1. The magnetic moment of an electron due to its orbital motion is equal to the orbital angular momentum times the gyromagnetic ratio, which is e/2me. According to quantum theory, the orbital angular momentum is an integer multiple of h/2π.
2. The magnetic moment of an electron due to its spin motion is equal to the spin angular momentum times the gyromagnetic ratio. The spin angular momentum is sħ/2π, where s is 1/2. Therefore, the magnetic moment due to spin is 1/2 Bohr magneton.
3. The Stern-Gerlach experiment established that the magnetic moment of an electron due to its spin is equal to one Bohr magnet
4. ANGULAR MOMENTUM &
GYROMAGNETIC RATIO
I. Magnetic moment of
electron due to its orbital:-
II.Magnetic moment of
electron due to spin
motion:-
5. ANGULAR MOMENTUM &MAGNETIC MOMENT OF REVOLVING OR
ROTATING CHARGE
Consider a charge ‘q’ or revolving in a circular path of
radius ‘r’ with a uniform angular velocity w.if the
charge takes a time T to complete one round, the
current produced due to motion of charge
I = charge /time
= q/T……………… …. (i)
but T = 2p/w where w = v/r
I = q/2pr/v
= qv/2p r…… (ii)
6. Thus a rotating charge behaves like a current loop of
area A = pr2. this current loop can be treated to
be equivalent to a magnetic dipole of magnetic
moment equal to the product of current in the
loop and the effective area of the loop.
Magnetic moment of the magnetic dipole
equivalent to current loop
Ml = (current in loop) * (effective area of loop)
= I*A
= (qv/2pr2) * (pr2)
= qvr/2
7. Since the direction of motion of charge ‘q’ is anticlockwise, hence the
upper face of current loop will behaves like north pole & its lower face as
south pole , due to which the direction of magnetic moment Ml will be
outwards along the axis of loop .
If mass of the charged particle is ‘m’, the angular momentum of charge about the axis of
rotation is
L = mr*v
|L|or L = mvr
the direction of angular momentum L will be normal to the plane of paper
outwards,
Ml = qvr*m/2m
= q/2m L
8. The angular momentum and magnetic moment of the charged particle. For
the positively charged particle , both the magnetic moment and angular
momentum are in same direction , where asfor the negatively charged
particle , the direction of the magnetic moment is opposite to the direction of
angular momentum.
in the eq. (ii)
the quantity q/2m gives us the ratio of magnetic moment of a
charged particle to its angular momentum.It is called the Gyromagnetic
ratio.
It may be mentioned here that if a charge ‘q’ is uniformally distributed over a
conductor , its magnetic moment due to rotational or orbital motion can also
obtained from the eq. (ii).
If Iinertia be the moment of the inertia of the conductor about the axis of
rotation , then its angular momentum will be
L = Iinertia w
magnetic moment Ml = q/2m * Iinertia w..........................................(iii)
Now we can determine the magnetic moment produced due to orbital and spin motions of electron in an atom
9. MAGNETIC MOMENT OF ELECTRON DUE TO ITS
ORBITAL MOTION
The charge on electron is q=(-e) ,hence magnetic moment produced due
to orbital motion of electron
Ml = (-e/2me) * L………….(iv)
where L is the orbital momentum.
Here the negative sign show that the direction of magnetic moment Ml
due to orbital motion is opposite to the direction of orbitalangular momentum L.
According to quantum theory , orbital momentum L is integer multiple
of h/2p .
L = (lh/2p) where l is orbital quantum no.(0,1,…)
h is plank constant
10. therefore Magnitude of orbital magnetic moment of electron
Ml = (e/2me)*(lh/2p)
= l{(eh/4pme )}……………(v)
if you assume (eh/4pe) = mB called the Bohr magneton , then
Ml = lmB ………………(vi)
thus magnetic moment of electron due to orbital motion is an
integer multiple of Bohr magneton.
Obviously , Bohr magneton is the convenient unit of atomic
magnetic moment .
1 mB = eh/4pme
= 9.2 x 10
11. MAGNETIC MOMENT OF ELECTRON DUE TO SPIN MOTION
The magnetic moment of electron due to its spin motion
Ms =(-e/2me) x S
Where S is the spin angular momentum. But according to quantum
theory,
Spin angular momentum s = sh/2p ,where s is the spin quantum number
(-1/2) . Here the negative sign shows that the direction of spin magnetic
moment is opposite to the direction of spin angular momentum.
Therfore moment due to spin of electron Ms = (e/2me ) X (sh/2p)
= s(eh/4pme)
= smB
Ms = 1/2mB …………………(vii)
12. STERN-GERLACH EXPERIMENT
It has been established that due to spin motion of
electron , the magnetic moment is equal to one Bohr
magneton , hence in general equation (ii) can be
written as
M= g(q/2m)L with Direction
where g is a cofficient which is known as Lande g factor
. The value of g is 1 for orbital motion &for motion of
electron