12.2 Rutherford's Nuclear Model

At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden performed some experiments. In one of their experiments, as shown in Fig. 12.1, they directed a beam of 5.5 MeV α-particles emitted from a 214 83 Bi radioactive source at a thin metal foil made of gold. Figure 12.2 shows a schematic diagram of this experiment. Alpha-particles emitted by a 214 83 Bi radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 1\(0^{–7}\) m. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. These flashes may be viewed through a microscope and the distribution of the number of scattered particles may be studied as a function of angle of scattering.

A typical graph of the total number of α-particles scattered at different angles, in a given interval of time, is shown in Fig. 12.3. The dots in this figure represent the data points and the solid curve is the theoretical prediction based on the assumption that the target atom has a small, dense, positively charged nucleus. Many of the α-particles pass through the foil. It means that they do not suffer any collisions. Only about 0.14% of the incident α-particles scatter by more than 1°; and about 1 in 8000 deflect by more than 90°. Rutherford argued that, to deflect the α-particle backwards, it must experience a large repulsive force. This force could be provided if the greater part of the mass of the atom and its positive charge were concentrated tightly at its centre. Then the incoming α-particle could get very close to the positive charge without penetrating it, and such a close encounter would result in a large deflection. This agreement supported the hypothesis of the nuclear atom. This is why Rutherford is credited with the discovery of the nucleus.

In Rutherford’s nuclear model of the atom, the entire positive charge and most of the mass of the atom are concentrated in the nucleus with the electrons some distance away. The electrons would be moving in orbits about the nucleus just as the planets do around the sun. Rutherford’s experiments suggested the size of the nucleus to be about 10 –15 m to 10 –14 m. From kinetic theory, the size of an atom was known to be 10 –10 m, about 10,000 to 100,000 times larger than the size of the nucleus (see Chapter 11, Section 11.6 in Class XI Physics textbook). Thus, the electrons would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the size of the nucleus itself. Thus, most of an atom is empty space. With the atom being largely empty space, it is easy to see why most α-particles go right through a thin metal foil. However, when α-particle happens to come near a nucleus, the intense electric field there scatters it through a large angle. The atomic electrons, being so light, do not appreciably affect the α-particles.

The scattering data shown in Fig. 12.3 can be analysed by employing Rutherford’s nuclear model of the atom. As the gold foil is very thin, it can be assumed that α-particles will suffer not more than one scattering during their passage through it. Therefore, computation of the trajectory of an alpha-particle scattered by a single nucleus is enough. Alpha- particles are nuclei of helium atoms and, therefore, carry two units, 2e, of positive charge and have the mass of the helium atom. The charge of the gold nucleus is Ze, where Z is the atomic number of the atom; for gold Z = 79. Since the nucleus of gold is about 50 times heavier than an α-particle, it is reasonable to assume that it remains stationary throughout the scattering process. Under these assumptions, the trajectory of an alpha-particle can be computed employing Newton’s second law of motion and the Coulomb’s law for electrostatic force of repulsion between the alpha-particle and the positively charged nucleus.

The magnitude of this force is

\[ F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{r^{2}} \]

where r is the distance between the α-particle and the nucleus. The force is directed along the line joining the α-particle and the nucleus. The magnitude and direction of the force on an α-particle continuously changes as it approaches the nucleus and recedes away from it.

12.2.1 Alpha-particle trajectory

The trajectory traced by an α-particle depends on the impact parameter, b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus (Fig. 12.4). A given beam of α-particles has a distribution of impact parameters b, so that the beam is scattered in various directions with different probabilities (Fig. 12.4). (In a beam, all particles have nearly same kinetic energy.) It is seen that an α-particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the α-particle rebounds back ( θ ≅ π). For a large impact parameter, the α-particle goes nearly undeviated and has a small deflection ( θ ≅ 0).

The fact that only a small fraction of the number of incident particles rebound back indicates that the number of α-particles undergoing head on collision is small. This, in turn, implies that the mass and positive charge of the atom is concentrated in a small volume. Rutherford scattering therefore, is a powerful way to determine an upper limit to the size of the nucleus.

12.2.2 Electron orbits

The Rutherford nuclear model of the atom which involves classical concepts, pictures the atom as an electrically neutral sphere consisting of a very small, massive and positively charged nucleus at the centre surrounded by the revolving electrons in their respective dynamically stable orbits. The electrostatic force of attraction, F e between the revolving electrons and the nucleus provides the requisite centripetal force (\(F_{c}\) ) to keep them in their orbits. Thus, for a dynamically stable orbit in a hydrogen atom

\[ \begin{array}{c} F_{e}=F_{c} \\ \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}=\frac{m w^{2}}{r} \end{array} \]

Thus the relation between the orbit radius and the electron velocity is

\[ r=\frac{e^{2}}{4 \pi \varepsilon_{0} m v^{2}} \]

The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are

\[ K=\frac{1}{2} m v^{2}=\frac{e^{2}}{8 \pi \varepsilon_{0} r} \text { and } U=-\frac{e^{2}}{4 \pi \varepsilon_{0} r} \]

(The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus the total energy E of the electron in a hydrogen atom is

\[ \begin{aligned} E=K+U &=\frac{e^{2}}{8 \pi \varepsilon_{0} r}-\frac{e^{2}}{4 \pi \varepsilon_{0} r} \\ &=-\frac{e^{2}}{8 \pi \varepsilon_{0} r} \end{aligned} \]

The total energy of the electron is negative. This implies the fact that the electron is bound to the nucleus. If E were positive, an electron will not follow a closed orbit around the nucleus.