Question

The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.

Answer 1

Frequency of revolution, `f=(v)/(2pir)`
The moving charge is equilvalent to a current loop, given by
`" " I=fxxe or I=(ev)/(2pir)`
If A be the area of the orbit, then the magnetic moment of the orbiting electrons is
`" " M=IA=((ev)/(2pir))(pir^(2))=(evr)/(2)`
Putting the value, we get
`" " M=((1.6xx10^(-19))(2.2xx10^(6))(5.3xx10^(-11)xx10^(-2)))/(2)`
`" " M==9.3xx10^(-26)A-m^(2)`