calculate the frequency of revolution of the electron in the ground state of hydrogen atom.

Explanation

Let V be the velocity of the electron in the ground state of hydrogen atom r the radius of first orbit. We have,
`v=(e^(2))/(2hepsilon_(0)) and r=(epsilon_(0)h^(2))/(pime^(2))`
If T be the period of revolution of the electron in the orbit then `T=(2pir)/(v)`
`:.` Frequency of revolution `f=1/T=v/2pir`
Substituting v and r we get, `f=((e^(2))/(22hepsilon_(0)))/(2pi(epsilon_(0)h^(2))/(pime^(2)))=(e^(2))/(2hepsilon_(0))xx(pime^(2))/(2piepsilon_(0)h^(2))=(me^(4))/(4piepsilon_(0)h^(3))`
`f=(9.1xx10^(-31)xx(1.6xx10^(-19))^(4))/(4xx(8.854xx10^(-12))(6.625xx10^(-33)))=6.569xx10^(15)` rps