Question

In a simple model of the hydrogen atom, an electron revolves around a proton in a circular orbit of radius 5.29 x 10^-11 m with a speed of 2.18 x 10^⁶ m/s. What is the angular speed of the electron?​

Answer 1

 The force acting on the electron in the Bohr model of the hydrogen atom is directed radially inward and is equal to

F=rmv2=0.529×10−10m(9.11×10−31kg)(2.20×106m/s)2

=8.33×10−8N inward.

 a=rv2=0.529×10−10m(2.20×106m/s)2=9.15×1022m/s2 inward.

Answer 2

Answer:

The angular speed of the electron is 4.12*10^4 m/s.

Explanation:

let, radius of orbit be r

         linear speed of electron be v

here, radius of orbit, r = 5.29 x 10^-11 m.

         linear speed of electron, v = 2.18 x 10^⁶ m/s

we know, v=rω

               ω =v/r

               ω =2.18*10^6/5.29*10^-11

               ω =4.12*10^4 m/s.

so, The angular speed of the electron is 4.12*10^4 m/s.