Observation Questions:
-
Given the silver has a face centered cubic structure, and an atomic radius of 1.44x10-10m, determine the density of silver show detailed calculations of density. Compare this value with the density published in reference texts.
The density of silver with an atomic radius of 1.44x10-10m is 1.06x107 g/m3. The theoretical density of silver found in the text is1.04x107 g/m3. These numbers vary due to imperfections in silver. Real silver is not in a perfect crystalline structure it has some defections to it, which change its volume and inturn, its density. The % error found between these numbers is 1.89%. This minimal difference in the density will effect the strength and the physical properties of this solid greatly.
- Explain clearly the differences in stacking arrangements of the three crystal models constructed. How can the face centered and the hexagonal close packed structure be converted from one to the other?
The three arrangements FCC, BCC, and HCP have very different packing arrangements.
BCC:
The BCC crystalline structure when broken down to its simplest unit cell can be observed as a cube with an eight of an atom at each corner of the unit square. In the center there is a whole atom making a total atom count of 2.
Figure 1 BCC structure and close packed plane
FCC:
The FCC crystalline structure is similar to that of the BCC. It has an eight of an atom at each corner of the square. On each face there is half of an atom giving a total atom count of 4.
Figure 2. FCC structure and close packed planes
HCP:
Unlike the FCC and BCC structures the HCP is oriented in a hexagonal shape as opposed to a square. There is a sixth of an atom at each point of the hexagon at the top and the bottom with a half atom in the center of each hexagon. In the middle of the structure there is a triangle with three whole atoms centered in the hexagonal structure totaling 6 atoms in a given unit cell.
Figure 3. HCP structure and close packed planes
To convert the FCC to the HCP you must take the FCC and stack it in its (111) plane. With this orientation as shown above you can see the hexagonal shape as in the HCP unit cell this is why both the HCP and the FCC have approximately the same atomic packing factor.
- For each of the above crystal structures which plane is the plane of closest packed atoms?
As shown in figure 1 the close packed plane for the BCC is the (110) plane. As shown in figure 2 the close packed plane for the FCC is the (111) plane. As shown in figure 3 the close packed plane for the HCP is the (0001) plane.
- Determine in terms of the radius of the atom, the distance between adjacent close packed planes in each of the above structures.
The distance between close packed planes in BCC crystalline structure is 4R /sqrt(3) sqrt(2). The distance between close packed planes in FCC crystalline structure is 2R sqrt(2)/ sqrt(3). The distance between close packed planes in HCP crystalline structure is 1.633R.
- What differences in the mechanical properties (e.g., strength, ductility) of metals result from the differences in crystal structure? Use examples, iron, aluminum, copper, and zinc.
Various differences occur in the strength and the ductility of the metals due to the structures of their crystals. The space between the atomic arrangements of the planes plays an important part in providing the metal with the properties of ductility and strength.
The densely packed atomic layers of HCP and especially FCC are organized in such a way that they fill the valley spaces in between the layers created by the atoms. Generally the FCC and HCP have a greater APF (Atomic Packing Factor), which is .75 as compared to the .68 of the BCC.
In general metals with more valence electrons form FCC and HCP structures. Both the APF and the coordinate number also play a role in the physical attributes of the solid. Iron has a BCC structure with a lower APF than FCC and HCP. The atoms in iron have more room to move around due to this allowing for a more ductile solid. Aluminum and copper have an FCC structure which have a higher APF making them packed closer together and giving them a larger coordination number making them less ductile but more tensile. Zinc has an HCP structure, which has the same APF as the FCC structure, giving it similar characteristics to copper and aluminum.
Conclusion:
This lab shows how the crystalline structure can affect the strength and characteristics of the solid. Both the FCC and HCP have similar packing planes although their unit cells are much different. The FCC can in turn be converted into the HCP showing how similar these structures really are. This lab showed a visual representation of the molecular structure of a molecule allowing one to further understand why certain solids act the way they do. As an engineer it is very important to understand how the structures of molecules affect their characteristics. This will allow an engineer to develop stronger molecules and implement other molecules in designs, which demand specific tolerances for the solid.
Appendix: sample calculations
Calculations for Question 1:
n = number of atoms associated with the crystal structure
A = atomic mass
Vc = Volume of the unit crystal structure
Na = Avogadro’s number
ρ = the density of silver
There are 4 atoms in an FCC unit cell and silver has an atomic mass of 107.8682 g/mol. The volume of an FCC unit cell is equal to its unit cell length (a)
For an FCC:
a = 2R (2)1/2
Vc = a3
= (2R (2)1/2)3
= (4.0729x10-10 m)3
ρ = (n x A) / (Vc x Na)
= [4 atoms (107.8682 g/mol)] / [(4.0729x10-10 m)3 (6.023x1023 atoms/mol)]
= 1.06x107 g/m3
Percentage error :
error = {(1.06x107 – 1.04x107)/1.06x107} x 100%
= 1.89%
Calculations for Question 4:
For FCC:
d=?
Close packed plane for FCC is (111)
a= 2R sqrt(2)
dhkl = a/ sqrt(h2+k2+l2)
= 2R sqrt(2)/ sqrt(12+12+12)
= 2R sqrt(2)/ sqrt(3)
For BCC:
d=?
Close packed plane for bcc is (110)
a= 4R/ sqrt(3)
dhkl = a/ sqrt(h2+k2+l2)
= 4R /sqrt(3) sqrt(12+12+02)
= 4R /sqrt(3) sqrt(2)
For HCP:
For HCP the ratio for c/a = 1.633 (pg. 37 of Callister, D. William, Jr. “Materials Science and Engineering An Introduction”). The distance between the close packed planes is half the height c.
a = 2R
c = (2)R(1.633)
d = c/2
=1.633R
where:
c = height
a = sides of the hexagonal structure
.
References
(1)Ryerson University Department of Mechanical and Industrial Engineering MTL 200 Materials Sciences 1
Prof. A, Varvani winter 2004 lecture notes, for Engineering Students in the First Year.
Toronto, Ontario, Canada. (2002)
(2)Ryerson University. Materials Sciences Laboratory Manual for Engineering Students in the First Year.
Toronto, Ontario, Canada. (2002)
(3)Serway and Beichner, Physics: For Scientists and Engineers. 5th Edition (Orlando, Florida), (2003)
Construction of Atomic Models of Metallic Materials
Prepared by: Paul Mazzone
Marko Bilal
Dennice Yip
Section: 15
Experiment Performed:
Monday Jan 19
