In 1913, Neils Bohr proposed a model of an atom based on the Planck's quantum theory of radiation. The basic postulates of Bohr's theory are:

- An atom consists of a small, heavily positively charged nucleus around which electrons revolve in definite circular paths called orbits.

- These orbits are associated with definite energies called energy shells/energy levels. They are designated as K, L, M, N, …. etc. shells or numbered as 1, 2, 3, 4, …..etc. from the nucleus.

- As long as the electron remains in a particular orbit /energy shell its energy remains constant. This accounts for the stability of an atom.

- Only those orbits are permitted in which angular momentum of the electron is a whole number multiple of where h is Plancks constant. Any moving body taking a circular orbit has an angular momentum equal to the product of its mass (m), velocity of movement (v) and radius of orbit (r). In other words the angular momentum of an electron

Thus,

This postulate introduces the concept of quantization of angular momentum.

- Electrons can either lose or absorb energy abruptly, when they jump from one energy level to another. For instance when an electron moves from the 'normal or ground state - E
_{1}' of an atom i.e., the state of lowest energy as required by its 'n' and 'l' values, to a higher level, it causes the atom to be in its 'excited state - E_{2}' i.e., where electrons in an atom occupy energy levels higher than those permitted by its 'n' and 'l' values. The reverse is also true and the change in energy isDE,

DE = E_{2} - E_{1} = hn

Fig: 3.13 - Energy changes in an electron jump

### Bohr's atomic model explained successfully:

- The stability of an atom. Bohr postulated that as long an electron remains in a particular orbit it does not emit radiation i.e. lose energy. Hence it does not become unstable.

- The atomic spectrum of hydrogen was explained due to the concept of definite energy levels. The one electron of hydrogen being closest to the nucleus is in its lowest energy shell (n =1) or normal ground state. It can absorb a definite amount of energy and jump to a higher energy state. This excited state being unstable, the electron comes back to a lower energy level.

When the energy emitted during transition, strikes a photographic plate, it gives its impression in the form of a line. This difference is also the energy of photon expressed as E_{2} - E_{1} = hn.

##### Since E

_{2}and E

_{1}have only definite values and are characteristic of energy levels of atoms, the values of 'n' will also be definite and characteristic of the atoms. Thus each transition will produce a light of definite wavelength, which is observed as a line in the spectrum.

For example, if the electron jumps down from the third to the first energy level having energies E_{3} and E_{1} respectively, then the wavelength of the spectral line would be

Similarly, when the electron jumps down from the fourth to the first energy level having energies E_{4} and E_{1} respectively or from the fifth to the second i.e., E_{5} and E_{2}, then we have

These will give different lines in the spectrum of the atom corresponding to different transitions having definite wavelengths.

- The sample of hydrogen gas contains a large number of atoms and when energy is supplied, the electrons in different hydrogen atoms absorb different amounts of energies. These are raised to different energy states. For example, the electrons in some atoms may jump to second energy level (L), while in others it may be to the third (M), fourth (N) or fifth (O) and so on. These electrons come back from the higher energy levels to the ground state in one or more jumps emitting different amount of energies.

##### Fig: 3.14 - Different routes to the ground state from n = 4

Different lines depending upon the difference in energies of the levels concerned can be summarized in the form of series named after the scientists who have discovered them.

Lyman series from n = 2, 3, 4, 5……to n = 1Balmer series from n = 3, 4, 5, 6……to n = 2

Paschen series from n = 4, 5, 6, 7……to n = 3Brackett series from n = 5, 6, 7, 8……to n = 4

Pfund series from n = 6, 7, 8, 9……to n = 5.- The energy of the electron in a particular orbit of hydrogen atom could be calculated by Bohr's theory. The energy of the electron in the 'n
^{t}^{h}' orbit has been found to be

where 'm' is the mass and 'e' is the charge of the electron. The energy expression for hydrogen like ions such as He, Li can be written as:

where 'Z' is the nuclear charge, which is equal to atomic number.

Problems

7. If the energy difference between the electronic states of hydrogen atom is 214.68 kJ mol^{-}^{1}, what will be the frequency of light emitted when the electron jumps from the higher to the lower energy state? (Planck's constant = 39.79 x 10^{-}^{1}^{4} kJ mol^{-}^{1})

### Solution

The frequency (n) of emitted light is related to the energy difference of two levels (DE) as

E = 214.68 kJ mol^{-}^{1}, h = 39.79 x 10^{-}^{1}^{4} kJ mol^{-}^{1}

= 5.39 x 10

^{1}

^{4}s

^{-}

^{1}

8. The wavelength of first spectral line in the Balmer series is 6561 Å units. Calculate the wavelength of the second spectral line in Balmer series.

### Solution

According to Rydberg equation:

For the first line in Balmer series, n_{1} = 2, n_{2} = 3

For the second line in Balmer series, n_{1} = 2, n_{2} = 4

Dividing equations (i) by (ii)

### Atomic emission spectra

When a substance is heated to a high temperature, the atoms in the vapours get energized. These energized atoms then return to the ground state by emitting electromagnetic radiations of certain definite wavelength. A series of bright lines separated from each other by dark spaces is obtained and this is called atomic emission spectra.

Fig: 3.9 - Mechanism of emission spectra

### Atomic absorption spectra

When the atomic vapours from a sample are placed in the path of white light from an arc lamp, it absorbs the light of certain characteristic wavelengths and the light of other wavelengths get transmitted. In such conditions a series of dark lines on a white background are formed. This is called an absorption spectrum.

The dark lines in the absorption spectrum and the bright lines in the emission spectrum of a given element appear at the same wavelength.##### Fig: 3.10 - Comparison of Absorption and emission spectra of sodium vapour

Since each element gives a definite pattern of lines at certain definite frequencies or wavelengths, the atomic spectra is used in chemical analysis to identify and estimate the elements present in any sample.

*Atomic spectra of hydrogen atom*

Hydrogen is the simplest element with its atom having only one electron. Hence, the atomic spectrum of hydrogen has played a significant role in the development of atomic structure. In the emission spectrum of hydrogen, when an electric discharge is passed through hydrogen gas, the molecules of hydrogen break into atoms. The hydrogen atoms get energized and go into an excited state. The excited atoms then return to the ground state by emitting light. Hydrogen atoms emit bluish light. On passing this light through a prism, a discontinuous line spectrum consisting of several sharp lines is obtained. This is the line spectrum of hydrogen.

Four sharp coloured lines were observed in the visible region of this spectrum by Balmer, in the ultra violet region by Lyman, in the infrared region by Paschen, Brackett and Pfund. These series of lines are named after these scientists who discovered them. Balmer expressed these lines in terms of inverse of their wavelength () by a mathematical relation, which was later modified by Rydberg.where 'R

_{H}' is the Rydberg's constant and 'n

_{1}', 'n

_{2}' are integers with values equal to or greater than 3 and 'l' is the wavelength.

##### Fig: 3.11 - Line spectrum of hydrogen atom

### Problem

6. Calculate the wavelength of the spectral line when the electron in the hydrogen atom undergoes a transition from 4^{t}^{h} energy level to 2^{n}^{d} energy level. What is the colour of the radiation?

### Solution

According to Rydberg's equation

here, n_{1} = 2, n_{2} = 4 and R = 109678 cm^{-}^{1}

l = 486 x 10

^{-}

^{9}m = 486 nm.

## No comments:

Post a Comment