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...

Assessment: WAEC SSCE - Physics - 2019 (Objective) Subject: Physics

Question 1 Report

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.

3.28 x 10\(^{16}\) rad s\(^{-1}\)

8.55 x 10\(^{3}\) rad s\(^{-1}\)

9.11 x 10\(^{13}\) rad s\(^{-1}\)

5.22 x 10\(^{15}\) rad s\(^{-1}\)

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)

Read lesson note on Position, Distance And Displacement (WAEC)

Position, Distance And Displacement

View Answer

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...

Related Questions

Download The App On Google Playstore