Calculate the frequency of revolution of the electron in this second Bohr orbit of the hydrogen atom. The radius of the orbit is `2.14 Å` and the speed of the electron in the orbit is `1.09 xx 10^(6) m//s`.

Explanation

Data: `r=2.14 Å= 2.14 xx 10^(-10) m, v=1.09 xx 10^(6) m//s`
`v=omega r =(2pif)r`
`therefore` The frequency of revolution of the electron in the second Bohr orbit,
`f=v/(2pir)`
`=(1.09 xx 10^(6))/(2 xx 3.142 xx 2.14 xx 10^(-10)) = 8.205 xx 10^(14) Hz`