A electron of mass, `m_(e)` revolves around a nucleus of charge `+Ze`. Show that it behaves like a tiny magnetic dipole. Hence, prove that the magnetic moment associated with it is expressed as `vec(mu) = - (e )/(2m_(e)) vec(L)`, where `vec(L)` is the orbital angular momentum of the electron. Give the signification of negative sign.

Explanation

In an atom electron revolves around the nucleus in circular obrit hence, behaves as a current loop, (magnetic dipole).
Consider an electron of mass `m_(c)` and charge e is revolving around a nucleus in a circular orbit of radius r with velocity V.
Current in the loop `I = - e/T =` (T is time period).
`:' T = (2pir)/(V) , I = (eV)/(2pir)`
Area of loop `A = pir^(2)`
`:'` Magnetic moment `= IA = - (eV)/(2pir) xx pir^(2) = -(eVr)/(2)`
`:. mu = ((m_(e)Vr)e)/(2m_(e))`
Now, `m_(e)Vr = L= " Angular momentum"`
`:. mu = -e/(2m_(e)) vec(L^(-))`
From Bohr's quantum condition `L = (nh)/(2pi) rArr mu = e/(2m_(e)) (nh)/(2pi) = n((eh)/(4m_(e)h))`
`(eh)/(4pim_(e)=` Bohr's magneton `= a "const"`.
Here, the -ve sign signfies direction of conventional current is opposite t to the motion of electron. Dipole moment and angular moment are oppositely directed.