Question

The electron in the hydrogen atom is moving with a speed at 2.3 x 10⁶ m/s in an orbit of radius 0.53Å. Calculate the magnetic moment of the revolving electron.

Answer 1

Given: v = 2.3 x 10⁶ m/s, r = 0.53Å = 0.53 x 10^-10 m

Period of revolution of electron

`T = (2pir)/v=(2pixx0.53xx10^-10)/(2.3xx10^6)`

`=(2xx3.142xx0.53xx10^-10)/(2.3xx10^6)`

`T = 1.448xx10^-16`sec

Current,`I = e/T`

`=(1.6xx10^-19)/(1.448xx10^-16)`

`= 1.105xx10^-3A`

Magnetic moment of the revolving electron,

M = IA

= I.πr²

= 1.105 x 10^-3 x 3.142 x (0.53 x 10^-10)²

= 0.9753 x 10^-23

M = 9.753 x 10^-24