An atom consists of a dense stationary 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 on the moving electron.

Out of many ci rcular orbits around the nucleus. an electron can revolve only in those orbits whose angular momentum `(mvr)` is an integral multiple of factor `h//2 pi`

`mvr=(nh)/(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 such 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 least possible energy 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 photon. 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 its energy and come back to the ground state.

Energy absorbed or released in an electron jump, ( `D` `E`) is given by `=>` `D``E`= `E_2 - E_1` = `h nu`

Where `E_2` and `E_1` are the energies of the electron in the first and second energy levels, and `nu` is the frequency of radiation absorbed or emitted.
[ Note: If the energy supplied to hydrogen atom is Jess 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 different from what is required for a particular transition 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 appears as kinetic energy of emitted photo electron].

Consider an electron of mass 'm' and charge 'e' revolving around a nucleus of charge `Ze` as shown in fig (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 Coulombic force

`= (KZe^2)/r^2`

`K = 1/(4 pi epsi_0)` (where `hat i_0` 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`

The centrifugal force acting on the electron is `(mv^2/r)`
'
Since the electrostatic force balance the centrifugal force, for the stable electron orbit.

`(mv^2)/r = (KZe^2)/r^2` .......(1) or `v^2 = (KZe^2)/(mr) `....(2)

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

`mvr = (nh)/(2 pi) ; v = (nh)/(2pi n r) ; v^2 = (n^2h^2)/(4 pi ^2 m^2 r^2)` .............(3)

Equating (2) and (3) `(KZe^2)/(mr) = (n^2h^2)/(4 pi m^2 r^2)`

Solving for `r` we get `r = (n^2h^2)/(4pi^2 m K Ze^2)`, where 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_0` .

`r_0 = (n^2h^2 )/(4 pi ^2 m e^2 K) = (1^2 xx (6.626 xx 10^-24 )^2)/(4xx (3.14)^2 xx 9 xx 10^-21 xx (1.6 xx 10 ^-19 )^2 xx 9 xx 10^2) = 5.29 xx 10^-11` m

`= 0.529 overset o A`

Radius of `n^(th)` orbit for an atom with atomic number `Z` is simply written as

`r_n = 0.529' n^2/Z overset o A`

We know that `mvr = (nh)/(2pi) ; v = (nh) /(2pi mr)`

By substituting for r we are getting ; `v = 2 (2 pi KZ e^2)/(nh)`

Where excepting `n` and `z` all are constants ; `V = 2.18' 10^8` Z/n` cm/sec.

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

Kinetic energy of the electron `= 1/2 mv^2`

Potential energy = `(-KZe^2)/t`

Total energy ` =1/2 mv^2 - (KZe^2)/r` .....(4)

From equation (1) we know that `(mv^2)/r = (KZe^2)/r^2 ; 1/2 mv^2 = (KZe^2)/(2r)`

Substituting this in equation (4)

Total energy `(E) = (KZe^2)/(2r) - (KZe^2)/r = - (KZe^2)/(2r)`

Substituting for `r`. gives us `E = (2 pi^2 m Z^2 e^4 K^2)/(n^2h^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.

`E = -21.8 ' 10^-12 ' z^2/n^2` erg per atom `= -21.8 ' 10^-19 ' z^2/n^2` J

per atom `= - 13.6 ' Z^2/n^2` eV per atom

`(1 eV= 3.83 ' 10^-23 kcal, 1 eV= 1.602 ' 10^-12 erg, 1eV= 1.602 ' 10^-19J)`

`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).