Nuclear Magnetic-Dipole Moments:
The spin angular momentum is given as
I (I + 1) h / 2π
Where I is the nuclear spin quantum number
Also, magnetic dipole moment is given as
µ = IA
where I is current in amperes, or coulomb per second and
A is the enclosed area of loop.
Refer to Figure 12.42
For a circular path, we write area as
A = π r2
and
the current i.e. a charge per unit time interval, we can write for a charge of
+e and time interval of 2 π r / ν
therefore we can write current I as
I = e / (2πr / ν )
= e ν / 2πr
Thus, the magnitude of magnetic dipole moment is given as
µ=IA
= (e ν / 2πr ) (π r2)
= eνr/2
Let us introduce mp the mass of proton in above equation
Therefore,
We have
µ = m p e ν r / 2 mp
Rewriting the above equation as
= ( e / 2 mp ) × mp ν r
= Magnetic moment × Angular momentum
Thus, the magnetic dipole moment is given as ( e / 2 mp ) (i.e. the
magnetic moment) times its angular momentum (mp ν r).
This is known as a reference nuclear magnetic moment, called the nuclear
magneton, µ N by
µ N = ( e / 2 mp ) (h / 2π)
= 5.051 × 10-27 JT-1
where the unit relation 1 T = 1 N C-1 m-1 S
is used.
The nuclear magnetic moments are expressed in terms of the nuclear
magneton µ N by introducing a factor called nuclear g factor, symbol gN.
Thus, for any nucleus, we write
µ = g N I (I + 1) µ N
Values of gN are different for different nuclei.
Nuclei in Magnetic Field:
When a nuclei is placed in a magnetic field
The magnetic dipole of nuclei try to line up in the direction of applied field
and
The energy of interaction between a magnetic field and a magnetic dipole
moment is given as the product of the applied field and the magnetic
moment in the direction of field.
Thus,
Energy of interaction = µB
and
1. When the magnetic moment is in the direction of applied magnetic
field,
Then the energy is lowered by an amount of µB
2. When the magnetic moment is in the opposite direction of applied
magnetic field,
Then the energy is increased by an amount of µB
Thus, for 1H and 13C where I = 1/2
There will be two orientations or energy levels with energies +µB and -µB
+µB
No Field
B
-µB
Applied Field
and
The difference between these energy levels is = 2µB
and is shown to be equal to
= gN µ N
B
(I = 1/2)
Therefore
∆ε = gN µ N B
and the frequency associated with this quanta of energy is
∆ε = h ν
ν = ∆ε / h
= gN µ N B / h
Problem 1:
Calculate the difference in the energies of protons oriented with and against
a magnetic field of strength 10 T. What is the frequency of radiation that has
photons with this energy? Given gN = 5.5857 and µ N = 5.0508 × 10-27 C S-1
m2.
1 T = N C-1 m-1 S
= J m-1 C-1 m-1 S
= J C-1 m-2 S
The energy spacing between the two spin states = ∆ε
= gN µ N B
= (5.5857) (5.0508 × 10-27 C S-1 m2) (10 J C-1 m-2 S)
= 2.8 × 10-25 J
The frequency of radiation with photons of this energy is given by
ν = ∆ε / h = 2.8 × 10-25 J / 6.626 × 10-34 Js
= 4.2 × 108 S-1
= 420 Mhz
Nuclear State Population:
For protons in a magnetic field of 10 T, radiation of frequency of about 420
MHz can produce transition between the two orientations.
The photos of this radiations have an energy of about 2.8 × 10-25 J = 0.00028
× 10-21 J and this is nothing but the difference between the two nuclear-spin
states.
The value of kT at room temperature is 4.1 × 10-21 J,
Thus, according to Boltzmann distribution, the excess population in the
lower state is given as
N (lower) / N (upper) = e -∆ε/kT
= e -0.00028 × 10-21 J/4.1 × 10-21 J
= 1.0007
Thus, both the nuclear-spin levels are equally populated.
Nuclear-Magnetic-Resonance Spectroscopy
The difference in the energies of nuclear-spin states that results from
magnetic moment of the nuclei and an applied magnetic field are studied by
nuclear-magnetic-resonance spectroscopy.
There are two ways with which we can study nuclear states which are
observed by applying external magnetic field.
1.Keeping applied magnetic field B constant and adjusting the frequency ν.
2.Keeping the frequency ν constant and adjusting the applied magnetic field
B to cause the transition.
Second way of doing NMR is experimentally more satisfactory.
Continuous-wave NMR spectroscopy:
In technique referred to as continuous-wave (CW) technique a continuos
beam of fixed radio-frequency (rf) radiations is sent into sample while the
magnetic field is varied through appropriate range.
Drawbacks:
Very Small Population difference between the nuclear-states
The absorption of radiation which depends upon the excess of nuclei in the
lower state produces very weak signal.
The system can became saturated easily as by exciting enough nuclei to the
higher energy states and producing equally populated states and making
absorption signal to zero.
In order to over come these difficulties, CW method is largely replaced by
pulsed Fourier transform-nuclear-magnetic-resonance, or FT-NMR
spectroscopy.
Pulsed Fourier-transform-NMR spectroscopy:
In this technique, pulses of rf radiations are applied to a sample in a
magnetic field.
Advantages
Since every time a pulse is applied to the sample, we get a spectral
information about the sample in the same as we get in CW NMR
spectroscopy.
After this the nuclei comes back to its initial state and system is not
saturated.
Refer to Figure 12.43