Bohr’s model of hydrogen atom

Niels Bohr atomic model of the hydrogen atom

In 1913, Neils Bohr adopted the Rutherford model for the explanation of the atomic spectra of the hydrogen atom by Bohr’s theory. But according to classical mechanics, when a charged particle subjected to the acceleration it emits radiation and loses energy to hit the nucleus.

Thus according to Rutherford atomic theory, the energy of the electron loses continuously and the observed spectra should be continuous spectra.

But the actual electromagnetic spectrum consists of well-defined lines of definite frequencies. To resolve these anomalous Niels Bohr’s proposed a new atomic model of the hydrogen atom.

Postulates of Bohr atomic model

  • An atom possesses several stable circular orbits in which an electron can stay. Thus an electron stays in a particular orbit where no emission or absorption of energy occurs. These orbits are called energy levels of an atom.

The energy levels designated by numbers 1, 2, 3, …….. or capital letters, K, L, M, …….. starting from the nucleus of the atom. Thus the energy associated with a certain energy level increases with the increase of its distance from the nucleus.

If E1, E2, E3 …… denotes the number of energy levels of 1 (K – Shell), 2 (L – Shell), 3 (M – Shell) ……. then the order of energy levels are

E1ㄑE2ㄑE3ㄑ……

  • When an electron can jump from one orbit to another higher energy on the absorption of energy but one orbit to another lower energy with the emission of energy.
Bohr atomic model of the hydrogen atom
Hydrogen energy levels
  • The angular momentum of an electron moving in an orbit is an integral multiple of h/2π. This integral multiple is known as the principal quantum energy level of an atom.

\therefore mvr=\frac{nh}{2\pi }

where m = mass of an electron,
v = tangential velocity of an electron,
r = radius of an atom and
n = principal quantum number.

The attraction between nucleus and electrons

Let the nucleus has a mass m′ and the electron has mass m, the radius of the circular orbit = r, and the linear velocity of the electron = v.

Niels Bohr atomic model theory
Bohr model

Thus on the revolving electron, two types of forces acting,

  1. Centrifugal force
  2. The electric force of attraction

Centrifugal force = mv2/r

The electric force of attraction between two opposite charges is given from Coulomb’s law of electrostatic attraction.

∴ Electrostatic force = e2/r2

Centrifugal force and electric force acted in the opposite direction. Thus the electron may keep on revolving in its orbit if these two forces act in the opposite direction must balance by each other.

\therefore \frac{mv^{2}}{r}=\frac{e^{2}}{r^{2}}

Bohr radius of the hydrogen atom

Remarkable suggestion from Niels Bohr, the angular momentum of the electron is an integral multiple of h/2π.

mvr=\frac{nh}{2\pi }

or,v=\left ( \frac{nh}{2\pi } \right )\times \left ( \frac{1}{mr} \right )

where mvr = angular momentum and n = quantum number having the values 1,2,3, ……∞.

Thus putting the value of v on the force equation

\frac{e^{2}}{r}=m\times \left (\frac{nh}{2\pi } \right )^{2}\times \left ( \frac{1}{mr} \right )^{2}=\frac{n^{2}h^{2}}{4\pi ^{2}mr^{2}}

{\color{DarkBlue} \therefore r=\frac{n^{2}h^{2}}{4\pi^{2}me^{2}}}

The radius of the first orbit of the hydrogen atom

From the above equation, we can easily find out the radius of the permitted energy levels of the hydrogen in terms of the quantum number. Thus when n = 1, the radius of the first stationary orbit of hydrogen.

r_{1}=\frac{1^{2}\times h^{2}}{4\pi ^{2}me^{2}}

=\frac{1^{2}\times \left (6.627\times 10^{-27} \right )^{2}}{4\times \left ( 3.14 \right )^{2}\times \left ( 9.108\times 10^{-28} \right )\times \left ( 4.8\times 10^{-10} \right )^{2}}

∴ r1 = 0.529 × 10-8 cm = 0.529 Å

The ratio of first Bohr orbit and nth orbit of the hydrogen atom

\frac{r_{n}}{r_{1}}=\left ( \frac{n^{2}h^{2}}{4\pi ^{2}me^{2}} \right )\times \left ( \frac{4\pi ^{2}me^{2}}{1^{2}h^{2}} \right )=n^{2}

∴ rn = n2 × r1

Question
Calculate the radius of the second orbit of hydrogen if the radius of the first orbit of hydrogen = 0.529 Å.

Answer
The radius of the second orbit of the hydrogen atom

r2 = n2 × r1 = 2.12 Å

The velocity of an electron in Bohr model

From the principle of quantization to the revolving electron in the hydrogen atom

mvr=\frac{nh}{2\pi }

or, v=\left ( \frac{nh}{2\pi } \right )\times \left ( \frac{1}{m} \right )\times \left (\frac{1}{r} \right )

Putting the values of r in the above equation, we have

v=\frac{nh}{2\pi m}\times \frac{4\pi ^{2}me^{2}}{n^{2}h^{2}}=\frac{2\pi e^{2}}{nh}

Thus the velocity of the second orbit will be one half of the first orbit and one-third of the first orbit and so on.

Question
Calculate the velocity of the hydrogen electron in the first and third energy levels of an atom. How to calculate the number of rotation of an electron per second in the third energy level?

Answer
The velocity of an electron nth level of the hydrogen atom

vn = 2πe2/nh
where n = 1, 2, 3, ……..

Thus the velocity of an electron in the first energy level

v1 = 2πe2/12 × h
= 2πe2/h

=\frac{2\times \left ( 3.14 \right )\times \left ( 4.8\times 10^{-10} \right )^{2}}{6.627\times 10^{-27}}

= 2.188 × 108 cm sec-1

The velocity of the third energy level
v3 = v1/3
∴ v3 = (2.188 × 108 cm sec-1)/3
= 7.30 × 107 cm sec-1

The radius of the third energy level
= 32 × 0.529 × 10-8 cm

Circumference of the third energy level
= 2πr = 2 × 3.14 × 0.529 × 10-8 cm

Thus the rotation of an electron per second in the third energy level of the hydrogen

= (7.30 × 107)/(2 × 3.14 × 9 × 0.529 × 10-8 )
= 2.44 × 1014 sec-1

The energy of an electron in Bohr’s model

The energy of an electron moving in one particular energy level can be calculated by the total energy or the sum of the kinetic energy and the potential energy of an electron.

Kinetic\, energy=\frac{1}{2}mv^{2}

Potential\, energy=\int_{0}^{\infty }\frac{e^{2}}{r^{2}}dr=-\frac{e^{2}}{r}

Thus the total energy for the nth level of the hydrogen atom

E_{total}=\frac{1}{2}mv^{2}+\left ( -\frac{e^{2}}{r} \right )=-\frac{1}{2}mv^{2}

where mv2 = e2/r

Putting the value of r in the above equation

{\color{DarkBlue} \therefore E_{total}=-\frac{e^{2}}{2}\times \frac{4\pi ^{2}me^{2}}{n^{2}h^{2}}=-\frac{2\pi ^{2}me^{4}}{n^{2}h^{2}}}

When n = 1, the energy of the first orbit of the hydrogen

E_{1}=\frac{2\pi ^{2}me^{4}}{1^{2}\times h^{2}}

{\color{DarkBlue} \therefore E_{n}=\frac{E_{1}}{n^{2}}}

As n increases the energy becomes less negative and hence the system becomes less stable. Because with increasing n, r also increases and orbit makes less stable.

Let the energies associated with 1st, 2nd, 3rd,…., nth orbits are E1, E2, E3 … En.

∴ E1ㄑE2ㄑE3ㄑ……..ㄑEn

The energy of an electron in first Bohr orbit

The energy of the moving electron in the first energy level obtained by putting n=1 in the energy expression of the hydrogen.

E_{1}=\frac{2\times \left ( 3.14 \right )^{2}\times 9.109\times 10^{-28}\left (4.8\times 10^{-10} \right )^{4}}{1^{2}\times \left (6.6256\times 10^{-27} \right )^{2}}

= – 21.79 × 10-12 erg = – 13.6 eV
= – 21.79 × 10-19 Joule = – 313.6 Kcal

Question
H, H+, He+ and Li+2 – for which of the species Bohr’s model of the hydrogen atom is not applicable?

Answer
From the above species H, He+, Li+2 contain one electron but H+-ion has no electron. But the Bohr’s model of hydrogen atom applicable for one electronic system. Thus for H+-ion, this model is not applicable.