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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, ms where n = 1, 2, 3,•••;/ = 0, 1, 2, • • • , n - 1; m, = -/,-/ + 1, ■ ■ •, l — l, 1; ms = — s, s with s = 1/2. The orbital angular momentum is L with eigenvalues km/, the spin angular momentum is S with eigenvalue hms. The orbital magnetic moment is M, = —xBL, and the spin magnetic moment is Ms = —where g = 2.0023 ~ 2 (Lande factor or electron spectroscopic factor). The total magnetic moment of an electron is thus Mf = —mb(L + gS).

Problem 2. Zeeman effect:


(a) The number of Fe atoms in 1 m3: IV = 7970x6.o25xio* _ 8.58 x 1028 (the Avogadro number per kilogram is 6.025 x 1026)

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

JA/m = 2.14/cfi (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)

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

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 = моН: magnetic field). We have Д£ = 0.9273 x 10~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 - u0 = 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-

Fermi-Dirac distribution function at T Ф 0 versus energy E. The scale of E is arbitrary, pt is taken equal to 1

Figure 18.2 Fermi-Dirac distribution function at T Ф 0 versus energy E. The scale of E is arbitrary, pt is taken equal to 1.

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


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 - рвВ and Ej = E + pBB. The

resulting magnetic moment is M = p#(iVT — Nt) which is written as

where we changed E -*■ E ± pBB. For weak fields, we can expand the density of states around E (energy in zero field): p(E ± рвВ) ~ p(E) ± pBB [p'(E)]£. We get M ~ 2В2вВ S™ dEp'(E)f{E, T, p). We deduce

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

where we have used (A.44): p[E) = AE1/2. Using (A.60), we obtain

The first term is independent of T as we have found in (1.19). The second term depends on T2.

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


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 §дцв)

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 = nr2, we have the orbital magnetic moment written as m = evr/2.
  • (c) The variation of the magnetic flux ф induced by В gives rise to an electric field


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 = NAp/M (NA Avogadro number). The resulting susceptibility is

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 д0> 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 = Nn0m2H/(ЗквТ), leading to the Curie’s law у = M/H = 0т2/[ЪквТ) > О (paramagnetism).

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