12.4 Bohr Model

The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates. However, there are some fundamental differences between the two situations. While the planetary system is held by gravitational force, the nucleus-electron system being charged objects, interact by Coulomb’s Law of force. We know that an object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus (Fig. 12.7). Thus, such an atom can not be stable. Further, according to the classical electromagnetic theory, the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of revolution. As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously, and so will the frequency of the light emitted. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. Clearly Rutherford model tells only a part of the story implying that the classical ideas are not sufficient to explain the atomic structure.

It was Niels Bohr (1885 – 1962) who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model. Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra. Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

These are:

1. Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom.
2. Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 1\(0^{–34}\) J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is \[ L=n h / 2 \pi \]
3. Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by \[ h v=E_{i}-E_{f} \] where \(E_{i}\) and \(E_{f}\) are the energies of the initial and final states and \(E_{i}\) > \(E_{f}\) .

For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The angular momentum L is given by

\[ L=m w r \]

Bohr’s second postulate of quantisation [Eq. (12.11)] says that the allowed values of angular momentum are integral multiples of h/2π.

\[ L_{n}=m v_{n} r_{n}=\frac{n h}{2 \pi} \]

where n is an integer, \(r_{n}\) is the radius of n th possible orbit and \(v_{n}\) is the speed of moving electron in the n th orbit. The allowed orbits are numbered 1, 2, 3 ..., according to the values of n, which is called the principal quantum number of the orbit.

From Eq. (12.3), the relation between \(v_{n}\) and \(r_{n}\) is

\[ v_{n}=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m r_{n}}} \]

Combining it with Eq. (12.13), we get the following expressions for \(v_{n}\) and \(r_{n}\)

\[ v_{\mathrm{n}}=\frac{1}{n} \frac{e^{2}}{4 \pi \varepsilon_{0}} \frac{1}{(h / 2 \pi)} \]

\[ r_{n}=\left(\frac{n^{2}}{m}\right)\left(\frac{h}{2 \pi}\right)^{2} \frac{4 \pi \varepsilon_{0}}{e^{2}} \]

Eq. (12.14) depicts that the orbital speed in the \(n^{th}\) orbit falls by a factor of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be obtained as

\[ r_{1}=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}} \]

This is called the Bohr radius, represented by the symbol \(a_{0}\) . Thus,

\[ a_{0}=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}} \]

Substitution of values of h, m, \(ε_{0}\) and e gives \(a_{0}\) = 5.29 × 1\(0^{–11}\) m. From Eq. (12.15), it can also be seen that the radii of the orbits increase as \(n^{2}\).

The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as

\[ \begin{array}{l} E_{\mathrm{n}}=-\left(\frac{e^{2}}{8 \pi \varepsilon_{0}}\right)\left(\frac{m}{n^{2}}\right)\left(\frac{2 \pi}{h}\right)^{2}\left(\frac{e^{2}}{4 \pi \varepsilon_{0}}\right) \\ E_{\mathrm{n}}=-\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}} \end{array} \]

Substituting values, Eq. (12.17) yields

\[ E_{n}=-\frac{2.18 \times 10^{-18}}{n^{2}} \mathrm{~J} \]

Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 1\(0^{–19}\) J, Eq. (12.18) can be rewritten as

\[ E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV} \]

The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus (or proton in hydrogen atom).

The derivation of Eqs. (12.17) – (12.19) involves the assumption that the electronic orbits are circular, though orbits under inverse square force are, in general elliptical. (Planets move in elliptical orbits under the inverse square gravitational force of the sun.) However, it was shown by the German physicist Arnold Sommerfeld (1868 – 1951) that, when the restriction of circular orbit is relaxed, these equations continue to hold even for elliptic orbits.

12.4.1 Energy levels

The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits. The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0 . The energy of this state (n = 1), E 1 is –13.6 eV. Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the ionisation energy of the hydrogen atom. This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy.

At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states. The atom is then said to be in an excited state. From Eq. (12.19), for n = 2; the energy \(E_{2}\) is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an energy equal to \(E_{2}\) – \(E_{1}\) = –3.40 eV – (–13.6) eV = 10.2 eV. Similarly, E 3 = –1.51 eV and E 3 – E 1 = 12.09 eV, or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3), 12.09 eV energy is required, and so on. From these excited states the electron can then fall back to a state of lower energy, emitting a photon in the process. Thus, as the excitation of hydrogen atom increases (that is as n increases) the value of minimum energy required to free the electron from the excited atom decreases.

The energy level diagram* for the stationary states of a hydrogen atom, computed from Eq. (12.19), is given in Fig. 12.8. The principal quantum number n labels the stationary states in the ascending order of energy. In this diagram, the highest energy state corresponds to n =∞ in Eq, (12.19) and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases.