Solids, cont.
Bragg Equation
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We can determine the distances between layers in a repeating structure using x-rays. The relevant distances can be seen in Figure 10.11, Zumdahl p 433, resulting in the Bragg equation:
n
= 2d sin
STEM: a newer method with equal resolution is Scanning Tunneling Electron Microscopy. This technique givse a highly detailed view of the surface of a crystal (or other object). Distances between surface atoms can be calculated etc.
Crystal Structure (overheads)
Lattice
Unit Cell (demo with cork ball cells)
Crystal lattice
3 kinds of cubic lattice: simple cubic, body-centered cubic (bcc), and face-centered cubic (fcc). (Figure 10.9, p 460 of Zumdahl)
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Questions on number of "atoms" in the three different cubic unit cells:
Model: For Simple Cubic corner atoms touch each other and have 8 corners with 1/8 atom at each corner = 1.
For Body-centered cubic corner atoms and center atom atoms touch each other; have 8 corners with 1/8 atom at each corner + one in center = 2.
For Face-centered cubic, corner atoms and face atoms atoms touch each other; have 8 corners with 1/8 atom at each corner + 1/2 atom on each face (6 x 1/2) = 4.
Example: What is the density of sodium? Need to find the mass and volume.
Know the mass of one Na atom in amu, can convert to grams using Avogadro's number:
From the Periodic Table mass of Na = 23.00 amu/atom, and we know that there are Avogado's number of amu/g,
so mass odf Na = (23.00 amu/atom)/(6.02 x 1023 amu/g) = 3.82 x 10–23 g/atom
Can calculate the volume of a unit cell for sodium. Na has an atomic radius of 186 pm and a body-centered cubic unit cell.
First need to find the length of an edge of the cell. The trouble is the sodium atoms on the corners don't touch each other, but they do touch the center atom. Thus the diagonal of the cube (dc) is two Na diameters or four radii = 186 + 2(186) + 186 = 744 pm
Need to use Pythagorean theorem twice:
dc2 = df2 + a2 where df = the face diagonal and a = the cube edge length
and df2 = a2 + a2
So dc2 = 3 a2 and a = (dc)/(31/2) = (744 pm)/(1.732) = 429.6 pm
So the volume of the unit cell is then = (429.6)3 = 7.927 x 107 pm3 = 7.927 x 10–2 nm3, since 1 pm = 10–3 nm,
so 1 pm3 = 10–9 nm3
(Try calculating volumes of the three cells using simple geometry and atom radii for edge (simple) or diagonal lengths. We have done two, try the face-centered using the face diagonal for face.)
For identical spheres can get two kinds of close packing: (Figure 10.13, p 466 of Zumdahl)
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cubic close packing, which turns out to be a face-centered cubic lattice (Figure 10.15, p 466 of Zumdahl)
hexagonal close packing (Figure 10.14, p 466 of Zumdahl)
The images below show the so-called cannon ball stacking in close-packing. Note that the stack is a direct result of the HCP lattice shown in the image above, where just the top ball and the next layer of three balls are darkened. Can you discern the next (triangular) layer in the diagram which cooresponds to the third layer down in the pictures below?
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public domain image via Wikipedia Creative Commons
Metallic solids: these are different in that they have ions at the lattice points in a "sea" of shared electrons (Figures 10.18–20, p 469–70 of Zumdahl). In metals that are very hard, such as chromium, the ions are also covalently bonded to each other.
Network (or covalent) solids - Carbon as example: carbon has allotropes (different physical forms of the same element) all of which have many atoms linked together in covalent networks, and two of which are covalent solids: (Figure 10.22, 24, 24, p 471–4 of Zumdahl)
Diamond (covalent solid) - each atom in the bulk solid is covalently bonded to four others in a tetrahedral arrangement. This is the secret to diamond's great hardness: to break a piece off many strong covalent bonds must be broken.
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© R A Paselk
Last modified 4 May 2015