Bohr's theory of hydrogen atom

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Bohr's theory of hydrogen atom


(1). Radius of permitted orbit

           
According to diagram, 

          Centripetal force on electron = m × v ×v/r

          Electrostatic force of attraction = k×z×e×e/r×r

         As this attraction provided centripetal force
         Then, it means
          
         Centripetal force = electrostatic force of                   attraction
  
         M×v×v/r = k×z×e×e/r×r   ..... (1) 
         
        M×v×v = k×z×e×e/r   ..... (2) 

        r = k×z×e×e/m×v×v   ....... (3) 
 
        W. K. T
         According to bohr's model
         L = m×v×r = n×h/2×π
         
         v = n×h/2×π×m×r
          From (2) 
    
           r = k×z×e×e/m×n×n×h×h/4π×π×m×m×r×r

           r = k×z×e×e × 4×π×π×m×m×r×r/m×n×n×h×h

           r = n×n×h×h/4×k×z×π×π×e×e×m


                                   Note

          Bohr's radius - it is defined by the radius of hydrogen atom denoted by ro. 

We have, 
 r = n×n×h×h/4×π×π×k×z×e×e×m
      
  For hydrogen atom z = 1 and n = 1
 r = ro

 ro = h×h/4×π×π×e×k×e×m

ro = 0.53 Angstrom


         

(2). Velocity of electron in different orbits

 We have, 
              v = n×h/2×π×m×r
And
              r = n×n×h×h/4×k×z×π×π×e×e×m
Substituting value of r in v

v = n×h/2×π×m×n×n×h×h/4×π×z×k×π×e×e×m

v = 2×k×z×π×e×e/n×h

v = (2×k×z×π×e×e/c×h) (c/n) 

v = alpha×(c/n) 

 Where for hydrogen (z = 1) 

Alpha = 2×k×z×π×e×e/c×h = 1/137 = constant

v = (1/137) (c/n) 

For first orbit n=1

v = c/137


Thus, it's is clear that the speed of electron is 1/137 times of rhe speed of light. 

(3). Total energy of the electron during revolution in an orbit

The total energy of orbiting electron is the sum of its KE and PE

TE = KE + PE   .... (1) 

KE of an electron =(1/2) (m×v×v) 

                       = (1/2) (k×z×e×e/r)    from equation (1) 

PE of an electron = k(z×e) (-e) /r
                                 = -k×z×e×e/r

From TE = KE + PE
TE = (1/2) (k×z×e×e/r) + (-k×z×e×e/r) 
      = k×z×e×e/2r - k×z×e×e/r = -k×z×e×e/2r

TE of orbit taken as energy of the nth orbit denoted by En

En = -k×z×e×e/2×r
But r = n×n×h×h/4×π×z×π×k×e×e×m

En = -k×z×e×e/2×n×n×h×h/4×k×z×π×π×e×e×m

En = -2×k×k×z×z×π×π×e×e×e×e×m/n×n×h×h

(4). Different spectral series of hydrogen atom 

We have, 

 En = -2×k×z×z×e×e×e×e×m×π×π/n×n×h×h

Consider two orbits having energies En1 and En2  (En2 is greater than En1) 

En1 = -2×k×k×z×z×e×e×e×e×m×π×π/n1 × n1 × h × h

And En2 = -2×k×k×z×z×π×π×e×e×e×e×m/n2 × n2 × h × h
Now, according to bohr's frequency conditions
En2 - En1 = hv

h×v = [2×k×k×z×z×π×π×e×e×e×e×m/h×h] × [(1/n1 × n1) - (1/n2 × n2) ]

v = [2×k×k×z×z×π×π×e×e×e×e×m/h×h×h] × [(1/n1×n1) - (1/n2×n2) 

W. K. T

c = (lembda) × v
. 
Lembda = c/v

1/lembda = v/c

But 1/lembda = wave no. 

Wave no. = v/c

Wave no. = [2×k×k×z×z×e×e×e×e×m×π×π/c×h×h×h]×[(1/n1×n1) - (1/n2×n2) ]

1/lembda = v/c = wave no. = R[(1/n1×n1) - (1/n2×n2)]


Where, R = Rydberg constant

Value of R = 1.09 × 10(of power of 7) m(of power of -1) 

As electron jumps from one orbit to another, it emits radiations of different wavelength and frequencies therefore we will get different in hydrogen atom. 



(1). Lymen series

        When electron jumps from orbits n2 = 2,3,4,......infinity to n1 = 1 it emits set of spectral lines lies in ultraviolet region
Then, the lines are given by

Wave no. = 1/lembda = R[(1/1×1) - (1/n2×n2) ]
                                         Where, n2 = 2,3,4, ...... Infinity

(2). Balmer series

         When electron jumps from orbits n2 = 3,4,5,....infinity to n=2
It emits set of spectral lines in visible region
Then the lines are given by

Wave no. = 1/lembda = n[(1/2×2) - (1/n2×n2) ]
                          Where, n2 = 3,4,5,....infinity

(3). Paschen series

         When electron jumps from orbit n2 = 4,5,6,....infinity to n1 = 3
It emits lines of spectral lines in infrared region
Then, the lines are given by

Wave no. = 1/lembda = R[(1/3×3) - (1/n2×n2) ]
                       Where, n2 = 4,5,6,.... Infinity

(4). Bracket series

         When electron jumps from n2 = 5,6,7 ..... Infinity to n1 = 4 it emits lines of spectral lines in infrared region then, the lines are given by

Wave no.  = 1/lembda = R[(1/4×4) - (1/n2×n2) ]
                     Where, n2 = 5,6,7,... Infinity

(5). Pfund series

         When electron jumps from orbit n2 = 6,7,8,... Infinity to n=5 it emits lines of spectral lies in infrared region then the lines are given by

Wave no. = 1/lembda = R[(1/5×5) - (1/n2×n2) ]
                     Where, n2 = 6,7,8,... Infinity

Energies of different orbits and energy level diagram

We have, 
En = -2×k×k×z×z×e×e×e×e×m×π×π/n×n×h×h
For hydrogen atom 
Z= 1
En = -2×k×k×π×π×e×e×e×e×m/n×n×h×h

En = -13. 3/n×n  ev

For (1) orbit n= 1
E1 = -13. 3 eV
For (2) orbit n= 2
E2 = -3.51eV
E3 = -13. 3/9 eV = -1.41eV
E4 = -13. 3/16 eV = -0.84eV
Einfinity = -13.3/infinity eV
                  = 0


Energy level diagram




                                        Note

To calculate shortest wavelength for all spectral series n2 taken as infinity. 


     

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