Physical Chemistry Third Edition

1008 24 Magnetic Resonance Spectroscopy

Exercise 24.4

Find the frequency and wavelength of photons with energy equal to the energy difference in

Example 24.7.

Nuclear Magnetic Dipoles

Like the electron, the proton does not obey nonrelativistic mechanics. The proton’s

magnetic dipole operator is

̂μgp

e

2 mp

̂I (24.2-7)

wherêIis the spin angular momentum operator of the proton, andmpis the mass of

the proton. The factorgpis analogous to the anomalousgfactor of the electron, and is

called thenucleargfactorof the proton. Its value to six significant digits is 5.58569.

The proton has the same spin angular momentum properties as the electron, with a

spin quantum numberIequal to 1/2. The magnitude of the spin angular momentum

of a proton is

|I|h ̄

√

I(I+1)h ̄

√

1

2

(

1

2

+ 1

)

h ̄

√

3

4

(24.2-8)

Thezcomponent of the spin angular momentum is

IzMIh ̄±

1

2

h ̄ (24.2-9)

whereMIis the quantum number for thezcomponent ofI. The magnitude of the

magnetic dipoleμfor a proton is

|μ|μgp

e

2 mp

h ̄

√

1

2

(

1

2

+ 1

)



√

3

4

gpβN 2. 44324 × 10 −^26 JT−^1 (24.2-10)

and itszcomponent is

μz±

1

2

gpβN± 1. 41061 × 10 −^6 JT−^1 (24.2-11)

The constantβNis analogous to the Bohr magneton and is called thenuclear magneton:

βN

eh ̄

2 mp

 5. 050787 × 10 −^27 JT−^1 (24.2-12)

In some tabulations (such as CODATA 63) the value given for the magnetic moment

of the proton is the magnitude of thezcomponent, equal to 1. 41061 × 10 −^26 JT−^1.

EXAMPLE24.8

Find the difference in the energies of the two spin states of a proton in a magnetic field of

0.500 T. Compare with the result of Example 24.7 for the electron.