Chapter 1st:- Atomic Structure and Periodic Table
B.sc 1st year Book
(Page 1)

# Bohr Atomic model

## What is Bohr’s Atomic Model?

In 1913, Niels Bohr’s proposed a theory for the hydrogen atom which is based on the quantum theory that some physical quantities only take discrete values. In an atom, The electrons(e) revolve around the positively charged nucleus(+) in a defined circular path called shells or orbits. Each shell or Orbit has an examined fixed energy and these circular shell or orbits are called orbitals shell.

## Bohr Atomic model :

After the discovery of protons, electrons, and neutrons, ventures were made to give a physical picture of an atom. Rutherford’s model of  atom supremacy by solar system suffered the following setbacks:
(i) Newtonian laws of motion can not be applied to charged particles and macroscopic.
(ii) According to Clark Maxwell when at any time a charged particle-like electron is subjected to. acceleration, it losses energy and emits radiation.
Bohr argued therefore that if an electron moves in a circular orbit it should be subjected to acceleration due to the continuous changes in its direction of motion. The electrons, therefore, continuously lose energy and emit radiation. As a result, of this, the electron should be moving in a spiral path with direction and ultimately collide with the nucleus. In other words, the atom should collapse. However, this does not happen.
(iii) it fails to explain the atomic spectra of elements.
Neils Bohr a Danish Physicist (1913) put forward a new model of atoms combining the ideas of J. J. Thomson, Rutherford, and Planks’s quantum theory. He was awarded the Nobel prize for his distinguished work modal of atom in 1922.

## Bohr’s postulates :

1. The electron revolves in circular orbits around the nucleus only in certain allowed energy states called stationary states or energy levels (as the earth revolves around the sun).

2. The electron did not radiate energy regularly during the motion and therefore it did not slow down. Thus, the motion of the electron is restricted in such a manner that angular momentum (mv) is

quantized and is an integral multiple of
$$\frac{h}{2 \pi}$$
$$m v r=\frac{n h}{2 \pi}$$

where ‘m’ is the mass of the electron, ‘v’ is its tangential velocity, ‘r’ is the radius, ‘h’ the Plank’s constant, and ‘n’ is an integer having values 1,2,3…… for the first, second, third, etc of Bohr’s orbits.
3. When an electron jumps from one stationary state to another stationary i.e. during an electronics transition the energy is being absorbed or emitted. The energy changes occur in a fixed amount called an energy pocket or quanta.
If E1 and E2 are the energies of two states, then
E2 – E1 = hv (where E2 > E1 and v is the frequency of emited radiation)
$$v = \frac{E2 – E1}{h}$$

## Related Topic | Atomic Structure and Periodic Table

### Bohr’s Theory of Hydrogen and Hydrogen-like atoms (e.g. He*, Ll2+, Be” etc.)

According to Bohr’s atomic model when an electron revolves in a circular orbit, the

electrostatic force of attraction between the electron and nucleus is balanced by Centrifugal force which results in the throw of the electron occurring out of the orbit. It is the mass of the electron moving with velocity in a circular orbit of radius as shown in figure 1.01.

Since centrifugal force = coulombic force for attraction
$$\frac{mv^2}{I}\:=\:\frac{Ze^2}{r^2}$$
Hence,
$$v^2\:=\:\frac{Ze^2}{mr}$$
According to Bohr’s second postulate
Or $$mvr=\:n\frac{h}{2\pi }$$
Or $$v^2=\:\frac{n^2h^2}{4\pi ^2m^2r^2}$$
On comparing the equation (v) and (vi) we get
$$\frac{2e^2}{mr}=\:\frac{n^2h^2}{4\pi ^2m^2r^2}$$
$$n_n=\:\frac{n^2h^2}{4\pi ^2m^2e^2}\:\left(vii\right)$$
For hydrogen atom n =1 and Z = 1
Hence
$$r=\:\frac{h^2}{4\pi ^2m^2e^2}}\:\left(viii\right)$$
Substituting the values h = 6.627 x 10-27 erg sec, m = 9.1 x 10-28g and e = 4.80 x 10-10 e.s.u in equation (viii) we have
r = 0.529 A0
Thus, atomic radius of the H-atom is 0.529 A0.
Calculation of total energy of an electron: The total energy of an electron E is given by –
E = K.E + P.E
Or
$$E\:=\:\frac{1}{2}mv^2\:-\:\frac{Ze^2}{r}$$
( $$K.\:E\:=\:\frac{1}{2}mv^2\:since\:and\:P.E\:=\:-\:\frac{Ze^2}{r}\:$$ )
From Equation (v)
$$\:\frac{1}{2}mv^2\:=\:-\:\frac{2e^2}{2r}\:…..\:\left(x\right)$$
$$E\:=\:-\:\frac{Ze^2}{2r}\:$$
Now putting the value of r from equation (vii) we get
$$E\:=\:\frac{Ze^24\pi ^2mZe^2}{2n^2h^2}\:$$
$$E\:=\:\frac{Ze^24\pi ^2mZ^{ }e^2}{2n^2h^2}$$
$$E\:=\:\frac{2\pi ^2mZ^2e^4}{n^2h^2}\:….\:\left(xii\right)$$
In the above equation for a particular atom all the quantities except n’ are constant. Therefore
$$E\:\infty \:-\:\frac{1}{n^2}\:….\:\left(xiii\right)$$
Since there is a negative sign, the value of ‘E’ increases as ‘n’ increases.
Explanation of Hydrogen spectrum on the basis of Bohr’s theory If electron. jumps from one energy level to another then,
$$E\:_1\:=\:\frac{2\pi ^2me^4}{n\frac{2}{1}h^2}\:and\:E_2=\:\frac{2\pi me^4}{n\frac{2}{2}h^2}\:\:\:\left(Z=1\right)$$
Therefore,
$$E_2\:-\:E\:_1\:=\:-\frac{2\pi \:me^4}{n\frac{2}{2}h^2}+\frac{2\pi ^2me^4}{n\frac{2}{1}h^2}\:$$
Or $$E_2\:-\:E\:_1\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right)$$
We know that E2 – E1 = hv Hence,
$$hv\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right)$$
Or $$v\:=\:-\frac{2\pi \:me^4}{n^2}\left(\frac{1}{n\frac{2}{1}}\:-\frac{1}{n\frac{2}{2}}\right)\left(xv\right)$$
$$Since\:c\:=\:cλ\:or\:v=\frac{c}{λ}=cv\::\:where\:’c’\:is\:velocity\:of\:light\:and\:λ\:is\:the\:wavelength.\:\:$$
Since
Or $$\overline{v}\:=\:\frac{2\pi ^2me^4}{eh^3}\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)….\:\left(xvi\right)$$
On comparing equation (xvi) with Rydberg’s empirical equation, we get Rydberg’s empirical equation $$\overline{v}\:=R\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)$$ (R = Rydberg constant)
I.e.

$$\overline{v}\:=R\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)\:=\:\frac{2\pi ^2me^4}{h^3}\left(\frac{1}{n\frac{2}{1}}-\frac{1}{n\frac{2}{2}}\right)$$ Hence $$k\:=\:\frac{2\pi ^2me^4}{eh^3}….\:\left(xvii\right)$$

A comparison of these two equations shows that Rydg-berg constant R is equal to and its value comes out to be 109740 cm(-1) which shows an excellent agreement with the Rydg-berg constant found empirically by Balmer. This agreement offers excellent proof of the correctness of Bohr’s theory.
If the electron in the hydrogen atom occupies the lowest orbit, the atom is said to be in the ground state. But when the electron is excited by the absorption of energy supplied from the external sources, jumps to higher energy levels. After excitation, the electron returns to the lower energy levels, releasing excess energy in discrete amounts. As a result, different spectral lines are observed. All the spectral lines with the region in which they appear are given in the table shown below.
Table 1.1: Electronic transitions and values of n1 and n2  for various spectral lines of the hydrogen spectrum.
Fig. 1.02: Energy level diagram for observed hydrogen spectrum.

$$E_1\:=-\:\frac{2\pi ^2me^4}{h^2}$$
Putting the value of p, ‘m’: ‘e’ and ‘h’ in the above expression of E1 we get the
Numerical value of E1 as :
$$E_1\:=-\:\frac{2\:\times \:\left(3.14\right)^2\times \:\left(9.1\times 20^{-28}\right)\:\times \:\left(4.8\times 10^{-10}\right)^4}{\left(6.627\:\times \:\:10^{-2}\right)^2}$$
= -21.79 x 10-12 ergs atom
= -13.6 eV/atom
= – 313.8 Kcal atom
(Since, 1 erg = 6.24 x 1011eV) and 1eV = 23.06Kcals)
Similarly, the values of E2, E3, E4 …. Bto in eV in terms of E1 are calculated by using the following expression:
$$E_n\:=E_1\times \frac{1}{n^2}$$
$$E_n\:=-13.6\times \frac{1}{n^2}$$

and so on.

## Limitations of Bohr’s Theory :

1- According to Bohr, radiation results when an electron jumps from one energy level to another but how this radiation occurs is not discussed in Bohr’s theory,
2- Bohr’s theory only explained the spectra of H-atom and H-like particles such as He’, etc. but failed the explain the spectra of other atoms having a larger number of electrons.
3- The pictorial concept of electrons jumping from one orbit to another is not justified because of the uncertainty of their position and velocities.
4- This theory has no explanation for the splitting of spectral lines in an electric field (Stark effect and magnetic effect (Zeeman effect).
5- The magnitude of energies of the stationary states in any atom containing more than one electron can not be accurately calculated by the application of Bohr’s Theory.
6- It does not explain the dual nature of electrons.
Example 1: What will be the energy of the electron in a hydrogen atom when the atom is excited and its electron revolves in the 4th orbit.
Solution According to the equation.

$$E_n\:=-\frac{2\pi ^2me^4}{n^2h^2}$$
For 4th orbit, n = 4
$$E_4\:=-\frac{2\pi ^2me^4}{n\frac{2}{4}h^2}$$
$$=-\frac{2\times \left(3.14\right)^2\times \left(9.109\times 10^{-28}\right)\times \left(4.8\times 10^{-10}\right)^4}{4^2\times \left(6.626\times 10^{-2}\right)}$$
$$=-1.343\times 10^{-12}+\:rg$$
Or in electron volts
$$E_A\:=\:-13.6\:\times \:\frac{1}{4^2}$$
$$=\:-13.6\:\times \:\frac{1}{^{16}}$$
= – 0.85eV

Example 2: Calculate the radius of the second orbit of the hydrogen atom.
Solution: According to the equation.

$$r_2=\:\frac{2^2\times \left(6.626\times 10^{27}\right)\:erg^2}{4\times \left(3.14\right)^2\times \left(9.109\times 10^{28}\right)\times 1\times \left(4.8\times 10^{-10}esu\right)^2}$$

= 4 x 0.529 x 10-8cm = g1/2cm3/2 s-1
(erg = g cm2s-2 and e.s.u = g1/2cm3/2 s-1)