The angular speed of the electron in the first orbit in a H atom is `4.103 xx 10^(16)` rad/s. Hence the angular speed of the electron in the third orbit is

Options

A. `2.5xx10^(15)` rad/s
B. `1.5xx10^(15)` rad/s
C. `5.1xx10^(15)` rad/s
D. `5.5xx10^(15)` rad/s

Answer

B

Explanation

`omega_(n)=omega_(1)/n^(3)`
`omega_(3)=omega_(1)/27`
`=(4.103xx10^(16))/27=1.5xx10^(15)` rad/s