An electron of mass 9.1 x 10\(^{-31}\) kg moves with a speed of 2.0 x 10\(^6\) ms\(^{-1}\) round the nucleus of an atom in a Circular path of radius 6.1 x 1...
An electron of mass 9.1 x 10\(^{-31}\) kg moves with a speed of 2.0 x 10\(^6\) ms\(^{-1}\) round the nucleus of an atom in a Circular path of radius 6.1 x 10(^{11}\) m. Determine the angular speed of the electron.
Answer Details
The angular speed of the electron can be determined using the formula:
ω = v / r
where ω is the angular speed, v is the velocity of the electron, and r is the radius of the circular path.
Substituting the given values, we get:
ω = (2.0 x 10^6 m/s) / (6.1 x 10^11 m)
ω = 3.28 x 10^(-5) rad/s
Therefore, the angular speed of the electron is 3.28 x 10^(-5) rad/s.
Hence, the correct option is: 3.28 x 10^16 rad s^(-1)