A hard sphere model of the unit cell of the FCC crystal structure is shown below: where aFCC represents the length of each of the edges of the cubic unit cell (since the unit cell is cubic, all of these edges are equal in length), and RFCC represents the radius of an atom within the unit cell (since all the atoms within a unit cell of an elemental solid are made up of the same element, they all possess the same radius).

From this diagram, the number of atoms in the unit cell of the FCC crystal structure can be calculated. There are eight corner atoms in the above unit cell, and each of these corner atoms is shared among eight unit cells.Therefore, each unit cell contains one-eighth of each of these corner atoms. There are also six face centered atoms in the above unit cell, and each of these face centered atoms is shared among two unit cells. Therefore, each unit cell contains one-half of each of these face centered atoms. Mathematically, therefore, the number of atoms in the unit cell of the FCC crystal structure can be determined in the following manner: From the above hard sphere model of the FCC unit cell, we can see that the atoms of the unit cell touch each other along the diagonals of each of the faces. This diagonal has a length of 4*RFCC.Therefore, we can use the pythagorean theorem to relate the edge length of the cube, aFCC, and the atomic radius, RFCC.

We can then rearrange the above equation to solve for aFCC: The atomic radius of platinum is given in the question as 0. 1387 nm. Similarly, the atomic radius for copper is given as 0.

1278 nm. We can substitute these values into the above equation for aFCC to determine the edge lengths of the unit cells of platinum and copper: with significant figures applied with significant figures applied where aPlatinum is the edge length of the unit cell of platinum, and aCopper is the edge length of the unit cell of copper.Since the three edges of a cubic unit cell are all equal in length, the volume of a cubic unit cell can be calculated by the following formula: where acubic_unit_cell is the edge length of a cubic unit cell. Substituting our values for edge lengths of the unit cells of platinum and copper into the above formula, we can obtain the volumes of the unit cells of platinum and copper: with significant figures applied with significant figures applied where (Volume/unit_cell) Platinum is the volume of a unit cell of platinum and (Volume/unit_cell) Copper is the volume of a unit cell of copper.Using the conversion factor of 1 nm = 10-9 m, we can convert the volumes of the unit cells of platinum and copper to meters: with significant figures applied with significant figures applied As shown in question #1, the mass of a unit cell of a material can be calculated by the following formula: The density of a material can be defined by the following formula: Therefore, the density of a material can be defined in terms of the above two equations: The (Mass/mole) of a substance is defined as its atomic weight or molar mass.According to Appendix B of Van Vlack, the mass/mole (or molar mass) of platinum is given as 195. 1 g/mole.

The mass/mole (or molar mass) of copper is given as 63. 54 g/mole. As stated previously, for atoms, Avogadro’s number is equal to 6.

022×1023 atoms/mole. Using these values in the above formula for density, the densities of platinum and copper can be determined. with significant figures applied with significant figures applied According to the question, copper forms a substitutional solid solution for concentrations of up to approximately 6 wt% copper at room temperature.Therefore, under the assumption that the alloy was formed at room temperature, an alloy containing 5 wt% copper and 95 wt% platinum will form a substitutional solid solution, with copper acting as the solute and platinum acting as the solvent. According to equation 4. 10a on page 72 of Callister, where ? ave is the density of a binary alloy (an alloy composed of two elements), C1 is the concentration of the first element in the alloy (in wt%) C2 is the concentration of the second element in the alloy (in wt%), ? 1 is the density of the first element in the alloy, and ?2 is the density of the second element in the alloy. Let us define platinum to be the first element in our alloy, and copper to be the second element in our alloy. Since the platinum-copper alloy in this question is defined as containing 5 wt% copper and 95 wt% platinum, this implies that CCopper = 5 and CPlatinum = 95.

Using these values, and the densities of copper and platinum, we can apply the above formula for the density of a binary alloy to determine the density of the platinum-copper alloy: with significant figures applied where densityalloy is the density of our copper-platinum alloy.According to equation 4. 11a on page 73 of Callister, where Aave is the atomic weight (molar mass) of a binary alloy (an alloy composed of two elements), C1 is the concentration of the first element in the alloy (in wt%) C2 is the concentration of the second element in the alloy (in wt%), A1 is the atomic weight (molar mass) of the first element in the alloy, and A2 is the atomic weight (molar mass) of the second element in the alloy. Note that according to page 73 of Callister, the above equation for atomic weight of a binary alloy does not always yield exact results.This is largely due to the fact that the above equation is derived based on the assumption that the alloy volume is exactly equal to the sum of the volumes of each of the elements that compose the alloy. For the majority of alloys, this assumption is typically false, but the assumption does not create significant errors in most practical situations. Please refer to page 73 of Callister for more details.

Using our values for the molar masses and concentrations (in wt%) of copper and platinum, we can apply the above formula for atomic weight of a binary alloy to determine the atomic weight of our copper-platinum alloy.with significant figures applied where Aalloy is the atomic weight (molar mass) of our copper-platinum alloy. We have previously defined the density of a material in the following manner:We can then rearrange this formula to solve for the volume of a unit cell. We have previously calculated the values for the atomic weight (molar mass) of our copper-platinum alloy, Aalloy, and the density of our alloy, ? alloy.

According to an email message I received from Adam Stefanski, “a solution of two elements with the same structure will have the same structure as the individual elements”.Therefore, since both copper and platinum have the FCC crystal structure, the copper-platinum alloy will also have the FCC crystal structure. As has previously been shown, the FCC crystal structure has 4 atoms per unit cell. Additionally, we know from our previous definition that Avogadro’s number is equal to 6.

022×1023 atoms/mole. Applying these values to the above formula for the volume of a unit cell of our copper-platinum alloy: with significant figures applied I have used the previously given formula for the volume of a cubic unit cell:Rearranging this formula to solve for acubic_unit_cell, the length of each of the edges of the cubic unit cell, yields: Using the volume of the unit cell of our copper-platinum alloy, we can use this formula to solve for the edge length of the cubic unit cell of our alloy. with significant figures applied 3).

According to Appendix B in Van Vlack, lithium has a BCC crystal structure, while aluminum has a FCC crystal structure. In question #2, the number of atoms in the unit cell of the FCC crystal structure were calculated:Additionally, the edge length of the unit cell of the FCC crystal structure, aFCC, was related to the atomic radius of an atom in the FCC crystal structure, RFCC, by the following equation: A hard sphere model of the unit cell of the BCC crystal structure is shown below: where aBCC represents the length of each of the edges of the cubic unit cell (since the unit cell is cubic, all of these edges are equal in length), and RBCC represents the radius of an atom within the unit cell (since all the atoms with a unit cell of an elemental solid are made up of the same element, they all possess the same radius).From this diagram, the number of atoms in the unit cell of the BCC crystal structure can be calculated.

There are eight corner atoms in the above unit cell, and each of these corner atoms is shared among eight unit cells. Therefore, each unit cell contains one-eighth of each of these corner atoms. There is also one body centered atom in the above unit cell, and this atom is completely enclosed within the unit cell.

Therefore, each unit cell contains the entirety of this body centered atom.Mathematically, therefore, the number of atoms in the unit cell of the BCC crystal structure can be determined in the following manner: From the above hard sphere model of the BCC unit cell, we can see that the body and corner atoms of the unit cell touch each other along the cube diagonals. This diagonal has a length of 4*RBCC. Therefore, we can use the pythagorean theorem to relate the edge length of the cube, aBCC, and the atomic radius, RBCC.

We can then rearrange the above equation to solve for aBCC: According to Appendix B in Van Vlack, the atomic radius of lithium is 0.1519 nm.

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