Find the frequency of revolution of an electron in Bohr's 2nd orbit, if the radius and speed of electron in that orbit is `2.14 xx 10^(-10)m and 1.09xx10^(6)m//s` respectively.
`[pi=3.142]`

Explanation

Given: Radius `(r_(2))=2.14 xx 10^(-10) m.`
Speed of electron `(v_(2))=1.09xx10^(6)m//s.`
`f_(2)=(v_(2))/(2pi r_(2))`
`=(1.09xx10^(6))/(2xx3.142xx2.14xx10^(-10))`
`=(1.09xx10^(16))/(13.44776)=8.105xx10^(14)Hz.`