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Given the type of lattice, the lattice constants of Fe, Ni, Co (look it up!), and the magnetization curves in chapter
4.3-2: How large are the magnetic moments of these atoms in terms of a Bohr magneton? |
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Simple - but still a bit tricky. |
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First we get the basic data:
- Lattice Fe: bcc; lattice constant a = 2.86 Å; atomic density rA(Fe)
= 2/0.2863 atoms/nm3 = 85.5 atoms/nm3
- Lattice Nifcc; lattice constant a = 3.52 Å; atomic density rA(Ni)
= 4/0.3523 atoms/nm3 = 91.7 atoms/nm3
- Lattice Co: bcc; lattice constant a = 2.51 Å, c = 4.0 7Å; atomic density rA(Co)
= 2/[½ · c · a2 · 3½ atoms/nm3 = 90.1 atoms/nm3
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Then we realize that the curves in chapter 4.3-2
give the maximum magnetization, i.e. the magnetization state for all magnetic moments perfectly aligned. From the figure
we can deduce the following numerical values for the saturation magnetization mSat:
- mSat(Fe) = 17 · 105 A/m
- mSat(Fe) = 5 · 105 A/m
- mSat(Fe) = 14 · 105 A/m
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However, the units shown are A/m, which are not what we would expect. Obviously we
must convert this to - well, what exactly? |
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If we look at a Bohr magneton, mBohr, we have |
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Obviously, the unit we need is Am2. We obtain that by multiplying the A/m by m3,
which makes clear that the mSat numbers given are per m3 - as they should be! |
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The magnetic moments mA per atom are thus |
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| What we obtain is |
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mA(Fe) = |
17 · 105 A/m 85.5 atoms/nm3 |
= | 17 · 105
85.5 |
A · 1027 m3 m |
= | 1.98 · 1023 A/m2 |
= 2.14 mB | |
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| mA(Ni) = |
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5.45 · 1024 A/m2 |
= 0.588 mB | |
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| mA(Co) = |
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1.55 · 1023 A/m2 |
= 1.67 mB |
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Now that is an interesting result! It's satisfying because we actually get sensible
numbers close to a Bohr magneton, and it's challenging because those numbers are not very close to 1, 2, or
possibly 3. |
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For example, how can a Ni atom have a magnetic moment of 0.588 mB,
and a Fe atom one of 2.14 mB, considering that the spins of the electrons carry exactly 1 mB? |
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There are two possibilities for this apparent discrepancy:
- Our calculation is somehow a bit wrong
- There are some effects not yet discussed that change the magnetic moment an atom in a crystal lattice carries around
with itself somewhat.
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The first possibility can be ruled out, because in standard textbooks, e.g. in the "Kittel" we find the following values for mA
- mA(Fe) = 2.22 mB
- mA(Ni) = 0.606 mB
- mA(Co) = 1.72 mB
Not identical, but close enough. In fact, looking more closely, the Kittel values are for T = 0 K, whereas
our values are for room temperature T = 300 K and thus should be a bit smaller. |
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Obviously, this leaves us with some effects not yet discussed. What these effects could be,
we can only guess at. Here is a short list:
- There might be some interaction between the spins of the electrons and the "orbits" of the electrons that
modifies the magnetic moment
- The free electrons of the electron gas in our metal also "feel" the ordered spins of the atoms and react to
some extent by adjusting their spins.
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This can lead to quite sizable effects. Dysprosium (Dy), for example, a rare earth
metal, is a ferromagnet below its Curie temperature of 88 K and its atoms than carry an mA(Dy) = 10.2mB. |
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© H. Föll (Electronic Materials - Script)