Download Homework 8 solution

Homework 8 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Second Edition ( 2001 McGraw-Hill)
Chapter 8
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
3
Apply,
m
g (E ) = (8π 2 ) 2e 
h
so that

9.109 × 10 −31 kg 
g (E F ) = (8π 2 )
2
−34
 (6.626 × 10 J ⋅ s) 
∴
g (E F ) = 9.197 × 10
Then,
χpara ≈ µo β2 g(EF)
46
2
E
-1
J m
(Equation 4.10)
3
2
(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 2 ) (9.199 × 10 46 J−1 m −3 )
2
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 8 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Obviously, there are 4 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
=
3
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 6 A/m
=
28
−3
−24
2 = 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 4 electrons that are paired into the "electron gas" in the solid. This would make
Dy+4 have 8 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 5 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 1 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=
5 × 10−2 m/s
= 2.5 × 10-6 m or 2.5 µm
20 × 103 Hz
This is a spatial wavelength of 2.5 µm on the tape.
8.2