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.
A
3.28 x 10\(^{16}\) rad s\(^{-1}\)
B
8.55 x 10\(^{3}\) rad s\(^{-1}\)
C
9.11 x 10\(^{13}\) rad s\(^{-1}\)
D
5.22 x 10\(^{15}\) rad s\(^{-1}\)
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Angular speed of the electron (W)
w = \(\frac{Velocity}{Radius} = \frac{V}{R}\)
= \(\frac{2.0 \times 10^6}{6.1 \times 10^{11}}\)
= 3.28 x 10\(^{16}\) rads\(^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #58963)