Bohr's Atomic Model


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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

 

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 - E1' 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 - E2' 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 is DE,

DE = E2 - E1 = hn

Energy changes in an electron jump

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 E2 - E1 = hn.

The frequency of the emitted radiation is:

frequency of the emitted radiation

Since E2 and E1 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 E3 and E1 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 E4 and E1 respectively or from the fifth to the second i.e., E5 and E2, 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.

Different routes to the ground state from n = 4

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 = 1

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

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

Brackett 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 'nth' orbit has been found to be
calculation of electron energy of  hydrogen using  bohrs theory

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:

calculation of electron energy of  hydrogen like ions using  bohrs theory

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

Although Bohr's model successfully explained the stability and the line spectrum of hydrogen, it had its limitations. They were:

Limitations and problems

  • It could not explain the line spectrum of multi electron atoms.
  • This model failed to explain the effect of magnetic field on the spectra of atoms (Zeeman effect).
  • The effect of electric field on the spectra could not be explained by Bohr's model (Stark effect).
  • The shapes of molecules arising out of directional bonding could not be explained.
  • The dual nature of electrons (both as wave and particle) and the path of motion of the electron in well defined orbits were not correct.

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 10-14 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 10-14 kJ mol-1

= 5.39 x 1014 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, n1 = 2, n2 = 3

For the second line in Balmer series, n1 = 2, n2 = 4

Dividing equations (i) by (ii)



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