a passing alpha particle would be far too small to cause large deflections — like firing a bullet through a sack of snowballs.

The Rutherford Scattering Experiment#

Alpha particles (\(\alpha\)): emitted by radioactive elements (e.g., radon); we now know they are helium nuclei — 2 protons + 2 neutrons, charge \(+2e\), mass \(\approx 7300\times\) electron mass.

Setup (Geiger–Marsden, 1911–1913):

  • Alpha source (radon gas) → thin metal foil (e.g., gold) → detector.

  • The detector is rotated to count alpha particles scattered through various angles \(\phi\).

Result: Most particles scatter through small angles, but a small fraction scatter through large angles, some approaching 180° (backscattering).

Rutherford’s reaction

“It was quite the most incredible event that ever happened to me in my life. It was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”

Why Large-Angle Scattering Implies a Nucleus#

To deflect an alpha particle backward, a large force is required. That force can arise only if the positive charge is concentrated in a small region — a nucleus — so the alpha particle can get very close to it.

  • Plum pudding: charge spread over the whole atom → weak force at any point → tiny deflections.

  • Nuclear model: charge concentrated at center → alpha can approach closely → strong Coulomb repulsion → large deflections.

Most alpha particles pass far from any nucleus and are barely deflected; only those whose paths pass very close to a nucleus are scattered through large angles.

Distance of Closest Approach#

For an alpha particle headed directly toward a nucleus, we can find the distance of closest approach \(d\) using energy conservation.

Initially: the alpha “sees” a neutral atom (nucleus + electron cloud); \(U_i = 0\), \(K_i\) = initial kinetic energy.

As the alpha enters the atom, it passes through the electron cloud. By Gauss’s law, a spherical shell of charge has no effect on a charge inside it, so the alpha effectively “sees” only the nuclear charge. The repulsive Coulomb force slows it.

At closest approach: \(K_f = 0\) (momentarily at rest), \(U_f = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_\alpha q_{\text{nuc}}}{d}\).

Conservation of energy: \(K_i + U_i = K_f + U_f\)

(434)#\[ K_i + 0 = 0 + \frac{1}{4\pi\varepsilon_0}\frac{q_\alpha q_{\text{nuc}}}{d} \quad \Rightarrow \quad d = \frac{1}{4\pi\varepsilon_0}\frac{q_\alpha q_{\text{nuc}}}{K_i} \]

For alpha (\(q_\alpha = 2e\)) and gold nucleus (\(q_{\text{Au}} = 79e\)):

(435)#\[ d = \frac{(2e)(79e)}{4\pi\varepsilon_0 K_i} = \frac{158\,e^2}{4\pi\varepsilon_0 K_i} \]

Worked example

An alpha particle with \(K_i = 5.30\) MeV is headed directly at a gold nucleus. Find \(d\).

Solution: Using \(e^2/(4\pi\varepsilon_0) \approx 2.31\times 10^{-28}\) J·m and \(K_i = 5.30\) MeV \(= 8.48\times 10^{-13}\) J:

\[ d = \frac{158\,e^2}{4\pi\varepsilon_0\,K_i} = \frac{158 \times 2.31\times 10^{-28}}{8.48\times 10^{-13}} \approx 4.3\times 10^{-14}\ \text{m} \]

This is much larger than the nuclear radius; the alpha reverses without “touching” the nucleus.

Size of the Nucleus#

From the scattering data, Rutherford concluded that the nuclear radius is smaller than the atomic radius by a factor of about \(10^4\). The atom is mostly empty space — the nucleus occupies a tiny central region.

Summary#

  • Rutherford scattering of alpha particles from metal foils revealed the nucleus: positive charge concentrated at the center.

  • Large-angle scattering contradicted the plum pudding model and required a small, dense, positively charged core.

  • Energy conservation gives \(d = q_\alpha q_{\text{nuc}}/(4\pi\varepsilon_0 K_i)\) for the distance of closest approach in a head-on collision.

  • The nucleus is \(\sim 10^4\) times smaller than the atom; the atom is mostly empty space.