# III: Solutions to Exercises and Problems

## Solutions to Exercises and Problems

### Solutions to Problems of Chapter 1

Problem 1. Orbital and spin moments of an electron:

Guide: The state of an electron in an atomic orbital is defined by four quantum numbers *n,* /, *mi, m _{s}* where

*n*= 1, 2, 3,•••;/ = 0, 1, 2, • • • ,

*n*- 1;

*m,*= -/,-/ + 1, ■ ■ •,

*l — l, 1; m*= —

_{s}*s, s*with s = 1/2. The orbital angular momentum is L with eigenvalues

*km/,*the spin angular momentum is S with eigenvalue

*hm*The orbital magnetic moment is M, =

_{s}.*—x*and the spin magnetic moment is M

_{B}L,_{s}= —where

*g*= 2.0023 ~ 2 (Lande factor or electron spectroscopic factor). The total magnetic moment of an electron is thus M

_{f}= —m

_{b}(L + gS).

Problem 2. Zeeman effect:

Guide:

(a) The number of Fe atoms in 1 m^{3}: IV = 7970x6.o25xio* _ 8.58 x 10^{28} (the Avogadro number per kilogram is 6.025 x 10^{26})

The magnetic moment of a Fe atom is thus M = ^{Am2} = ^{L98 x 10 -23} = ^{2 49 x 10-29}

JA/m = 2.14/c_{fi} (Bohr magneton = 9.27 x 10^{-24 }Joule/Tesla= 1.16 x 10^{-29} JA/m).

*Physics of Magnetic Thin Films: Theory and Simulation *Hung T. Diep

Copyright © 2021 Jenny Stanford Publishing Pte. Ltd.

ISBN 978-981-4877-42-8 (Hardcover), 978-1-003-12110-7 (eBook) www.jennystanford.com

Figure 18.1 Fermi-Dirac distribution function at *T **=* 0 versus energy *E. *The scale of *E* is arbitrary, *ц* is taken equal to 1.

(b) The energy due to the Zeeman effect is *AE **= ивВ = ИвИоН* (б = м*оН:* magnetic field). We have Д£ = 0.9273 x 10*~ ^{2г}цоН* Joules (/to, vacuum permeability, = 1.257 x 10-

^{6}H/m).

For *но **H =* 0.5 Tesla, Д *E* = 0.464 x 10^{-23} J For *но H =* 1 Tesla, *AE* = 0.927 x 10“^{23} J For *HoH* = 2 Tesla, *AE* = 1.85 x 10“^{23} J.

The variation of the frequency: *Av **= **v -* u_{0} = *AE/h.*

Problem 3. Fermi-Dirac distribution for free-electron gas:

Solution: At Г = 0, namely *p* = oc, we see that / = 1 for *E *< h, and / = 0 for E > h- The electrons occupy all energy levels up to h- Each energy level has two electrons, one with up spin and the other with down spin. One defines the Fermi level £> by £> *= и,* namely the highest energy level which is occupied at Г = 0. The function *f at T* = 0 is shown in Fig. 18.1.

At Г *Ф* 0, the Fermi-Dirac distribution function is shown in Fig. 18.2. Electrons occupy all levels with decreasing / for increasing *E.*

Problem 4. Sommerfeld’s expansion:

Solution: We show that

where *h^(E) _{E=lt}* is the n-th derivative of

*h[E)*at

*E*

*= и-*

Figure 18.2 Fermi-Dirac distribution function at * T Ф* 0 versus energy

*The scale of*

**E.***is arbitrary,*

**E***is taken equal to 1.*

**pt**Demonstration: We have

We define *g{E*) = *ff^dEh^E).* Integration of / by parts gives

where we have used / -> 0 when *E -*■* oo, and -> 0 when *E* -> —oo.

At low *T,* the function *-Щр* is significant only near /i (slope of /(Я)). This justifies an expansion of *g[E) *around *ц: *

Replacing this series in (18.3), we get

where only terms of even power are non-zero, odd terms being zero because the integrands are odd functions with symmetric limits [note that *Црр-* is an even function with respect to *(E* - д)]. Putting *x* = — д), we obtain

where

Integration by parts gives

where we have used the formula We have

2 4

with £(2) = ^-and<'(4) = ^.Rarely, we need to go beyond *2n =* 4 in the Sommerfeld’s expansion.

Problem 5. Pauli paramagnetism:

Solution: Energies of spin in a magnetic field В applied in the *z* direction: = *E - р _{в}В* and Ej =

*E + p*The

_{B}B.resulting magnetic moment is *M =* p#(iV_{T} — N_{t}) which is written as

where we changed *E -*■ E* ± *p _{B}B.* For weak fields, we can expand the density of states around

*E*(energy in zero field): p(E ±

*р*) ~ p(E) ±

_{в}В*p*[p'(E)]

_{B}B_{£}. We get

*M*~

*We deduce*

^{2}В^{2}вВ S™ dEp'(E)f{E, T, p).

At low *T,* we can make a Sommerfeld’s expansion [see (A.58)] for this integral

where we have used (A.44): *p[E) = AE ^{1/2}.* Using (A.60), we obtain

The first term is independent of *T* as we have found in (1.19). The second term depends on *T ^{2}.*

At high *T, f* ~ (18.10) becomes

where *N* is the total number of electrons of spins f and j. This results is called "Curie’s law.”

Remark: We have used (A.44) without factor 2 of the spin degeneracy because we distinguish in the calculation each kind of spin.

Problem 6. Paramagnetism of free atoms for arbitrary J:

Solution: The average total magnetic moment in the magnetic field В applied in the *z* direction is

where *N* is the total number of atoms. *J?* is the *z* component of the moment J, of the z'-th atom. The Zeeman energy of the magnetic moment of the z'-th atom in the field is *Hi =* -M, • В = *-MfB.*

The average value < *Jf >* is calculated by the canonical description (see Appendix A) as follows:

where *fi* = ^ and Z, the partition function defined by

where we have used the formula of geometric series. We obtain

Thus

*В)* (• • •) is the Brillouin function defined by

where We get

At high temperatures, one has *Bj* (x) ~ *— ■ ■ ■.* Thus,

This is the Curie’s 1/ Г law. At low *T,* one has *Bj* (x) ~ 1 - *j* exp(-x//). One gets

The value at *T =* 0 corresponds to the saturated value of *m, *namely *§дц _{в})* •

Problem 7. Langevin's theory of diamagnetism:

Solution: The first explanation of the diamagnetism has been given by the theory of Langevin using the classical mechanics:

- (a) We have the relation
*m = i A* - (b) We have for an electron,
*m = eA/r*where r is the period (time necessary to make a full circular motion),*e*electron charge. With r =*2лr/v*(r: radius, v: velocity) and*A*=*nr*we have the orbital magnetic moment written as^{2},*m = evr/2.* - (c) The variation of the magnetic flux
*ф*induced by В gives rise to an electric field

Acceleration:

Integrating this relation, we have

The variation of the magnetic moment of the electron is thus

The negative sign indicates the diamagnetic character.

(d) We project the orbit of radius r on a plane perpendicular to the field: the radius of the projected orbit *R = r cosd.* We replace, in the above result of *Am, r Ъуг cosd.* The average on all directions is obtained by integrating on *в:*

(e) If there are *Z* electrons in an atom:

where *N* is the number of atoms in a volume unit: *N* = *N _{A}p/M* (

*N*Avogadro number). The resulting susceptibility is

_{A}

Note: See Section 1.4 for a quantum treatment.

Problem 8. Langevin's theory of paramagnetism:

Solution: The case of discrete spins of magnitude 1/2 has been studied in Section 1.2. Here we study the case of continuous spins (Langevin's theory).

Langevin’s theory: The Maxwell-Boltzmann's probability for a state of energy *E* = -m В (Zeeman energy) is

where *C* is the normalization constant.

In an isotropic material, magnetic moments m are distributed in random directions. The number of moments in an elementary volume is

The total number of moments in a volume unit is thus

The component along the *z* axis of the total resulting magnetic moment, namely magnetization, is

where we used *x **= **cos в (dx* = - sinfldfl) for integration, and *C[y~)* = coth(y) - *К* д_{0}> vacuum permeability, is equal to 1, 257 x 10^{-6} H/m. For weak fields, an expansion of the Langevin function *C[y*) gives *M* = *Nn _{0}m^{2}H/(ЗквТ), *leading to the Curie’s law у =

*M/H*=

*Nц*О (paramagnetism).

_{0}т^{2}/[Ък_{в}Т) >