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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:

Guide:

(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.

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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

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 - 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

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 §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

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 = 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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