# Bohr’s model of hydrogen atom

### Bohr’s theory of hydrogen atom

Explanation of the hydrogen spectrum, Neils Bohr’s in 1913 adopted the Rutherford model of the hydrogen atom in which an electron revolves around the single proton at the central nucleus.

According to classical mechanics, when a charged particle is subjected to the acceleration it emits radiation and loses energy and the orbital radius must also be changed.

But an electron revolving around the nucleus would, therefore, be continually accelerated towards the center of the orbit and consequently emitting radiation.

Thus the radius of curvature of its path would go on decreasing and due to spiral motion, the electrons will finally fall on the nucleus when all its rotational energy spent on the electromagnetic radiation and the atom would collapse.

If the energy of the electron loses continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other. But the observed emission spectrum consists of well-defined lines of definite frequencies.

To resolve the anomalous position Niels Bohr’s proposed a new atomic model of the hydrogen atom.

#### Bohr’s model of hydrogen atom postulates

An atom possesses several stable circular orbits in which an electron can stay. So long as an electron can stay in a particular orbit there is no emission or absorption of energy. These orbits are called energy levels of an atom.

The energy levels are numbered as 1, 2, 3, …….. starting from the nucleus of an atom and are designated as capital letters, K, L, M, …….. respectively.

The energy associated with a certain energy level increases with the increase of its distance from the nucleus.

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

E1ㄑE2ㄑE3ㄑ……

An electron can jump from one energy level to another higher energy level on the absorption of energy and one energy level to another lower energy orbit with the emission of energy.

The angular momentum of an electron moving in an orbit or energy level is an integral multiple of h/2π. This integral multiple is known as the principal quantum number of an atom.

$\therefore&space;mvr=\frac{nh}{2\pi&space;}$

where m = mass of an electron, v = tangential velocity of an electron in an energy level, r = distance between the electron and nucleus of an atom and n = whole number which has been given the principal quantum number of an atom.

The amount of energy (ΔE) emitted or absorption in this type of jump of the electron is given by Plank’s equation.

ΔE = hν

where ν is the frequency of the radiation emitted or absorbed by an electron and h is the Plank constant.

#### Ground state energy of hydrogen atom formula

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.

Evidently, on the revolving electron, two types of forces are acting, centrifugal force and electric force of attraction.

#### Centrifugal and electric force of an atom

Orbitals are stable when the centrifugal force exerted by the moving electron must equal to the attractive force between the electron and nucleus of the hydrogen atom.

Centrifugal force = mv2/r

Two attractive forces are in the operation, one being the electric force of attraction between the nucleus and the electron

But the other force is the weak gravitational force, which can be neglected.

So the electric force of attraction between two opposite charges is given by Coulomb’s law.

${\color{DarkBlue}&space;e\times&space;\frac{e}{r^{2}}=\frac{e^{2}}{r^{2}}}$

Centrifugal force and electric force has acted in the opposite direction. Though the electron may keep on revolving in its orbit, these two forces, which act in the opposite direction must balance each other.

$\therefore&space;\frac{mv^{2}}{r}=\frac{e^{2}}{r^{2}}$

$or,mv^{2}=\frac{e^{2}}{r}$

#### How do you calculate the atomic radius of hydrogen?

Bohr’s remarkable suggestion for the angular momentum of the electron is an integral multiple of h/2π. If angular momentum of an electron = mvr.

$\therefore&space;mvr=\frac{nh}{2\pi&space;}$

where n = integer called quantum number indicating hydrogen energy levels having the values 1,2,3, ……∞.

$v=n\times&space;\frac{h}{2\pi&space;}\times&space;\frac{1}{mr}$

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

Putting the value of v on this equation,

$\frac{e^{2}}{r}=m\times&space;n^{2}\times&space;\left&space;(\frac{h}{2\pi&space;}&space;\right&space;)^{2}\times&space;\left&space;(&space;\frac{1}{mr}&space;\right&space;)^{2}$

$or,\frac{e^{2}}{r}=\frac{n^{2}h^{2}}{4\pi&space;^{2}mr^{2}}$

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

#### The radius of orbit of the hydrogen atom

Solution for the radius of the permitted energy levels of the hydrogen atom in terms of the quantum number. When n = 1, the radius of the first stationary orbit of hydrogen.

$\therefore&space;r_{1}=1^{2}\times&space;\frac{h^{2}}{4\pi&space;^{2}me^{2}}$

$=\frac{1^{2}\times&space;6.627\times&space;10^{-27}}{4\times&space;\left&space;(&space;3.1416&space;\right&space;)^{2}\times&space;\left&space;(&space;9.108\times&space;10^{-28}&space;\right&space;)\times&space;\left&space;(&space;4.8\times&space;10^{-10}&space;\right&space;)^{2}}$

= 0.529 × 10-8 cm = 0.529 Å = a₀

Thus the radius of first orbit r1 = a₀, second orbit r2 = 4 a₀ and third orbit r3 = 9 a₀.

${\color{DarkBlue}&space;\therefore&space;r_{n}=\frac{n^{2}h^{2}}{4\pi^{2}me^{2}}}$

$or,&space;r_{n}=n^{2}\times&space;\frac{h^{2}}{4\pi^{2}me^{2}}$

∴ rn = n2 × r1 = r1 × a0

Question

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

The radius of the second orbit of a hydrogen atom

r2 = n2 × r1 = 2.12 Å

#### The velocity of an electron of the hydrogen atom

$mvr=\frac{nh}{2\pi&space;}$

where v = velocity of an electron.

$\therefore&space;v=\frac{nh}{2\pi&space;m}\times&space;\frac{1}{r}$

Putting the values of r

$v=\frac{nh}{2\pi&space;m}\times&space;\frac{4\pi&space;^{2}me^{2}}{n^{2}h^{2}}$

${\color{DarkBlue}&space;\therefore&space;v=\frac{2\pi&space;e^{2}}{nh}}$

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.

v2 = v1/2

v3 = v2/3

∴ vn = v1/n = velocity of first orbit/principal quantum shell.

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?

v = 2πe2/nh
where n = 1, 2, 3, ……..

∴ The velocity of an electron in the first energy level

v1 = 2πe2/12 × h
= 2πe2/h

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

∴ 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

∴ Rotation of an electron per second in the third energy level of the hydrogen atom

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

#### The kinetic and potential energy of an electron

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

∴ The kinetic energy of a moving electron

$=\frac{1}{2}mv^{2}$

But the potential energy due to electric attraction

$V=\int_{0}^{\infty&space;}\frac{e^{2}}{r^{2}}dr=-\frac{e^{2}}{r}$

$\therefore&space;E_{total}=\frac{1}{2}mv^{2}+\left&space;(&space;-\frac{e^{2}}{r}&space;\right&space;)$

Putting the value e2/r = mv2

$\therefore&space;E=\frac{1}{2}mv^{2}-mv^{2}$

${\color{DarkBlue}&space;=-\frac{1}{2}mv^{2}=-\frac{1}{2}\frac{e^{2}}{r}}$

#### The energy of an electron in hydrogen energy levels

The energy of an electron

$E=-\frac{1}{2}\frac{e^{2}}{r}$

$\because&space;r=\frac{n^{2}h^{2}}{4\pi&space;^{2}me^{2}}$

${\color{DarkBlue}&space;\therefore&space;E=-\frac{e^{2}}{2}\times&space;\frac{4\pi&space;^{2}me^{2}}{n^{2}h^{2}}=-\frac{2\pi&space;^{2}me^{4}}{n^{2}h^{2}}}$

When n = 1, that is the energy of the first orbit of the hydrogen atom

$E_{1}=\frac{2\pi&space;^{2}me^{4}}{1^{2}\times&space;h^{2}}$

${\color{DarkBlue}&space;\therefore&space;E_{n}=\frac{E_{1}}{n^{2}}}$

The energy being governed by the value of quantum number n. As n increases the energy becomes less negative and hence the system becomes less stable.

Also note that with increasing n, r also increases. Thus increasing r also makes the orbit less stable.

If energies associated with 1st, 2nd, 3rd,…., nth orbits are E1, E2, E3 … En, these will be in the order,

E1ㄑE2ㄑE3ㄑ……..ㄑEn

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

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

= – 21.79 × 10-12 = – 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 is not applicable?

From the above species H, He+, Li+2 contain one electron but H+-ion has no electron. Bohr’s model is applicable for one electronic system thus for H+-ion Bohr’s model is not applicable.

#### The kinetic energy of moving electron of an atom

The kinetic energy of a moving electron

$T=\frac{1}{2}mv^{2}$

${\color{DarkBlue}&space;=\frac{1}{2}m\times&space;\left&space;(\frac{2\pi&space;Ze^{2}}{nh}&space;\right&space;)^{2}=\frac{2\pi&space;^{2}mZ^{2}e^{4}}{n^{2}h^{2}}}$

where the charge of an electron (e) = 4.8 × 10-10 esu.

Plank’s constant (h) = 6.626 × 10-27 erg sec.

mass of an electron (m) = 9.1 × 10-28 gm.

∴ The kinetic energy of hydrogen atom in first energy level

= 13.6 eV.

Question

The energy of an electron in the first energy level of the hydrogen atom = -13.6 eV. What is the energy value of the electron in the excited level of lithium-ion?