What is the frequency of revolution of electron present in `2nd` Bohr's orbit of `H-` atom ?

Options

A. `1.016 xx 10^(16)s^(-1)`
B. `4.065 xx 10^(16) s^(-1)`
C. `1.626 xx 10^(15) s^(-1)`
D. `8.2 xx 10^(14) s^(-1)`

Answer

D

Explanation

Frequency of revolution .
`= ("velocity in second orbit" (V_(2)))/(2 pi r_(2))`
` = (1.082xx 10^(6) ms^(-1))/(2 xx pi xx (2.12 xx 10^(-10) )m) = 8.2xx 10^(14) s^(-1)`