Obtain an expression for orbital magnetic moment of an electron rotating about the nucleus in an atom and explain Gyromagnetic ratio.

Explanation

1. Hydrogen Bohr model is shown in figure.

2. The electron of charge (-e) performs uniform circular motion around a stationary heavy nucleus of charge (+ ze). This constitutes a current I, where,
`I=e/T" "...(1)`
and T is the time period of revolution.
3. Let r be the orbital radius of the electron and v the orbital speed.
Then, `v=(2pir)/T`
`thereforeT=(2pir)/v" "...(2)`
4. Substitute equation (2) in (1),
`I=(ev)/(2pir)" "...(3)`
(5) There will be a magnetic moment usually denoted by `mu_(l)` associated with this circulating current, generally `mu_(l)` is given by,
`mu_(l)=IA`
`mu_(l)=I(pir^(2))`
= `(ev)/(2pir)(pir^(2))`
`mu_(l)=(evr)/2" "...(4)`
6. The direction of this magnetic moment is into the plane of the paper in figure. (This follows from the right hand rule discussed earlier and the fact that the negatively charged electron is moving anticlockwise, leading to a clockwise current).
7. Multiplying and dividing the right hand side of the above expression by the electron mass `m_(e)` we have,
`mu_(l)=e/(2m_(e))(m_(e)vr)`
but `m_(e)vr` = angular momentum of orbital electron
`(becausevecl=vecrxxvecp=l=rp sintheta=rm_(e)vsin90^(@)=m_(e)vr)`
`thereforemu_(l)=e/(2m_(e))(l)" "...(5)`
8. Vectorially,
`vecmu_(l)=(-e)/(2m_(e))(vecl)" "...(6)`
9. The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.
10. Instead of electron with charge (-e), if we had taken a particle with charge (+ q) the angular momentum and magnetic moment would be in the same direction.
11. From equation (5), we can take ratio of `mu_(l)andl`, then
`mu_(l)/l=e/(2m_(e))" "...(7)`
This ratio is called gyromagnetic ratio which is constant having value `8.8xx10^(10)C//kg` for an electron which has been verified by experiment.