Find the maximum angular speed of the electron of a hydrogen atoms in a statoonary orbit

Answer

A, D

Explanation

We know `mv r =(nh)/(2pi)`
implies `mr^2omega = (nh)/(2pi)`
`implies omega = (nh)/(2pi xx m xx r^2)`
`=(1 xx 6.63 xx 10^-34)/(2 xx 3.14 xx 9.1 xx 10^-31 xx (0.53)^2 xx 10^-20)`
`=0.413 xx 10^17 rad//s = 4.13 xx 10^17 rad//s`