Question

Radius of nth Bohr orbit is directly proportional to square of orbit number.
( operatorname{Rn} alpha n^{wedge} 2 )
As ratio of orbits is given -
( R / R^{prime}=left(n / n^{prime}right)^{wedge} 2 )
( 4 / 9=left(n / n^{prime}right)^{wedge} 2 )
( n / n^{prime}=2 / 3 )
Frequency of revolution of electron in nth Bohr orbit is inversely proportional to cube of orbit number.
( f propto 1 / n^{wedge} 3 )
Therefore, ratio of frequencies of revolution is -
( f / f^{prime}=left(n^{prime} / nright)^{wedge} 3 )
( f / f^{prime}=(3 / 2)^{wedge} 3 )
( f / f^{prime}=27 / 8 )
Hence, ratio of frequencies of revolution of electrons is ( 27 / 8 ).

# 17.2 The ratio of radius of two different orbits in a H-atom is 4: 9. Then, the ratio of the frequency of revolution of electron in these orbits is: (A) 2:3 (B) 27:8 (D) 8:27 (C) 3:2

Solution