SCATTERING EXPERIMENTS AND THE THOMSON MODEL

An early model of the structure of the atom was proposed (in 1904) by J. J.

Thomson, who was known for his previous identification of the electron and measurement of its charge-to-mass ratio e/m. The Thomson model incorporates many of the known properties of atoms: size, mass, number of electrons, and electrical neutrality. In this model, an atom contains Z electrons that are embedded in a uniform sphere of positive charge (Figure 6.1). The total positive charge of the sphere is Ze, the mass of the sphere is essentially the mass of the atom (the electrons don’t contribute significantly to the total mass), and the radius R of the sphere is the radius of the atom. (This model is sometimes known as the “plum-pudding” model, because the electrons are distributed throughout the atom like raisins in a plum pudding.) As we will see, the Thomson model gives predictions that disagree with experiment, and so it is not the correct way of understanding the structure of atoms.

One way of studying atoms is by probing the distribution of electric charge in their interior, which we can do by bombarding the atom with charged particles and observing the angle by which particles are deflected from their original direction.

This type of experiment is called a scattering experiment. Ideally we would like to do this experiment with a single atom, such as is represented in Figure 6.2.

The scattering angleθ depends on the impact parameter b, which measures the distance from the center of the atom that a projectile would pass if it were not deflected. Each different value of the impact parameter results in a different value of the scattering angle.

y

x θ R

b b = 0

FIGURE 6.2 A positively charged par-ticle is deflected by an angle θ as it passes through a positively charged sphere, representing a Thomson model atom. The scattering angle depends on the value of the impact parameter b, which varies from 0 to R.

The particle is deflected from its original trajectory by the electrical forces exerted on the particle by the atom. For a positively charged particle, these forces are: (1) a repulsive force due to the positive charge of the atom, and (2) an attractive force due to the negatively charged electrons. We assume that the mass of the deflected particle is much greater than the mass of an electron but also much less than the mass of the atom. In the encounter between the projectile and an electron,

the forces exerted on each by the other are equal and opposite (by Newton’s third law), and so the principal victim of the encounter is the much less massive electron; the effect on the projectile is negligible. (Imagine rolling a bowling ball through a field of Ping-Pong balls!) We thus need consider only the positively charged atom as a cause of the deflection of the particle. By the same argument, we neglect any possible motion of the more massive atom caused by the passage of the projectile. The basic experiment, then, is the scattering of a positively charged projectile by the stationary positively charged massive part of the atom.

θ θ θ

θ

FIGURE 6.3 Scattering by a thin foil.

Some individual scatterings tend to increase θ, while others tend to decreaseθ.

In practice we cannot do the experiment with one atom. Instead, we bombard a thin foil, as in Figure 6.3. The scattering angleθ that we observe in the laboratory is the result of scattering by many atoms, with impact parameters that we do not know and cannot control. Let’s assume that for a single atom the average scattering angle is θav, which represents an average over all possible impact parameters from zero up to the atomic radius R. For a typical foil thickness of 1μm (10⁻⁶m), the projectile is scattered by about 10⁴atoms.

The total scattering angleθ is determined by statistical considerations, because some of the individual scatterings move the projectile toward larger scattering angles and some toward smaller angles, as represented in Figure 6.3. This is an example of a “random walk” problem— for N scatterings, the most likely observed net scattering angleθ is related to the average individual scattering angle by

θ √

Nθav (6.1)

According to the Thomson model, the average scattering angle for a single atom is on the order of 0.01^◦, and for a foil that is 10⁴ atoms thick the net scattering angle should be about 1^◦. This is consistent with experimental observations.

Ernest Rutherford (1871–1937, Eng-land). Founder of nuclear physics, he is known for his pioneering work on alpha-particle scattering and radioac-tive decays. His inspiring leadership influenced a generation of British nu-clear and atomic scientists.

The most critical test of the Thomson model, which it fails completely, occurs when we examine the probability for scattering at large angles. If each individual scattering deflects the projectile through an angle of around 0.01^◦, then to observe projectiles scattered through a total angle greater than 90^◦, we must have about 10⁴successive scatterings, all of which push the projectile toward larger angles.

Because the probabilities of individual scatterings toward either larger or smaller angles are equal, the probability of having 10⁴ successive scatterings toward larger angles, like the probability of finding 10⁴ successive heads in tossing a coin, is about(1/2)^10,000= 10⁻³⁰⁰⁰.

An experiment to observe this scattering was performed by Hans Geiger and Ernest Marsden in the laboratory of Ernest Rutherford at Manchester University in 1910. For projectiles they used alpha particles, which are nuclei of helium (of charge+2e) emitted in radioactive decay. Their results showed that the probability of an alpha particle scattering at angles greater than 90^◦ was about 10⁻⁴. This remarkable discrepancy between the expected value based on the Thomson model (10⁻³⁰⁰⁰) and the observed value (10⁻⁴) was described by Rutherford in this way:

It was quite the most incredible event that ever happened to me in my life.

It was as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.

The analysis of the results of such scattering experiments led Rutherford to propose that the mass and positive charge of the atom are not distributed uniformly over the volume of the atom, but instead are concentrated in an extremely small region, about 10⁻¹⁴m in diameter, at the center of the atom. In Section 6.3 we will see how this proposal is consistent with the large-angle scattering results.

Scattering in the Thomson Model (Optional) Let’s assume that a projectile of positive charge ze is incident on an atom of radius R that we represent according to the Thomson model as a uniform sphere of positive charge Ze. The force on the projectile when it is a distance r from the center of the atom can be computed using Gauss’s law (see Problem 2):

F= zZe²

4πε₀R³r (6.2)

Before discussing the scattering, we should note that this equation can also describe (if we put z= 1) the force on an electron embedded in the Thomson atom at a distance r from its center. This force can be written F= kr with k = Ze²/4πε₀R³. This linear restoring force permits the electrons to oscillate about their equilibrium positions just like a mass on a spring subject to the linear restoring force F = kx. We therefore expect the electrons in the Thomson atom to oscillate about their equilib-rium positions with a frequency f = (2π)⁻¹

k/m , where k is the force constant.

Because an oscillating electric charge radiates electromagnetic waves whose fre-quency is identical to the oscillation frefre-quency, we might expect, based on the Thomson model, that the radiation emitted by atoms would show this characteristic frequency. This turns out not to be true (see Problem 3); the calculated frequencies do not correspond to the frequencies observed for radiation emitted by atoms.

The exact calculation of the scattering angle for different values of the impact parameter in the Thomson model of the atom is fairly complicated, but for our purposes we want only an estimate of the average value of the angle. As we will find out later, it’s not very important if our estimate is off by a small factor.

Initially the projectile moves in the x direction in the geometry of Figure 6.2, but the atom exerts a force in the y direction that produces a small component of momentum p_y in that direction. Using Newton’s second law we can find the momentum from the impulse received by the projectile due to the electrostatic force:

p_y=



F_ydt (6.3)

Rather than carry out this complicated integral for a force that is changing in mag-nitude and direction as the projectile travels, we’ll estimate the average scattering angle by choosing an average value for the impact parameter, namely b= R/2 (rep-resenting the middle trajectory of Figure 6.2), and we’ll assume the force acts in the y direction for a timet determined by the projectile’s flight along a line of length roughly equal to R. This underestimates the amount of time during which the force acts but overestimates the effect of the force (which doesn’t act purely in the y direction along the entire trajectory), so to some extent these two effects should cancel one another.

Making these approximations, we obtain p_y∼= Ft ∼=zZe²(R/2)

4πε₀R³ R

v = zZe²

8πε₀Rv (6.4)

The angleθ is small, so we can make the approximation tan θ ∼= θ, and we can assume that p_xchanges very little from its initial value mv, and so the average scattering angle is

θ_av∼= tan θav= p_y p_x = p_y

mv = zZe² 8πε0Rv

1

mv = zZe²

16πε0RK (6.5)

using the nonrelativistic kinetic energy K= ¹₂mv². This gives an estimate of the scattering angle when the impact parameter b is equal to half the radius R. Smaller values of b will give smaller deflection angles, and larger values of b will give larger angles, so this is a reasonable estimate for the average scattering angle for a Thomson model atom.

Example 6.1

Using the Thomson model, estimate the average scatter-ing angle when alpha particles (z= 2) with kinetic energy 3 MeV are scattered from gold (Z= 79). The atomic radius of gold is 0.179 nm.

Solution

Using e²/4πε₀ = 1.44 eV·nm, we have

θav∼= zZe² 16πε₀RK = 1

4 e² 4πε₀

zZ RK

=0.25(1.44 eV · nm)(2)(79) (0.179 nm)(3 × 10⁶eV)

= 1 × 10⁻⁴rad= 0.01^◦

Even though this result represents a rough estimate of the average scattering angle in the Thomson model of the atom, its accuracy does not affect our conclusions about the failure of the model. Even if our estimate were too small by as much as a factor of 10 (which is highly unlikely), we would be comparing an expected probability of 10⁻³⁰⁰(instead of 10⁻³⁰⁰⁰) with the observed 10⁻⁴, still a spectacular disagreement. Any reasonable estimate shows the complete failure of the Thomson model to account for these scattering experiments.