Angular speed of an electron in a Bohr's orbit is given by

Explanation

The radius of the nth Bohr orbit is
`r=(epsilonh^(2)n^(2))/(pi me^(2)) " " ` …(1)
where `epsilon_(0)` is the permittivity of free space, h is Planck's constant, n is the principal quantum number, m is the mass of the electron and e is the charge on the electron. The linear speed of the electron in this orbit is
`v=(e^(2))/(2epsilon_(0)nh)`
Since angular speed `omega=(v)/(r)`. then from Eqs. (1) and (2), the angular speed of the electron in the nth Bohr orbit is
`omega=(v)/(r)=(e^(2))/(2epsilon_(0)nh)*(pime^(2))/(epsilon h^(2)n^(2))=(pime^(4))/(2epsilon_(0)^(2)h^(3)n^(3)) " " `...(3)
From Eq. (3), the frequency of revolution of the electron,
`f=(omega)/(2pi) =(1)/(2pi)xx(pime^(4))/(2epsilon_(0)^(2)h^(3)n^(3)) =(me^(4))/(4epsilon_(0)^(2)h^(3)n^(3)) " "`...(4)
[ Note : The period of revolution, `T=(1)/(f)=(4epsilon_(0)^(2)h^(3)n^(3))/(me^(4))`]