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Bohr’s Model explained how electrons travel in different circular orbits around the nucleus. The orbits are symbolized with the letter ‘n’, where the value of n is an integer. The transfer of electrons is possible by emission and absorption of energy.
Viewing the demerits of the Rutherford model, Neil Bohr concluded that classical mechanics and electromagnetism cannot be applied to the processes on the atomic scale. To overcome this, Neil Bohr combined classical ideas with the quantum concepts of Planck to give something, which is known as the Neil Bohr atomic model of Hydrogen. There are three Bohr’s postulates of atomic models, we will talk about these in detail.
Postulates of Bohr Theory
There are three Bohr’s Postulates in Neil Bohr Model, each of these are described in detail below:
First Postulate
The first postulate states that every atom has a positively charged central core called the nucleus in which the entire mass of an atom is concentrated. Negatively charged electron revolves about the nucleus in a circular orbit, the centripetal force required for revolution is provided by the electrostatic force of attraction between the nucleus and electrons.
If ‘m’ is the mass of an electron revolving around the nucleus with a velocity ‘v’ in a circular orbit, then the required centripetal force is:
F =
[frac{mv^{2}}{r}] …(1)
Also, the electrostatic force of attraction between the nucleus of charge (+Ze) and the electron is (- e) will be:
F =
[frac{1}{4πε₀}] [frac{Ze*e}{r^{2}}] …(2)
We know that K = [frac{1}{4πε₀}]
F = K
[frac{Ze*e}{r^{2}}] = K[frac{Ze^{2}}{r^{2}}]
Now, equating eq (1) & (2), we get:
[frac{mv^{2}}{r}] = K[frac{Ze^{2}}{r^{2}}]
Second Postulate
The second postulate talks about stable orbits. According to Bohr, electrons can revolve in a certain discrete (discontinuous) non-radiating orbits, called stationary (permitted) orbits, for which angular momentum (L) of the revolving electron is an integral multiple of
[frac{h}{2π}] . Thus the angular momentum of the orbiting electron is an integer number.
The angular momentum of an electron = mvr
For any stationary orbit:
mvr = [frac{nh}{2π}]
Where,
n = A principal quantum number, which has a positive integral value ranging from 1, 2, 3, to n.
h = Planck’s constant = 6.626 x 10-34 Js
Third Postulate
The emission or absorption of energy occurs only when an electron jumps from one non-radiating orbit to another. The difference between the total energies of electrons in the two stationary orbits is absorbed when the electron jumps from inner to the outer orbit, and emitted when electrons jump from outer orbit to the inner one.
If E1 & E2 = Total energy (T.E.) of an e– in an inner and outer stationary orbit respectively, then the frequency of radiation emitted on jumping of from outer to inner orbit is given by:
E = hf = E2 – E1….(3)
We know that most of the hydrogen atom is in the ground state and when this atom receives energy either electron collision or heat, it may require sufficient energy to raise the electron to higher energy state, i.e., from n = 1 to n = 2, 3,…,∞. The atom is said to be in excited state, and difference in their energies can be calculated from equation (3), which is:
hf = [frac{2πmK^{2}Z^{2}e^{4}}{n1^{2}h^{2}}] + [frac{2πmK^{2}Z^{2}e^{4}}{n1^{2}h^{2}}] [[frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}]]
[frac{1}{λ}] = = [frac{2πmK^{2}Z^{2}e^{4}}{ch^{3}}] [[frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}]]
Also,
[frac{2πmK^{2}Z^{2}e^{4}}{ch^{3}}]= R, a constant called the Rydberg constant
And, ⊽ = [frac{1}{λ}] = R [[frac{1}{n_{2}^{2}}-frac{1}{n_{1}^{2}}]]….(4)
Where
Eq (4) is the Rydberg formula
⊽ = wave number of emitted radiation.
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Radius of Bohr’s stationary orbit is:
r = [frac{n^{2}h^{2}}{4π^{2}mKe^{2}}]
We can see that r n2, if the radii of stationary orbit are in the ratio of 1: 22: 32, i.e., 1: 4: 9; this means the stationary orbits are unequally spaced.
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Velocity of an electron in Bohr’s stationary orbit is given by:
v = [frac{KZe^{2}}{nhr}]
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Total energy of an electron in Bohr’s nth stationary orbit is:
En = – [frac{2πmK^{2}Z^{2}e^{4}}{n^{2}h^{2}}] ….(5)
Or,
En = – [frac{13.6}{n^{2}}] …(6)
Here, TE of an e– in a stationary orbit is negative, which means the electron is tightly bound to the nucleus.
Energy Level Diagram
On putting value of n = 1, 2, 3, …we get the energies of electrons in various stationary orbits as:
E1 = – [frac{13.6}{1^{2}}]eV = 13.6 eV …..(a)
E2 = – [frac{13.6}{2^{2}}]eV = – 3.4 eV…..(b)
E3 = – [frac{13.6}{3^{2}}]eV = – 1.5111 eV…(c)
E4 = – [frac{13.6}{4^{2}}] eV = – 0.85 eV…(d)
E5 = – [frac{13.6}{5^{2}}]eV = – 0.544 eV…(e)
6= – [frac{13.6}{6^{2}}]eV = – 0.3778 eV…(f)
A pattern from eq (a) to (f) shows us that the value of E 1 n2. It means the value of E becomes less negative with the increase in the value of square of n. Also, the value of En becomes zero when n reaches to ∞.
En = – [frac{13.6}{∞^{2}}] eV = 0 eV