NMR THEORY-Orientation of a nuclei in presence of Magnetic Field

The nuclei of many elemental isotopes possess an Intrinsic spin, therefore they are associated with an angular momentum μ . The total angular momentum of a nucleus is given by

μ = h/2π . [I(I+1)]

Where

h= Planks Constant

I= Nuclear Spin or Spin Number having values as 0, 1/2, 1, 3/2.......

Note: If I = 0, then the nucleus does not posses a spin.

The numerical values of the spin number I is related to the mass number and the atomic number as follows

Atomic Number	Mass     number	No. of        Protons	No. of Neutrons	Spin Quantum no.	Examples
even	even	even	even	I=0	12C,16O
even	odd	even	odd	I= ½,3/2 , 5/2	13C
odd	odd	odd	Zero or even	I= ½,3/2 , 5/2	1H
odd	even			I = 1,2,…	14N,2H

Since atomic nuclei are also associated with an electric charge, the spin give rise to a magnetic field. Therefore a spinning nucleus may be considered as a minute bar magnet.

In absence of magnetic field the nucleus is oriented in random orientations and they have a degenerate energy level that is equal energies.

In a magnetic field the angular momentum of a nucleus (I>0) is quantized, the nucleus takes up one of (2I+1) orientations with respect to the direction of the applied field.

Each orientations corresponds to a characteristic potential energy of the nucleus equal to

E = μ . H₀ . Cos θ

Where

Ho = Strength of applied field

θ = Angle which the spin axis of the nucleus makes with the direction of the applied field.