194 5.4  NMR and Other Radio Frequency and Microwave Resonance Spectroscopies

a

What is the expected resonance frequency ν0 in this NMR device of a 1H atomic

nucleus?

A test sample of 300 μL of 1 mM ethanol dissolved in TMS was used in the device.

b

If a general magnetic sample consisting of identical atoms of nuclear spin quantum

number of I has a bulk magnetization M0 given by the sum of all magnetic moments

per unit volume, show that M0 is proportional to I(I +​ 1)B in an external magnetic field

of magnitude B, stating any assumptions. Assuming that all single proton atomic

nuclei in the ethanol sample have the same resonance frequency ν0, estimate its bulk

magnetization.

c

The average measured resonance frequency ν of 1H in the sample was slightly different

to ν0 by an amount of Δν. Explain why this is, and estimate Δν.

[You can assume that the vacuum permeability ≈ 1.3 × 10−6 H m−1; hint: the sum of n

natural squares is n(n +​ 1)(2 n +​ 1)/​6].

Answers

a

The number of wire turns n in the solenoid for tightly packed wires is given

roughly by

′ =

(

)

n

complete

6 0

5 0

0 85

11

.

.

/ .

turns

However, each wire contains ~500 filaments, so the number of total turns n in

solenoid n =​ 11 × 500 =​ 5500 turns. Assuming the long solenoid approximation,

the B-​field is given by

B =

×

(

)×

×

(

) =

1 3 10

5500

100

0 07

10 2

6

.

/

.

. T

1H resonance frequency is 400 MHz for a 9.4 T field; thus, here the resonance fre­

quency will be

ν =

×(

) =

400

10 2 9 4

434

. / .

MHz

b

If the magnetization is given by the sum of the magnetic moments per unit

volume, this is the same as the total number of all magnetic moments per unit

volume multiplied by the expected magnetic moment. The expected value of the

magnet moment is given by 〈〉

µ the probability-​weighted sum over all possible

μ values. The probability pm of a general spin quantum number m is given by

the Boltzmann factor for that energy state normalized by the sum of all possible

Boltzmann factors of all energy states. Therefore,

〈〉

=

( ) =

(



=−

=−

=

=−

=

µ

µ

m

I

I

m

m

I

m I

m

m

I

m I

p

m

mh

E

k T

γ

π

/

)exp[

/

exp[

B

2

=

(



=−

=

=−

=

E

k T

mh

mhB

k T

mhB

m

m

I

m I

m

I

m I

/

/

)exp[

/

exp[

B

B

]

γ

π

γ

π

γ

2

2

/

exp[–

exp[–

B

2

2

π

γ

π

k T

h

m

m

m

m

I

m I

m

I

m I

]

]

]

=

=−

=

=−

=