Rutherford realized that he could use these high-speed particles to probe inside other atoms. In 1909, Rutherford and his students set up the experiment shown in FIGURE 37.16 to shoot alpha particles through very thin metal foils. It was observed that some alpha particles were slightly deflected, but this was not surprising. The positive and negative charges of the atoms exert forces on the positively charged alpha particles as they pass through the foil, but Thomson’s raisin-cake model of the atom suggested that repulsive and attractive forces should be roughly balanced, causing only small deflections.

FIGURE 37.16 Rutherford’s experiment to shoot high-speed alpha particles through a thin gold foil.

At Rutherford’s suggestion, his students then set up the apparatus to see if any alpha particles were deflected at large angles. It took only a few days to find the answer. Not only were alpha particles deflected at large angles, but a very few were reflected almost straight backward toward the source!

How can we understand this result? FIGURE 37.17a shows that only a small deflection is expected for an alpha particle passing through a Thomson atom. But if an atom has a small, positive core, such as the one in FIGURE 37.17b, a few of the alpha particles can come very close to the core. Because the electric force varies with the inverse square of the distance, the very large force of this very close approach can cause a large-angle scattering or a backward deflection of the alpha particle.

FIGURE 37.17 Alpha particles interact differently with a concentrated positive nucleus than they would with the spread-out charge in Thomson’s model.

Thus the discovery of large-angle scattering of alpha particles led Rutherford to envision an atom in which negative electrons orbit an unbelievably small, massive, positive nucleus, rather like a miniature solar system. This is the nuclear model of the atom. Notice that nearly all of the atom is empty space—the void!

EXAMPLE 37.4 A nuclear physics experiment

An alpha particle is shot with a speed of $2.0\times {10}^7\text{m}\text{/}\text{s}$ directly toward the nucleus of a gold atom. What is the distance of closest approach to the nucleus?

MODEL Energy is conserved in electric interactions. Assume that the gold nucleus, which is much more massive than the alpha particle, does not move. Also recall that the exterior electric field and potential of a sphere of charge can be found by treating the total charge as a point charge at the center.

VISUALIZE FIGURE 37.18 is a pictorial representation. The motion is in and out along a straight line.

FIGURE 37.18 A before-and-after pictorial representation of an alpha particle colliding with a nucleus.

FIGURE 37.18 Full Alternative Text

SOLVE We are not interested in how long the collision takes or any of the details of the trajectory, so using conservation of energy rather than Newton’s laws is appropriate. Initially, when the alpha particle is very far away, the system has only kinetic energy. At the moment of closest approach, just before the alpha particle is reflected, the charges are at rest and the system has only potential energy. The conservation of energy statement $K_{\text{f}}+U_{\text{f}}=K_{\text{i}}+U_{\text{i}}$ is

$$0+\frac{1}{4\pi {\epsilon }_0}\frac{q_{\alpha }{q}_{\text{Au}}}{r_{\text{min}}}=\frac{1}{2}m{v}_{\text{i}}^2+0$$

where $q_{\alpha }$ is the alpha-particle charge and we’ve treated the gold nucleus as a point charge $q_{\text{Au}}$. The mass m is that of the alpha particle. The solution for $r_{\text{min}}$ is

$$r_{\text{min}}=\frac{1}{4\pi {ϵ}_0}\frac{2q_{\alpha }{q}_{\text{Au}}}{m{v}_{\text{i}}^2}$$

The alpha particle is a helium nucleus, so $m=4\text{ u}=6.64\times {10}^{-27}\text{ kg}$ and $q_{\alpha }=2e=3.20\times {10}^{-19}\text{C}$. Gold has atomic number 79, so $q_{\text{Au}}=79e=1.26\times {10}^{-17}\text{C}$. We can then calculate

$$r_{\text{min}}=2.7\times {10}^{-14}\text{m}$$

This is only about $1\text{/}10\text{,}000$ the size of the atom itself!

REVIEW We ignored the atom’s electrons in this example. In fact, they make almost no contribution to the alpha particle’s trajectory. The alpha particle is exceedingly massive compared to the electrons, and the electrons are spread out over a distance very large compared to the size of the nucleus. Hence the alpha particle easily pushes them aside without any noticeable change in its velocity.

Rutherford went on to make careful experiments of how the alpha particles scattered at different angles. From these experiments he deduced that the diameter of the atomic nucleus is $\approx 1\times {10}^{-14}\text{ m}=10\text{ fm}(1\text{ fm}=1\text{ femtometer}={10}^{-15}\text{ m})$, increasing a little for elements of higher atomic number and atomic mass.

It may seem surprising to you that the Rutherford model of the atom, with its solar-system analogy, was not Thomson’s original choice. However, scientists at the time could not imagine matter having the extraordinarily high density implied by a small nucleus. Neither could they understand what holds the nucleus together, why the positive charges do not push each other apart. Thomson’s model, in which the positive charge was spread out and balanced by the negative electrons, actually made more sense. It would be several decades before the forces holding the nucleus together began to be understood, but Rutherford’s evidence for a very small nucleus was indisputable.