Theory of Hydrogen atom 

 The  orbital electron in the hydrogen atom  and the central proton are in the same plane which is perpendicular to the spin of the proton.  In the orbital motion of the electron ,it cann't move parallel to the magnetic field. Inorder to  have continuous motion in the  nuclear magnetic field , all the circumnuclear electrons move is a plane perpendicular to the magnetic field.

      For stability of the orbital electron in the Hydrogen atom, the electrical force of attraction between the nucleus and the  electron in the circular orbit  must be equal to the centrifugal force.  

                             electrostatic force of attraction      =    centrifugal force

                                              e^2 / [4πεo] r^2                           =  m v^2 / r                                      ....(1)

                                                e^2 / [4πεo] m   = constant    =  v^2 r

r = radius of the electronic orbit, m = mass of the electron, v = linear velocity of motion in the circular orbit and  εo = permittivity of free space. It shows that v^2  is inversely proportional to the radius of the orbit. Greater the radius  smaller will be the value of v. If it is so, the angular velocity  (ω = v/r) of the circumnuclear electron would be different in different orbits. (If  ω is different , the relative position of the circumnuclear electrons in many electron system will vary continuously so that the mutual interactions on different electrons  will be different  which may have some impact on the stability of the atom. Infact the relative distribution of the electrons in different orbits must be such that the force experienced by every individual electrons will be same .The electric charge in the particles and the magnetic field produced by virtue of its spin play an important role in keeping up the disciplined distribution.) 

 The electron in an atom will take up certain selective , priviledged orbits  it will not stay in any intermediate orbits between any two non-radiative orbits.The reason is the micro particles have dual nature. Its wave characteristics demand that the circumference of any electronic orbits must be equal to some multiples of electron wavelenth ( λ= h/mv).  2π r = nλ  It proves that the angular momentum of the orbital electron is some integral multiples of h/2π. 

                                                  2π r = nλ = n [h/mv]

                                        mvr = n [h/ 2π]                                                       ....   (2)

The wave characteristics of the electron make the electronic orbits to be discrete.

    From Equ (1)     m v^2 r =  e^2 / [4πεo] 

                             mvr = n [h/ 2π]   

Dividing one by the other we have   v =   [e2 x 2π]/n [4πεo] h   ,which indicates that v is inversely proportional to n .i.e, in outer orbits electrons move slow.

Radius of the electronic orbits 

                                   mvr = n [h/ 2π] = n ħ

 ħ is called reduced Planck's constant . Its value is 1.054571817×10−34 joule seconds.

Squaring both sides, m2 v2 r2 = n2   ħ2  or v2  =   n2   ħ2  /m2  r2 .Substituting this value in equ (1)

                                     e2 / [4πεo] m     =  v2 r =  n2   ħ2  /m2  r

                                                                             rn =   [n2   ħ2  /m2] [4πεo] m/ e2 ] =  [4πεo] n2   ħ2  / m e2      ....... (2)

It shows that the radius of the orbital electron is directly proportional to n2 , as we go out from the centre the orbits are widely separated.

Radius of the first orbit in the hydrogen atom

     It is called Bohr's radius and is denoted by ao.  When n = 1, rn = ao

                               ao  = [4πεo]  ħ2  / m e2

                      =[4 x22 x 8.85 x 10-12 x (1-54589 x 10-34)2 ] / 7 x (1.602 x 10-19)2 x 9.11 x 10-31 

                      = 866.1486184077 x 10-80 / 163.65958308 x 10-69 

                            = 5.2923 x 10-11 m

The orbits of higher radius are 4 ao , 9 ao , 16 ao .....     n2 ao

Total energy of the orbital electron          

The orbital electron has both kinetic energy due to its orbital motion and potential energy by virtue of its position in the electric field.

Kinetic energy of the single electron in the hydrogen atom = (1/2) m v2

Since  e2 / [4πεo]r = m v2 , K.E = (1/2) [e2 / [(4πεo) r] = e2 / 8 πεor

Potential energy of the electron = - e2 / 4 πεor =  - 2 e2 / 8 πεor

Total energy of the electron = K.E + P.E = -   e2 / 8 πεor.                           ......(3)

It is negative as it binds with the nucleus 

Knowing the practical value of binding energy of the electron in the hydrogen atom (ionization energy) , one can estimate Bohr's radius.

Binding energy = - 13.6 e.V   = 13.6 x 1.602 x 10-19  = 2.1787 x 10-18 j      [ 1 e.V = 1.602 x 10-19 joule ]

         e2 / 8 πεor. =  13.6 e.V = 2.1787 x 10-18 j

or  r = [1.602 x 10-19]2 / 8 x (22/7) x 8.85 x 10-12 x 2.1787 x 10-18

         = [7 x (1.602)2  x 10-38] / 8 x 22 x 8.85 x 2.1787 x 10-30

       = [17.964828 / 3393.54312] x 10-8 

     = 5.3 x 10-11 m 

It has same value as that of Bohr's radius.

Energy of the orbital electron in various orbits in hydrogen atom

Total energy of the electron  in a privileged orbit denoted by n is En.  From equ (3) 

                En  = -  e^2 / 8πεorn

Substitute the value of rn  from equ (2)   En  = - [ e^2 / 8πεo] x [ m e^2 / [4πεo] n^2   ħ^2 

                                                                                                                                    = - [ e^4 x m] / 2 (4πεo)^2  ħ^2] x (1/n^2) 

When the electron is in the first stable orbit  the system is said to be in the ground state. The energy of the system  becomes E1  =  - [ e4 x m] / 2 (4πεo )2  ħ2]  =  -  [ e4 x m] / 8 εo2 h2

                       = - [(1.602 x 10-19)^4 x 9.11 x 10^-31 x 7^2] / [2 x 16 x 22^2 x (8.85 x 10^-12)^2 x(1.054589x 10^-34 )^2]

                                    =  - (1.602)^4  x 9.11x 49 x 10^-15 / 2 x 16 x484 x (8.85)^2  x (1.0545879)^2

                                   = - 2.1792956 x 10-18  j = 2.1792956  x 10-19 / 1.602 x 10^-19   = -13.6 e.V

Energy of the excited states 

When the electron is in the higher orbits with n ≥ 2, the system is said to be in the excited state.The system has less binding energy.

The energy of the electron in the excited states En =  - [ e^4 x m] / 2 (4πεo )^2  ħ^2] x (1/n^2)  where n = 2,3,4..... or  En = - 13,6 / n^2    eV 

When the electron is in any one of the excited states, automatically without any external influence, it turns into its ground state, where the potential energy is minimum . During this transition, the difference in total energy is emitted out as em waves.

                        ΔE = E excited /initial  - E ground / final

                                  = [ e^4 x m] / 2 (4πεo )^2  ħ^2] [(1/nf^2 - 1/ni^2)] =[m e4  / 8 εo^2 h^2]  [(1/nf^2 - 1/ni^2)]

The frequency of the emitted radiation ν =  ΔE / h = [m e^4  / 8 εo^2 h^3] [(1/nf^2 - 1/ni^2)]

The wavelength of the emitted radiation  λ = C/ ν=  8 c εo^2 h^3 / (1/nf^2 - 1/ni^2) m e^4

                                                                                                                               ___                                                                                                                                                           The wave number of the emitted radiation   ν = 1/  λ  = ν / c= [(m e4 )/ 8 c εo^2 h^3 ] x [(1/nf^2 - 1/ni^2)].  = R H  (1/nf^2 - 1/ni^2) where nf  > n and RH  is called Rydberg constant 

RH  = [(m e4 )/ 8 c εo^2 h^3 ] 

     = [(1.602 x 10-19)4  x 9.109 x 10-31] / [8 x 2.998 x 108 x (8.854x 10-12)2 x (6.626  x 10-34)3 ]  

     = 1.09737 x 107 m-1

 This relation explains why hydrogen atom has a spectrum with only certain selective wave-lenths, which are the characteristic feature of the hydrogen atom

When nf  = 2,3,4... and ni  = 1 -  we have Lyman series

           nf  = 3,4,5... and ni  = 2 -  we have Balmer series

          nf  = 4,5,6... and ni  = 3 -  we have Paschen series

          nf  = 5,6,7.... and ... and ni  = 4 -  we have Brackett  series

nf  = ∞ , gives the limit of the  each series

The hydrogen spectrum exhibits distinct wavelengths within different spectral series. The Lyman series lies in the ultraviolet range (91 nm - 121 nm), the Balmer series in the visible light range (364 nm - 656 nm), the Paschen series in the infrared range (820 nm - 1875 nm), and the Brackett series in the infrared range (1458 nm - 4051 nm).    

 Lyman Series 

These are the spectral lines due to transition of electron from the excited state with n =2 or > 2 to its ground state with n = 1. The wavelength of the emitted radiation   λ =  8 c εo2 h3 / (1/nf2 - 1/ni2)  m e4 =  1/ [RH  (1/nf2 - 1/ni2)] = 10-7 / 1.09737[ 1 - 1/ni2 ]   m  

 First line is due to the transition of electron from 2 nd to 1 st orbit.  λL1 = 10-7 / 1.09737 x (3/4)  

= 0.91127 x 10-7 x 4/3 = 1.215 x 10-7  m = 121.5 nm  

The second line is due to the transition of the electron from 3 --> 1 . λL2  =  0.91127 x10-7   x 9/8 = 102.56 nm

The third line is due to the transition of the electron from 4--> 1 .λL3  =  0.91127 x10-7   x 16/15 =  97.25 nm

The limit of the series  is due to the transition of the electron from infinity to 1 st orbit. λL∞ =  0.91127 x10-7   x 1 =  91.127 nm

Balmer series

The first line of the Balmer series is due to the transition of electron from 3--> 2. λB1 = 0.91127 x 10-7  x  36/5 = 656.28 nm

The second line of the Balmer series is due to the transition of electron from 4--->2 . λB2 = 0.91127 x 10-7  x  16/3 = 486.0 nm

The third line of the Balmer series is due to the transition of electron from 5 ---> 2. λB3 = 0.91127 x 10-7  x  100/21 = 434.0 nm

The limit of the series  is due to the transition of the electron from infinity to 2 nd orbit. λB∞ =  0.91127 x10-7   x 4 =  364.51 nm

Paschen series

The first line of the Paschen series is due to the transition of electron from 4--> 3. λP1 = 0.91127 x 10-7  x  144/7 = 187.46 nm

The second line of the Paschen series is due to the transition of electron from 5--->3 . λP2 = 0.91127 x 10-7  x  225/16 = 128.15  nm

The third line of the Paschen series is due to the transition of electron from 6 ---> 3. λP3 = 0.91127 x 10-7  x  12 = 109.35 nm

The limit of the series  is due to the transition of the electron from infinity to 3 nd orbit. λP∞ =  0.91127 x10-7   x 9 =  82.02 nm