Atoms

By the nineteenth century, enough evidence had accumulated in favour of atomic hypothesis of matter. In 1897, the experiments on electric discharge through gases carried out by the English physicist J. J. Thomson (1856 – 1940) revealed that atoms of different elements contain negatively charged constituents (electrons) that are identical for all atoms. However, atoms on a whole are electrically neutral. Therefore, an atom must also contain some positive charge to neutralise the negative charge of the electrons. But what is the arrangement of the positive charge and the electrons inside the atom? In other words, what is the structure of an atom? The first model of atom was proposed by J. J. Thomson in 1898. According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. This model was picturesquely called plum pudding model of the atom. However subsequent studies on atoms, as described in this chapter, showed that the distribution of the electrons and positive charges are very different from that proposed in this model. We know that condensed matter (solids and liquids) and dense gases at all temperatures emit electromagnetic radiation in which a continuous distribution of several wavelengths is present, though with different intensities. This radiation is considered to be due to oscillations of atoms

and molecules, governed by the interaction of each atom or molecule with its neighbours. In contrast, light emitted from rarefied gases heated in a flame, or excited electrically in a glow tube such as the familiar neon sign or mercury vapour light has only certain discrete wavelengths. The spectrum appears as a series of bright lines. In such gases, the average spacing between atoms is large. Hence, the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules. In the early nineteenth century it was also established that each element is associated with a characteristic spectrum of radiation, for example, hydrogen always gives a set of lines with fixed relative position between the lines. This fact suggested an intimate relationship between the internal structure of an atom and the spectrum of radiation emitted by it. In 1885, Johann Jakob Balmer (1825 – 1898) obtained a simple empirical formula which gave the wavelengths of a group of lines emitted by atomic hydrogen. Since hydrogen is simplest of the elements known, we shall consider its spectrum in detail in this chapter. Ernst Rutherford (1871–1937), a former research student of J. J. Thomson, was engaged in experiments on $\alpha$-particles emitted by some radioactive elements. In 1906, he proposed a classic experiment of scattering of these $\alpha$-particles by atoms to investigate the atomic structure. This experiment was later performed around 1911 by Hans Geiger (1882–1945) and Ernst Marsden (1889–1970, who was 20 year-old student and had not yet earned his bachelor’s degree). The details are discussed in Section 12.2. The explanation of the results led to the birth of Rutherford’s planetary model of atom (also called the nuclear model of the atom). According to this the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus with electrons revolving around the nucleus just as planets revolve around the sun. Rutherford’s nuclear model was a major step towards how we see the atom today. However, it could not explain why atoms emit light of only discrete wavelengths. How could an atom as simple as hydrogen, consisting of a single electron and a single proton, emit a complex spectrum of specific wavelengths? In the classical picture of an atom, the electron revolves round the nucleus much like the way a planet revolves round the sun. However, we shall see that there are some serious difficulties in accepting such a 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

New Zealand born, British physicist who did pioneering work on radioactive radiation. He discovered alpha-rays and beta-rays. Along with Federick Soddy, he created the modern theory of radioactivity. He studied the ‘emanation’ of thorium and discovered a new noble gas, an isotope of radon, now known as thoron. By scattering alpha-rays from the metal foils, he discovered the atomic nucleus and proposed the planetary model of the atom. He also estimated the approximate size of the nucleus.

A typical graph of the total number of $\alpha$-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 $\alpha$-particles pass through the foil. It means that they do not suffer any collisions. Only about 0.14% of the incident $\alpha$-particles scatter by more than $1^{\circ }$; and about 1 in 8000 deflect by more than $90^{\circ }$. Rutherford argued that, to deflect the $\alpha$-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 $\alpha$-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 $10^{-15}\mathrm{m}$ to $10^{-14}\mathrm{m}$. From kinetic theory, the size of an atom was known to be $10^{-10}\mathrm{m}$, about 10,000 to 100,000 times larger than the size of the nucleus. 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 $\alpha$-particles go right through a thin metal foil. However, when $\alpha$-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 $\alpha$-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 $\alpha$-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 $\alpha$-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

$$F=\frac{1}{4\pi {\epsilon }_0}\frac{(2e)(Ze)}{r^2}$$ (12.1)

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

The trajectory traced by an $\alpha$-particle depends on the impact parameter, $b$ of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the centre of the nucleus (Fig. 12.4). A given beam of $\alpha$-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 $\alpha$-particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the $\alpha$-particle rebounds back ($\theta \approx \pi$). For a large impact parameter, the $\alpha$-particle goes nearly undeviated and has a small deflection ($\theta \approx 0$). The fact that only a small fraction of the number of incident particles rebound back indicates that the number of $\alpha$-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.

Example 12.1 In the Rutherford's nuclear model of the atom, the nucleus (radius about $10^{-15}\mathrm{m}$) is analogous to the sun about which the electron move in orbit (radius $\approx 10^{-10}\mathrm{m}$) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth's orbit is about $1.5\times 10^{11}\mathrm{m}$. The radius of sun is taken as $7\times 10^8\mathrm{m}$.

Solution The ratio of the radius of electron's orbit to the radius of nucleus is $(10^{-10}\mathrm{m})/(10^{-15}\mathrm{m})=10^5$, that is, the radius of the electron's orbit is $10^5$ times larger than the radius of nucleus. If the radius of the earth's orbit around the sun were $10^5$ times larger than the radius of the sun, the radius of the earth's orbit would be $10^5\times 7\times 10^8\mathrm{m}=7\times 10^{13}\mathrm{m}$. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.

EXAMPLE 12.1

Example 12.2 In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a $7.7\mathrm{MeV}$ $\alpha$-particle before it comes momentarily to rest and reverses its direction?

Solution The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an $\alpha$-particle and a gold nucleus is conserved. The system's initial mechanical energy is $E_i$, before the particle and nucleus interact, and it is equal to its mechanical energy $E_f$ when the $\alpha$-particle momentarily stops. The initial energy $E_i$ is just the kinetic energy $K$ of the incoming $\alpha$-particle. The final energy $E_f$ is just the electric potential energy $U$ of the system. The potential energy $U$ can be calculated from Eq. (12.1). Let $d$ be the centre-to-centre distance between the $\alpha$-particle and the gold nucleus when the $\alpha$-particle is at its stopping point. Then we can write the conservation of energy $E_i=E_f$ as

$$K=\frac{1}{4\pi {\epsilon }_0}\frac{(2e)(Ze)}{d}=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}d}$$