Question 2: (a) What are the postulates of Bohr’s atomic model?
(b) How can Bohr’s model of atom be applied to hydrogen atom to calculate the radius of the nth shell?
(c) Derive expressions using Bohr’s model, for energy difference (ΔE), frequency ν, and wave number in hydrogen atom.
(d) How does Bohr’s model explain the hydrogen spectrum?
(e) What are the short comings of Bohr’s atomic model?
Answer
Bohr’s Postulates of Atomic Model
Bohr said the electron could not have just any value as we assume classically. He assumed the energy of electron is quantized, i-e, having some specific values. The postulates of his theory are summarized as follow;
- Electrons revolve around the nucleus in fixed circular paths, called orbits or shells. Each orbit is associated with a definite amount of energy. (It means electron cannot revolve in any path in-between two orbits as its energy is not continuous.)
- When electron is in its own particular orbit, it neither radiates nor absorbs energy. However, when it goes from one orbit to another, it emits or absorbs energy so that its energy is equal to the energy of its new orbit.
- When the electron jumps from a lower orbit to a higher orbit it absorbs energy (as the energy of the higher orbit is more) and when it jumps from a higher orbit to a lower orbit it emits energy (as the energy of lower orbit is less than the higher one.) The emission or absorption of energy is equal to the difference of energies, ΔE, of both the orbits. Mathematically, ΔE = E2 – E1 = hν, where h is Planck’s constant (= 6.6262 × 10-34 Js), E2 is the energy of the orbit from where it jumps E1 is the energy of the orbit where it goes.
- The angular momentum (=mvr) of electron in the hydrogen atom is quantized and its value is integral multiple of h/2π. Thus, mvr = n(h/2π), where n = 1, 2, 3, … . (This means 0.5h/2π or 0.7h/2π etc are not possible).
(b) Calculation of the radius of electron of Bohr’s orbit of H
Let charge on electron (= charge on a proton in magnitude) is e. If the Z is the number of protons in the nucleus, then charge of the nucleus is Ze. Let the electron revolves around the nucleus in a circular path of radius r; the distance between electron and nucleus is r. There must be an electrostatic force between the electron and nucleus which is given by the Coulomb’s law as,
Here k is a constant an its value is ( 
). ϵ0 is called the permittivity of free space and its value is 8.85×10-12 C2 J-1 m-1. So put the values in equation (1),
This force acts as the centripetal force for the circular motion of electron and equal to the centrifugal force of the system, . Hence,
Simplifying for ‘r’,
Velocity of the electron cannot be easily calculated. However, if Bohr postulate that the angular momentum of the electron is quantized is applied, the velocity can be measured in measurable quantities. Hence,
Now put this value of v2 in equation (3).
Divide both sides by r2,
Take the reciprocal of both sides,
We know for hydrogen atom, Z = 1. Therefore, for the nth orbital of hydrogen atom,
Now the quantity in brackets is constant as all the values are constant. If we calculate this constant by putting the values of ϵ0 = 8.85 × 10-12 C2 J-1 m-1, h = 6.6262 × 10-34 J s, π = 3.1428, m = 9.11 × 10-31 kg and e= 1.6021 × 10-19 C, we get,
This is the radius of the first Bohr’s orbit for hydrogen atom.
(c) Expressions for energy difference, ΔE, frequency ν and wave number ν̅ for hydrogen atom
We know the energy of the electron in the nth orbit is given by
Let us suppose E1 and E2 are the energies of the electron in the lower orbit n1 and higher orbit n2 respectively. Then
And the difference in the energies (ΔE) can be found by subtracting one energy (here E1) from the other (here E2). So put the values,
For hydrogen atom, Z = 1, put this value and rearrange the above equation,
The quantity out of the bracket on RHS is constant. Its value is 2.18 × 10-18 J. Put this value,
Frequency: When an electron jumps from higher energy level to a lower energy level, it emits the excess of energy as radiation. The emitted energy is equal to the difference in the energy of the two levels and given by the Planck’s equation as ΔE = hν. We know that
See equation (C) above. So equating with Planck’s equation,
Given the number of energy level between which the transition of electron takes place, one can find the frequency in Hz of the emitted radiation.
Wave number: Relation between frequency and wave number is given by ν = cν̅. Where ν is the frequency of the emitted radiation, c is the speed of radiation and ν̅ is the wave number. Put the value of frequency from the above equation, we have
This is the wave number of the emitted radiation for hydrogen atom.
(d) Shortcomings of Bohr’s model
Bohr’s theory was very successful in predicting and accounting the energies of line spectra of hydrogen and other one electron systems like….. However, there are some drawbacks which surfaced with due course of time. Some of them are summarized as follow.
- Though Bohr Theory was successful in simple single electron atoms and ions (hydrogen, He+, Li++ etc) but failed to explain the spectrum of more complicated multi electrons systems.
- In hydrogen system, some spectral lines are not just single lines but groups of much closed lines when seen through a powerful spectrometer. These lines belong to slightly different frequencies and called fine structure. Bohr’s theory could not explain this fine structure of the spectrum of hydrogen.
- Zeeman effect is the splitting of a spectral line into several components in the presence of a static magnetic field. Bohr’s theory cannot explain this effect.
- Stark effect is the shifting and splitting of spectral lines of atoms in the presence of an external electric field. Bohr’s theory was unable to explain this phenomenon.
- It does not explain why the orbits are circular (2-dimensional). Recent investigation showed the
- In Bohr’s equation the position and momentum of the electron are well defined. However, according to Heisenberg’s principle it is impossible to measure the position and momentum of electron precisely at a time. Therefore, Bohr model of atom goes against the Heisenberg’s uncertainty principle.
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