Physical Chemistry Third Edition

24.2 Electronic and Nuclear Magnetic Dipoles 1007

Electron Spin Magnetic Dipole

The electron spin angular momentum also produces a magnetic dipole. This magnetic

dipole is found experimentally to be slightly more than twice as large as predicted by

Eq. (24.2-1). The correct operator for this magnetic dipole is

̂μ−g

e

2 me

̂S−g e

2 me

(îSx+ĵSy+k̂Sz) (24.2-2)

wherêSis the spin angular momentum operator and wherei,j, andkare the unit vectors

in thex,y, andzdirections. The quantitygis called theanomalousgfactor of the

electron, and corrects for the failure of the electron to obey nonrelativistic mechanics.

Its value is 2. 0023 ....

The only eigenvalue of the operator̂S

2

for an electron ish ̄^2 (1/2)(3/2), so that the

magnitude of the spin magnetic dipole of an electron is

|μ|g

e

2 me

h ̄

√

1

2

(

1

2

+ 1

)

g

e

2 me

h ̄

√

3

4

gβe

√

3

4

(24.2-3)

and itszcomponent is

μz±

1

2

gβe (24.2-4)

The constantβeis called theBohr magneton:

βe

eh ̄

2 me

 9. 2740 × 10 −^24 JT−^1 (24.2.-5)

For a magnetic fieldBzin the direction of thezaxis, the magnetic energy of an

electron is

Emag−μzBz±

gβeBz

2

(24.2-6)

The difference between the two energies is exploited in ESR spectroscopy.

EXAMPLE24.7

a.For a magnetic field of 0.500 T, find the difference in the energies of the two electron spin

states.

b.Find the ratio of the populations of these two states at 298.15 K.

Solution

a. ∆EmaggβeBz(2.0023)(9.^2740 ×^10 −^24 JT−^1 )(0.500 T)^9.^28 ×^10 −^24 J

b.

pup

pdown

exp

(

− 9. 28 × 10 −^24 J

(1. 3807 × 10 −^23 JK−^1 )(298 K)

)

e−^0.^00226  0. 99775