__Planck’s quantum theory :__

When a black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, Max Planck put forward a theory known as Planck’s quantum theory.

**The main points of quantum theory are :**

**(i)** Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy.

**(ii)** The smallest packet of energy is called quantum. In case of light the quantum is known as photon.

**(iii)** The energy of a quantum is directly proportional to the frequency of the radiation .

**E ∝ ν**

**E = hν**

were ν is the frequency of radiation and h is Planck’s constant having the value 6.626 × 10^{-27} erg – sec or 6.626 × 10^{-34} J-sec.

**(iv)** A body can radiate or absorb energy in whole number multiples of a quantum hν , 2hν , 3hν ….. nhν where n is the positive integer.

**Neils Bohr used this theory to explain the structure of atom.**

### Bohr’s Atomic Model:

Bohr developed a model for hydrogen atom and hydrogen like one-electron species (hydrogenic species). He applied quantum theory in considering the energy of an electron bound to the nucleus.

__Important postulates :__

∎ An atom consists of a dense nucleus situated at the centre with the electron revolving around it in circular orbits without emitting any energy. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron.

∎ Of the finite number of circular orbits around the nucleus, an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor

$\Large m v r = n \frac{h}{2\pi} $

where, m = mass of the electron

v = velocity of the electron

n = orbit number in which electron is present

r = radius of the orbit

∎ As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary states. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. The greater the distance of the energy level from the nucleus, the more is the energy associated with it. The different energy levels are numbered as 1 , 2 , 3 , 4, ( from nucleus onwards) or K , L , M , N etc.

∎ Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state.

∎ If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2 , 3 , 4, etc.) by absorbing one or more quanta of energy. This new state of electron is called as excited state. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels.

Since the excited state is less stable, atom will lose it’s energy and come back to the ground state.

Energy absorbed or released in an electron jump, (ΔE) is given by

**ΔE = E _{2} – E_{1} = hν**

where E_{2} and E_{1} are the energies of the electron in the first and second energy levels, and ν is the frequency of radiation absorbed or emitted.

**[Note: **

If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those quanta which can take it to a certain higher energy level i.e.,

all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appear as kinetic energy of emitted photo electron ]

__Radius of hydrogen atom__

Consider an electron of mass ‘ m ‘ and charge ‘ e ‘ revolving around a nucleus of charge Ze

(where, Z = atomic number and e is the charge of the proton) with a tangential velocity v .

r is the radius of the orbit in which electron is revolving.

By Coulomb’s Law , the electrostatic force of attraction between the moving electron and nucleus is

Coulomb force $\large F = K \frac{ Z e^2}{r^2} $

$\large K = \frac{1}{4\pi \epsilon_0} $

(where ε_{o} is permittivity of free space)

K = 9 × 10^{9} Nm^{2} C^{-2}

In C.G.S. units, value of K = 1 dyne cm^{2} (esu)^{-2}

For the stable electron orbit.

$\large \frac{m v^2}{r} = K \frac{ Z e^2}{r^2} $ …………..(i)

According to Bohr’s postulate of angular momentum quantization, we have

$\large m v r = n \frac{h}{2\pi} $

$\large v = n \frac{h}{2\pi m r} $ ………(ii)

Solving (i) and (ii)

$\Large r = \frac{n^2 h^2}{4 \pi^2 m K Z e^2} $

where n = 1 , 2 , 3 – – – – – ∞

Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of n, i.e., farther the energy level from the nucleus the greater is the radius.

The radius of the smallest orbit (n = 1) for hydrogen atom (Z = 1) is r_{o}

$\large r_0 = \frac{n^2 h^2}{4 \pi^2 m K e^2} $

On putting the standard values

r_{o} = 5.29 × 10^{-11} m = 0.529 A°

Radius of nth orbit for an atom with atomic number Z is simply written as

$\Large r_n = 0.529 \frac{n^2}{Z} A^o$

__Calculation of energy of an electron__

The total energy, E of the electron is the sum of kinetic energy and potential energy.

Kinetic energy of the electron $\large = \frac{1}{2} m v^2$

From equation (i)

$\large \frac{m v^2}{r} = K\frac{Z e^2}{r^2} $

$\large \frac{1}{2}m v^2 = K\frac{Z e^2}{2r} $

Potential energy $\large = – K\frac{Z e^2}{r} $

Total energy $\large E = P.E + K.E $

Total energy $\large E = – \frac{ K Z e^2}{r} + \frac{ KZ e^2}{2r} $

$\Large E = -K\frac{Z e^2}{2r} $ ………. (iii)

Substituting for r, gives us

$\large E = \frac{2\pi^2 m K^2 Z^2 e^4}{n^2 h^2} $ ; where n = 1,2,3 ……….

This expression shows that only certain energies are allowed to the electron. Since this energy expression consist of so many fundamental constant, we are giving you the following simplified expressions

$\large E = – \frac{21.8 \times 10^{-12} Z^2}{n^2}$ erg per atom

$\large E = – \frac{21.8 \times 10^{-19} Z^2}{n^2}$ joule per atom

$\Large E = – \frac{13.6 Z^2}{n^2}$ eV per atom

= – 21.8 × 10^{-19} × Z^{2}/n^{2} J per atom = – 13.6 × Z^{2}/n^{2} eV per atom

(1eV = 3.83 × 10^{-23} kcal

1eV = 1.602 × 10^{-12} erg

1eV = 1.602 × 10^{-19} J )

E = – 313.6 × Z^{2}/n^{2} kcal / mole (1 cal = 4.18 J)

The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n = 1, and as the quantum number increases, E becomes less negative.

When n = ∞ , E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated.

H → H^{+} + e^{–} (ionisation)

**Exercise : **Find out the value of electrostatic potential energy of two electrons separated by 3.0A° in vacuum. Express your answer in joules and electron volt.