Question

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

JEE/Engineering Exams

Chemistry

Solution

191

3.7 (3 ratings)

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 ).