Question

Obtain an expression for the frequency of radiation emitted when ahydrogen atom de-excites from level n to level (n–1). For large n,show that this frequency equals the classical frequency of revolutionof the electron in the orbit.

Answer 1

Consider, a hydrogen atom is de-excited from an upper level ( n ) to a lower level ( n−1 ).

The relation for radiated energy ( E 1 ) at level n is given as,

E 1 =h ν 1 = hm e 4 ( 4π ) 3 × ε 0 2 × { h 2π } 3 ×( 1 n 2 ) (1)

where, the frequency of radiation at level n is ν 1 , the Planck’s constant is h, mass of the hydrogen atom is m, the charge on electron is e , and permittivity of the free space is ε 0 .

Now, the relation for radiated energy ( E 2 ) at level ( n−1 ) given as,

E 2 =h ν 2 = hm e 4 ( 4π ) 3 × ε 0 2 × { h 2π } 3 ×( 1 ( n−2 ) 2 ) (2)

Where, the frequency of radiation at level ( n−1 ) is ν 2 .

Energy released as a result of de-excitation can be given as,

E= E 2 − E 1 hν= E 2 − E 1 (3)

Substitute the values from equation (1) and (2) in the equation (3).

ν= m e 4 ( 4π ) 3 × ε 0 2 × { h 2π } 3 [ 1 ( n−1 ) 2 − 1 n 2 ] = m e 4 ( 2n−1 ) ( 4π ) 3 × ε 0 2 × { h 2π } 3 × n 2 × ( n−1 ) 2 ν= m e 4 32 π 3 ε 0 2 ( h 2π ) 3 × n 3 (4)

Where, ( 2n−1 )=n and ( n−1 )=n for larger value of n.

According to classical relation of frequency of revolution of an electron,

ν c = v 2πr (5)

where, the velocity of electron is v and radius of orbit is r.

The velocity of electron of n th orbit is given as,

v= e 2 4π ε 0 ( h 2π )n (6)

The radius of n th orbit is given as,

r= 4π ε 0 ( h 2π ) 2 m e 2 × n 2 (7)

Substitute the value from equation (6) and (7) in the equation (5).

ν c = v 2πr = e 2 4π ε 0 ( h 2π )n 2π×[ 4π ε 0 ( h 2π ) 2 m e 2 × n 2 ] ν c = m e 4 32 π 3 ε 0 2 ( h 2π ) 3 × n 3 (8)