Determine the frequency of revalution of an electron in the second Bohr orbit in hydrogen atom

Explanation

The frequency of revolation of an electron is given by

frequency = `(1)/("Time period")`
Time period = `("Total distance converd in one revolation")/("velocity")`
`= (2pi r)/(v)`
:. Frequency= `(v)/(2pi r)`
Calculate the velocity `(v_(2))` and radius `(r_(2))` of the electron in the second Bohr orbit in H atom
`Z = 1` for H atom
using `r_(n) = 0.529 (n^(2))/(Z) Å`
We get
`r_(2) = 0.529 xx 10^(-10)((2)^(2))/(1)m = 1.12 xx 10^(-10) m`
`v_(n) = 2.18 xx 10^(6)m s^(-1)`
`v_(2) = 2.18 xx 10^(6) (1//2)= 1.09 xx 10^(6) ms^(-1)`
`Frequency = (v_(2))/(2pir_(2)) = (1.09 xx 10^(6))/(2(pi)(2.12 xx 10^(-10)))`
`v = 8.18 xx 10^(14) Hz`