An electron moving with velocity `2.2 xx 10^(6)` m/s, revolving in circular orbit of radius 0.53 Å. Calculate its angular velocity.

Explanation

`v= 2.2 xx 10^(6)` m/s
`r= 0.53 A= 0.53 xx 10^(-10)`m
Angular velocity `omega= (v)/(r)= (2.2 xx 10^(6))/(0.53 xx 10^(-10))`
`omega= 4.15 xx 10^(16) rad s^(-1)`