Chapter 8 EE145 Spring 2002 Homework 8 Solution Prof Ali Shakouri Second Edition ( 2001 McGraw Hill) Chapter 8 8 3 Pauli spin paramagnetism Paramagnetism in metals depends on the number of conduction[.]
Homework Solution EE145 Spring 2002 Prof Ali Shakouri Second Edition ( 2001 McGraw-Hill) Chapter 8.3 Pauli spin paramagnetism Paramagnetism in metals depends on the number of conduction electrons that can flip their spins and align with the applied magnetic field These electrons are near the Fermi level EF, and their number is determined by the density of states g(EF) at EF Since each electron has a spin magnetic moment of β, paramagnetic susceptibility can be shown to be given by χpara ≈ µo β2 g(EF) Pauli spin paramagnetism where the density of states is given by Equation 4.10 (in the textbook) The Fermi energy of calcium, EF, is 4.68 eV Evaluate the paramagnetic susceptibility of calcium and compare with the experimental value of 1.9 × 10-5 Solution Apply, m g (E ) = (8π ) 2e h so that 9.109 × 10 −31 kg g (E F ) = (8π ) 2 −34 (6.626 × 10 J ⋅ s) ∴ g (E F ) = 9.197 × 10 Then, χpara ≈ µo β2 g(EF) 46 E -1 J m (Equation 4.10) (4.68 eV)(1.602 × 10 −19 J/eV ) −3 χ para ≈ (4π × 10−7 Wb A -1 m-1 )(9.273 × 10 −24 A m ) (9.199 × 10 46 J−1 m −3 ) χpara ≈ 0.994 × 10-5 ∴ This is in reasonable agreement within an order of magnitude with the experimental value of 1.9 × 10-5 8.4 Ferromagnetism and the exchange interaction Consider dysprosium (Dy), which is a rare earth metal with a density of 8.54 g cm-3 and atomic mass of 162.50 g mol-1 The isolated atom has the electron structure [Xe] 4f106s2 What is the spin magnetic moment in the isolated atom in terms of number of Bohr magnetons? If the saturation magnetization of Dy near absolute zero of temperature is 2.4 MA m−1, what is the effective number of spins per atom in the ferromagnetic state? How does this compare with the number of spins in the isolated atom? What is the order of magnitude for the exchange interaction in eV per atom in Dy if the Curie temperature is 85 K? Solution In an isolated Dy atom, the valence shells will fill in accordance with the exchange interaction: 4f10 ↑ ↓ ↑ ↓ ↑ ↓ ↑ 6s2 ↑ ↑ ↑ ↓ ↑ 8.1 Homework Solution EE145 Spring 2002 Prof Ali Shakouri Obviously, there are unpaired electrons Therefore for an isolated Dy atom, the spin magnetic moment = 4β Atomic concentration in dysprosium (Dy) solid is (where ρ is the density, NA is Avogadro’s number and Mat is the atomic mass): nat = ρN A Mat (8.54 × 10 = kg/m3 )(6.022 × 1023 mol-1) 162.50 × 10 −3 kg/mol = 3.165 × 1028 m −3 Suppose that each atom contributes x Bohr magnetons, then Msat = nat xβ x= Msat 2.4 × 10 A/m = 28 −3 −24 = 8.18 nat β (3.165 × 10 m )(9.273 × 10 Am ) This is almost twice the net magnetic moment in the isolated atom Suppose that the Dy atom in the solid loses all the electrons that are paired into the "electron gas" in the solid This would make Dy+4 have unpaired electrons and a net spin magnetic moment of 8β (this is an oversimplified view) Exchange interaction ∼ kTC = (8.617 × 10−5 eV/K)(85 K) = 0.00732 eV The order of magnitude of exchange interaction ~ 10-2 eV/atom for Dy (small) 8.15 Magnetic storage media a Consider the storage of video information (FM signal) on a video tape Suppose that the maximum signal frequency to be recorded as a spatial magnetic pattern is 10 MHz The heads helically scan the tape, and the relative velocity of the tape to head is about 10 m s-1 What is the minimum spatial wavelength of the stored magnetic pattern (information) on the tape? b Suppose that the speed of an audio cassette tape in a cassette player is cm s-1 If the maximum frequency that needs to be recorded is 20 kHz, what is the minimum spatial wavelength on the tape? Solution a Consider a video tape The maximum frequency in the signal is fvideo = 10 MHz The speed of the video tape is 10 m/s The spatial wavelength is λ = v/fvideo = (10 m/s) / (10 × 106 s-1) = 10-6 m or µm By comparison, typically, a page of a book is 50 - 100 µm thick and so is a typical human hair b Consider an audio tape The maximum frequency in the signal is faudio = 20 kHz The speed of the video tape is 0.05 m/s The spatial wavelength is λ= × 10−2 m/s = 2.5 ì 10-6 m or 2.5 àm 20 ì 103 Hz This is a spatial wavelength of 2.5 µm on the tape 8.2