In a hydrogen atom, the binding energy of the electron in the ground state is `E_(1)` then the frequency of revolution of the electron in the nth orbit is

Options

A. `(2E_(1))/(n^(3)h)`
B. `(2E_(1)n^(3))/(h)`
C. `sqrt((2mE_(1))/(n^(3)h))`
D. `(E_(1)n^(2))/(h)`

Answer

A

Explanation

`E_(1)=(me^(4))/(8epsi_(0)^(2)h^(2))`
`f_(n)=(v_(n))/(2pir_(n))=((e^(2)//2epsi_(0)nh))/((2pi)epsi_(0)n^(2)h^(2)//pime^(2))=(me^(2))/(4epsi_(0)^(2)n^(3)h^(3))=(8E_(1)epsi_(0)^(2)h^(2))/(4epsi_(0)^(2)n^(3)h^(3))=(2E_(1))/(n^(3)h)`