NMR is a spectroscopic method based on the absorption of electromagnetic

radiation (EMR) by atom nuclei in the radio-frequency range (4 MHz to 900

MHz). NMR is helpful for determining the structures of molecules, particularly
in the case of organic compounds. The interaction of nucleus with an external
magnetic field is the basis of NMR spectroscopy.

Nuclear Magnetic Resonance

• Nucleus • Magnetic Field • Matching of

two
frequencies

Discovery (1946): Independently Discovered by-

• F. Bloch (Stanford University)
• E.M. Purcell (Harvard University)

➢ The first application of NMR (Chemistry,1951): Recorded NMR

spectrum of ethanol.

Atomic nucleus contains protons and neutrons. Proton & Neutron Both these
particles spin around their own axis with the same spin quantum number 1/2.

Charge Spin about own axis

1
Proton: +𝑒 ± 1
2 (𝑆𝑝𝑖𝑛 𝑄𝑢𝑎𝑛𝑡𝑢𝑚 𝑛𝑜 = )
1 2
Neutron: 0 ±
2

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 Each nucleus possesses a total spin quantum number (𝐼)
which results simply from the vector addition of the spins
of its protons and neutrons.
✓ Magnetic inactive or NMR inactive Nuclei : 𝐼 = 0
✓ Magnetic active or NMR active Nuclei: 𝐼≠0

All magnetic active atomic nuclei possess nuclear spin, 𝐼,

which may be:
▪ Integral (𝑖. 𝑒. , 𝐼 = 1, 2, 3, 𝑒𝑡𝑐. ) or
1 3 5 7
▪ Half- integral (𝑖. 𝑒. , 𝐼 = , , , 𝑒𝑡𝑐. ).
2 2 2 2

It is difficult to predict the actual spin of a nucleus. Some empirical rules based
on experimental facts regarding the total spin quantum number are available. Viz-

Atomic Mass
Spin Quantum No Example
No No
𝐼 = 0 (NMR Inactive nuclei) 4 12 16
Even Even 2𝐻𝑒 , 6𝐶 , 8𝑂 𝑒𝑡𝑐
Integral, 𝐼 = 1, 2, 3, 𝑒𝑡𝑐. 2 14 10
Odd Even 1𝐻 , 7𝑁, , 5𝐵 𝑒𝑡𝑐
1 3 5 7 13 17
Even Odd Half Integral, 𝐼 = , , , 𝑒𝑡𝑐. 6𝐶 , 8𝑂 𝑒𝑡𝑐
2 2 2 2
1 3 5 7 1 15 19 35
Odd Odd Half Integral, 𝐼 = , , , 𝑒𝑡𝑐. 1𝐻 , 7𝑁 , 9𝐹 , 17𝐶𝑙 𝑒𝑡𝑐
2 2 2 2

A nucleus with spin quantum number 𝐼 has

1 1
𝟐𝑰 + 𝟏 spin states; if 𝐼 = , there are {(2 × ) + 1} = 2 allowed spin states.
2 2

1 2 13 14 15 17 19 35 10
Nuclei 1𝐻 1𝐻 6𝐶 , 7𝑁 7𝑁 8𝑂 9𝐹 17𝐶𝑙 5𝐵
Spin Quantum 1 1 1 5 1 3
1 1 3
no(𝐼) 2 2 2 2 2 2
Spin State(2𝐼 +
2 3 2 3 2 6 2 4 7
1)

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As electric charge present in a nucleus, the spinning nucleus generates a magnetic
field whose axis is coincides to the nuclear spin axis. Therefore, each nucleus
may be compared to a tiny magnet with a magnetic moment.𝜇.
▪ Each nucleus with 𝐼 > 0 has magnetic moment.
However,
▪ 126𝐶 and 168𝑂 having 𝐼 = 0 do not possess a magnetic moment, i.e., they
are non-magnetic.

▪ Nuclei possessing, 𝐼 ≥ 1 ⟹ Quadrupolar nuclei

Which have nuclear electric quadrupole moment, Q, which is a measure of the
deviation of the nuclear charge distribution from spherical symmetry.
5
e.g., 21𝐻 (Deuteron) (𝐼 = 1), 147𝑁 (𝐼 = 1), 178𝑂(𝐼 = ) etc.
2
Quadrupolar nuclei not only 'spoil' their own NMR spectra but also the NMR
spectra of nuclei attached to them in a molecule. Thus, the NMR spectra of
quadrupolar nuclei are generally very broad and not sharp at all.

When a magnetic nucleus with spin (𝐼) is placed in a uniform magnetic field(𝐵𝑧 )
applied along Z direction
ℎ
The angular momentum, 𝐿⃗ = √𝐼(𝐼 + 1) is Quantized, as well as the direction
2𝜋
of magnetic moment (𝑚𝐼 ) also quantized and have certain discrete values.

𝑚𝐼 = 𝐼, (𝐼 − 1), (𝐼 − 2), … … . . , 0, …. , −(𝐼 − 2), −(𝐼 − 1), −𝐼

There are (2𝐼 + 1) values of 𝑚𝐼 i.e., (2𝐼 + 1) spin states.

▪ Magnetic dipole moment of nucleus(𝜇) is proportional to the angular

momentum 𝐿⃗ is given by –
𝑔𝑁 → Nuclear 𝑔 factor
𝜇 ∝ 𝐿⃗
⇒ 𝝁 = 𝒈 . ⃗𝑳. 𝝁 … …  𝜇𝑁 → Nuclear magneton
𝑵 𝑵
𝑚𝑃 → Mass of proton
𝑒 → electronic Charge
𝑒. ℎ (1.602 × 10−19 𝐶) × (6.626 × 10−34 𝐽. 𝑆)
𝜇𝑁 = =
4𝜋𝑚𝑃 4𝜋 × (1.673 × 10−27 𝑘𝑔)
= 5.05 × 10−27 𝐽𝑇 −1

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The magnitude of the nuclear magnetic moment is thus given by
⃗ |. 𝝁𝑵
𝝁 = 𝒈𝑵 . |𝑳

⇒ 𝝁 = 𝒈𝑵 . 𝝁𝑵 . √𝑰(𝑰 + 𝟏) … … . . … … 

Q. Calculate magnetic moment & spin angular momentum of 199𝐹 . Given g-

factor of the nucleus,5.2567.
⟹
1
i) For, 199𝐹 , 𝐼 =
2
We know, the spin angular momentum of nucleus is given by –
ℎ
𝐿⃗ = √𝐼(𝐼 + 1)
2𝜋
1 1 ℎ
⇒ 𝐿 = √ ( + 1)
2 2 2𝜋

3 6.626 × 10−34 𝐽. 𝑆
⇒𝐿= √ × = 9.13 × 10−35 𝐽. 𝑆
4 2𝜋

ii) By the problem,

𝑔𝑁 = 5.2567
We Know, nuclear magnetic moment, 𝜇 = 𝑔𝑁 . 𝜇𝑁 . √𝐼(𝐼 + 1)
1 1
⇒ 𝜇 = 5.2567 × √ ( + 1) . 𝜇𝑁
2 2

3
⇒ 𝜇 = 5.2567 × √ × 5.05 × 10−27 𝐽𝑇 −1 = 2.298 × 10−26 𝐽𝑇 −1
4

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NMR of a Free proton :

NMR of a free proton or a nucleus with 1 proton is an ideal concept only.

Interaction energy (𝐸) of nuclear magnetic moment (𝜇) with an external
magnetic field (𝐵𝑍 ), applied in the z-direction, is given by –

𝐸 = −𝜇𝑍 𝐵𝑍 … … … … . . 

where 𝜇 is the z-component of the magnetic momentum vector. Let 𝜃 be the angle
between the magnetic momentum vector and the z-axis, then
𝜇𝑍 = 𝜇 cos 𝜃
From eqn  we get –
𝜇𝑍 = (𝑔𝑁 . 𝜇𝑁 . √𝐼(𝐼 + 1)) cos 𝜃 … … … . 
The angle 𝜃 cannot have any arbitrary value, but only a few allowed discrete
values which satisfy the condition of quantization of the component of angular
momentum vector along the axis of magnetic field, i.e.
ℎ ℎ
(√𝐼(𝐼 + 1) ) cos 𝜃 = 𝑚𝐼
2𝜋 2𝜋
⇒ √𝐼(𝐼 + 1) cos 𝜃 = 𝑚𝐼 … … … … … … … . 

From eqn  &  we get –

𝜇𝑍 = 𝑔𝑁 . 𝜇𝑁 . 𝑚𝐼 …………….

Substituting value of 𝜇𝑍 in eqn  using eqn  we get –

𝑬 = −𝒈𝑵 . 𝝁𝑵 . 𝒎𝑰 . 𝑩𝒁

1 1
For a free proton ¹H with 𝐼 = so that 𝑚𝐼 = ±
2 2
1
When 𝑚𝐼 = + ,
2
1
The energy, 𝑬𝟏/𝟐 = − 𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍
2
1
& When 𝑚𝐼 = − ,
2
1
The energy, 𝑬−𝟏/𝟐 = + 𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍
2
.

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 Figure 1: Splitting of nuclear energy levels of a free proton in a magnetic field

In the absence of the external magnetic field, the two components of the proton
spin have the same energy (i.e., they are degenerate). However, the magnetic field
removes the degeneracy of the spin components. This is called Zeeman splitting
of energy levels. The energy difference is given by –
1 1
∆𝑬 = 𝑬−𝟏/𝟐 − 𝑬𝟏/𝟐 = + 𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍 − (− 𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍 )
2 2
⟹ ∆𝑬 = 𝒈𝑵 . 𝝁𝑵 . 𝑩𝒁

In order to transitions between the two energy levels of proton, we must apply
oscillating radiofrequency (energy= ℎ𝜈) perpendicular to the direction of B. The
nuclear spin flips, when ∆𝐸 becomes exactly equal to ℎ𝜈
𝑖. 𝑒. ∆𝐸 = ℎ𝜈
This is called the Bohr frequency condition. Thus, the NMR frequency of a bare
proton is given by –
∆𝑬 𝒈𝑵 . 𝝁𝑵 . 𝑩𝒁
𝝂= =
𝒉 𝒉

Page | 6
Page | 7
Q. What magnetic field strength is required for proton magnetic resonance at 220
MHz . Given: The factor g for proton is 5.585.

 By the problem,
𝜈 = 220𝑀𝐻𝑧 = 220 × 106 𝑆 −1
𝑔𝑁 = 5.585
We know Bohr frequency condition for NMR
𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍
𝜈=
ℎ
ℎ𝜈 (6.626 × 10−34 𝐽. 𝑆) × (220 × 106 𝑆 −1 )
⇒ 𝐵𝑍 = =
𝑔𝑁 . 𝜇𝑁 5.585 × (5.05 × 10−27 𝐽. 𝑇 −1 )

∴ 𝐵𝑍 = 5.168 T

Q. Calculate the precessional frequency of a proton in a field of 1.5 T.

 By the problem,
𝐵𝑍 = 1.5𝑇
For proton, 𝑔𝑁 = 5.585
We know Bohr frequency condition for NMR
𝑔𝑁 . 𝜇𝑁 . 𝐵𝑍 5.585 × (5.05 × 10−27 𝐽. 𝑇 −1 ) × 1.5𝑇
𝜈= =
ℎ 6.626 × 10−34 𝐽. 𝑆

∴ 𝜈 = 63.84 × 106 𝑆 −1 = 63.84𝑀𝐻𝑧

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Larmor Precession:

Classically, the phenomenon of magnetic resonance can also be Understood by

invoking Larmor precession. A nucleus having a magnetic moment 𝜇 behaves as
a bar magnet which spins on its axis. In the pence of the magnetic field 𝐵𝑍 , the
interaction of 𝜇 with 𝐵𝑍 produces a torque. This torque causes 𝜇 to precess about
𝐵𝑍 as shown below.

This phenomenon is called Larmor precession (in honour of the 19th century
British mathematician. Joseph Larmor) and the angular frequency of precession
is referred to as the Larmor frequency, 𝜔𝐿 (expressed in radians per second).
Mathematically, Larmor frequency given by

𝜔𝐿 = 𝛾𝐵𝑍
where 𝛾 is the gyromagnetic ratio of the nucleus. It is defined as –
Magnetic moment 𝒈𝑵 . 𝝁𝑵 . √𝑰(𝑰 + 𝟏)
𝜸= = 𝒓𝒂𝒅−𝟏 𝑻−𝟏
Angular Momentum 𝒉
√𝑰(𝑰 + 𝟏) 𝟐𝝅
𝒈𝑵 . 𝝁𝑵 −𝟏 −𝟏 𝒈𝑵 . 𝝁𝑵 −𝟏
∴𝜸= 𝑻 𝑺 𝜸= 𝑻 𝒓𝒂𝒅−𝟏
𝒉 ℏ
So,

Magnetic moment
𝝎𝑳 = ( )𝑩 𝒈𝑵 . 𝝁𝑵
Angular Momentum 𝒁 ∴ 𝝎𝑳 = ( ) 𝑩𝒁 𝑺−𝟏
𝒉

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Q. Calculate the value of gyromagnetic ratio 𝜸 The value of proton.

 Gyromagnetic ratio is,

𝒈𝑵 . 𝝁𝑵 .
𝜸=
𝒉
−27 −1
𝑔𝑁 . 𝜇𝑁 5.585 × (5.05 × 10 𝐽. 𝑇 ) 6 −1
𝛾= = = 42.566 × 10 𝑇 . 𝑠−1
ℎ −34
6.626 × 10 𝐽. 𝑆
2𝜋

Q. In a spectrometer operating at 1T. the NMR frequency of

19
9𝐹 is 40.06 MHz . Calculate the magnetogyric ratio of 199𝐹 .

 We Know,
𝜔𝐿 = 𝛾𝐵𝑍
𝜔𝐿 2𝜋𝜈 2𝜋 × 40.06 × 106 𝑆 −1
⇒𝛾= = =
𝐵𝑍 𝐵𝑍 1𝑇
∴ 𝛾 = 2.517 × 108 𝑇 −1 . 𝑠 −1

Q. The magnetic moment of 31𝑃 is equal to 1.1305 nuclear magnetons, i.e.,

1.1305 𝜇𝑁 . Calculate its magnetogyric ratio and the g-factor. Given 𝜇𝑁 =
5.05 × 10−27 𝐽. 𝑇 −1 and nuclear spin of P is 1/2.

The magnetogyric ratio or gyromagnetic ratio is defined by

Magnetic moment
𝜸=
Angular Momentum

1.1305 𝜇𝑁 1.1305 × 5.05 × 10−27 𝐽. 𝑇 −1

⇒𝛾= =
√𝐼(𝐼 + 1)ℎ √3 × 6.626 × 10−34 𝐽. 𝑆
4

∴ 𝛾 = 9.949 × 107 𝑇 −1 𝑆 −1
1.1305
𝒊𝒊) Magnetic moment=𝒈𝑵 . 𝝁𝑵 . √𝑰(𝑰 + 𝟏) ⇒ 𝑔𝑁 =
√𝐼(𝐼 + 1)
⇒ 1.1305 𝜇𝑁 =𝑔𝑁 . 𝜇𝑁 . √𝐼(𝐼 + 1)
1.1305
∴ 𝑔𝑁 = = 1.305
√ 3
4

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