The angular speed of the electron in the `n^(th)` Bohr orbit of the hydrogen atom is proportional to......

Options

A. n
B. `n^(3)`
C. `(1)/(n)`
D. `(1)/(n^(3))`

Answer

D

Explanation

Angular momentum `mvr=mr^(2)omega=(nh)/(2pi)`
Angular velocity `omega=(nh)/(2pimr^(2))`
`:.r^(2)=(n^(4)h^(4)epsi_(0)^(2))/(pi^(2)m^(2)e^(4))`
`omega=(pime^(4))/(2h^(3)in_(0)^(2)).(1)/(n^(3))`
`:.omegaprop(1)/(n^(3))`