For hydrogen-like sustems find the magnetic moment `mu_(n)` corresponding to the motion of an electron along the `n-th` orbit and the ratio of the magnetic and nechanical moments `mu_(n)//M_(n)`. Calculate the magnetic moment of an electron occuppying the first Bohr orbit.

Explanation

An electron moving in a circle with a time period `T` consitutes a current
`I=(e )/(T)`
and forms a current loop of area `pir^(2)`. This equivalent to magnetic a moment,
`mu= 1pir^(2)=(epir^(2))/(T)=(evr)/(2)`
on using `v=2pi r//T`. Thus
`mu_(n)=(emvr)/(2m)=(n e ħ)/(2m)`
for the `n^(th)` orbit. (In Gaussian units `mu_(n)=n e ħ//2mc`)
We see that `mu_(n)=(e )/(2.m)M_(n)`
where `M_(n)=n ħ= mvr` is the angular momentum
Thus `(mu_(n))/(M_(n))=(e )/(2m)`
`mu_(1)=(e ħ)/(2m)=mu_(B)= 9.27xx10^(-24)A.m^(2)`
(In `CGS` i nits `mu_(1)=mu_(B 9.27xx10^(-21)) erg//gauss)`