We can also calculate the interaxial angle between the x-axis and the z-axis, which is denoted by the symbol β, by considering 0.46 nm edge of the (101) plane along with the 0.35 nm and 0.30 nm edges of the unit cell. These three sides form a triangle, as shown in the above diagram of the unit cell. Since the lengths of the sides of the triangle are known, we can use the cosine law to determine the angles of the triangle. The cosine law is defined (The Cosine Law, 2003) as:
where side1, side2 and side3 represent the three sides of the triangle in question, lside1, lside2, and lside3 represent lengths of these three sides, respectively, and θ represents the angle between side2 and side3.
This formula can be rearranged to solve for θ.
The angles θ1, θ2 and β, as defined in the above diagram of the unit cell, can now all be determined.
with significant figures applied
with significant figures applied
with significant figures applied
Thus, using the appropriate number of significant figures, we currently have determined that β = 90o, α = 90o and γ = 90o.
These values for β, α, γ, θ1 and θ2 can then be substituted into our above diagram for the unit cell.
b).
From the above diagram of the unit cell of our hypothetical metal, we can see that all three edges of the unit cell have a different length. According to table 3.2 on page 39 of Callister, the only crystal system which satisfies the conditions of all three edges having different lengths and all interaxial angles being equal to 90o is the orthorhombic crystal system. Therefore, our unit cell must belong to the orthorhombic crystal system.
c).
For each of the given planes, the locations of the atoms residing within those planes has been provided. Those planes have been oriented so that the locations of the atoms in our unit cell are now fixed. In the above diagram of the unit cell of our hypothetical metal, we can observe the positions of each of the atoms within our unit cell. There are atoms at each of the corners of the unit cell. Additionally, the centre of the (101) plane and the centre of the (110) plane both lie at the centre of the unit cell of our hypothetical metal. Therefore, the atom at the centre of the (101) plane is coincident with the atom at the centre of the (110) plane. This produces an atom at the centre of the unit cell. Therefore, our unit cell must have a body centered crystal structure. The unit cell also belongs to the orthorhombic crystal system. The crystal structure of the unit cell would therefore be called body centered orthorhombic, or BCO.
d).
From the above diagram of our BCO unit cell, the number of atoms in this unit cell 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 this unit cell can be determined in the following manner:
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:
We can rearrange this formula to solve for (Mass/mole) in the following manner:
The (Mass/mole) of a substance is defined as its atomic weight. Therefore, the above formula can be rewritten:
Since the three edges of our unit cell all possess different lengths, (do I have to make a change for monoclinic) the volume of this unit cell be calculated by the following formula:
where lengthunit_cell_edge_1, lengthunit_cell_edge_2, and lengthunit_cell_edge_3 represent the three different edge lengths of our unit cell of the hypothetical metal.
Since the values for the edge lengths of our unit cell have been determined previously, we can obtain the volume of our unit cell:
with significant figures applied
Using the conversion factors of 1 nm = 10-9 m and 1 cm = 10-2 m , we can convert the volume of the unit cells to meters:
with significant figures applied
The number of atoms in one mole of an element (or molecules in one mole of a compound) is defined by Avogadro’s number. For atoms, Avogadro’s number is equal to 6.022x1023 atoms/mole. The density of the metal is given in the question to be 8.95 g/cm3. Using these values in the above formula for atomic weight, the atomic weight of the metal can be determined.
with significant figures applied
2).
According to the question, copper and platinum both have the FCC crystal structure. 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 6.022x1023 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 ρave 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%), ρ1 is the density of the first element in the alloy, and ρ2 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, ρalloy. 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 6.022x1023 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. Similarly, the atomic radius for aluminum is given as 0.1431 nm. We can substitute these values into the above equations for aFCC (for aluminum) and aBCC (for lithium) to determine the edge lengths of the unit cells of lithium and aluminum:
with significant figures applied
with significant figures applied
where aLithium is the edge length of a unit cell of lithium, and aAluminum is the edge length of a unit cell of aluminum. As shown in question #2 above, the volume of a unit cell of a cubic material can be calculated by the following formula:
Substituting our values for edge lengths of the unit cells of lithium and aluminum into the above formula, we can obtain the volumes of the unit cells of lithium and aluminum:
with significant figures applied
with significant figures applied
where (Volume/unit_cell) Lithium is the volume of a unit cell of lithium and (Volume/unit_cell) Aluminum is the volume of a unit cell of aluminum. Using the conversion factor of 1 nm = 10-9 m, we can convert the volumes of the unit cells of lithium and aluminum to meters:
with significant figures applied
with significant figures applied
The mass of a unit cell of a material can be calculated by the following formula, as demonstrated in question #2:
The density of a material can be defined by the following formula:
Therefore, as shown previously, the density of a material can be defined in terms of the above two equations:
According to Appendix B of Van Vlack, the mass/mole (or molar mass) of lithium is given as 6.94 g/mole. The mass/mole (or molar mass) of aluminum is given as 26.98 g/mole. As stated previously, for atoms, Avogadro’s number is equal to 6.022x1023 atoms/mole. Using these values in the above formula for density, the densities of lithium and aluminum can be determined.
with significant figures applied
with significant figures applied
According to the question, we desire an aluminum-lithium alloy with a density of 2.55 g/cm3. Using the conversion factor of 1 cm = 10-2 m, we can convert the density of our desired aluminum-lithium alloy to g/m3:
with significant figures applied
where densityAlloy is the density of our desired aluminum-lithium alloy.
According to equation 4.10a on page 72 of Callister,
where ρave 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%), ρ1 is the density of the first element in the alloy, and ρ2 is the density of the second element in the alloy.
Let us define lithium to be the first element in our alloy, and aluminum to be the second element in our alloy. Using these values, and the desired density of our lithium-aluminum alloy, we can apply the above formula for the density of a binary alloy in the following manner:
where densityalloy is the density of our desired lithium-aluminum alloy, CAluminum is the concentration of aluminum in our desired alloy (in wt%), and CLithium is the concentration of lithium in our desired alloy (in wt%).
Additionally, aluminum and lithium are the only two components of our desired alloy. Therefore, the total of the concentration of aluminum in our desired alloy (in wt%) and the concentration of lithium in our desired alloy (in wt%) must sum to 100. Mathematically, this relationship can be stated as:
Rearranging this relationship to isolate CAluminum yields:
This expression for CAluminum can then be substituted into the above formula for the density of our aluminum-lithium alloy.
This expression can now be rearranged to solve for CLithium.
with significant figures applied
4).
In this question, we desire the weight percentage of molybdenum that must be added to tungsten to create an alloy containing 1.0x1022 Mo atoms per cubic centimeter of the alloy. Let us denote the number of atoms of molybdenum per cubic centimeter of our alloy by the symbol NMolybenum. Then we can construct the following equation relating NMolybenum to the number of atoms of molybdenum per mole of molybdenum, to the number of grams of molybdenum per mole of molybdenum, and to the number of grams of molybdenum per cubic centimeter of our alloy.
The number of grams of molybdenum per mole of molybdenum is defined to be the atomic weight (molar mass) of molybdenum. Let us denote this atomic weight by the symbol AMolybdenum. As previously defined, the number of atoms in one mole of an element (or molecules in one mole of a compound) is defined by Avogadro’s number. Therefore, let us define Avogadro’s number by the symbol NA. The number of grams of molybdenum per cubic centimeter of our alloy represents the concentration of the mass of molybdenum in a unit volume of our alloy. Let us define the number of grams of molybdenum per cubic centimeter of our alloy by the symbol ((CMolybdenum)’’)g/cm3, where the double-primed notation is used to distinguish this concentration from concentration by weight percent. We can rewrite the above equation for NMolybdenum using this notation in the following manner:
According to equation 4.9a on page 72 of Callister,
where (C1)’’ is the concentration of the mass of one element (denoted element 1) in a unit volume of a binary alloy (an alloy composed of two elements), C1 is the concentration of element 1 in the alloy (in wt%) C2 is the concentration of the second element (denoted element 2) in the alloy (in wt%), ρ1 is the density of element 1 in the alloy, and ρ2 is the density of element 2 in the alloy.
Note that according to page 73 of Callister, the above equation for C1’’ does not always yield exact results. This is due to the fact that the 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.
According to page 72 of Callister, if densities ρ1 and ρ2 are provided in units of g/cm3, the above equation yields a value of C1’’ in kg/m3. Using the conversion factors of 1 kg = 103 g and 1cm = 10-2 m, we can convert this equation to yield a value of C1’’ in g/cm3.
where ((C1)’’)g/cm3 denotes the concentration of the mass of one element (denoted element 1) in a unit volume of a binary alloy, in g/cm3.
Let us define molybdenum to be element 1 in our alloy, and tungsten to be element 2 in our alloy. The above equation can therefore be slightly modified to represent the concentration of the mass of molybdenum in a unit volume of our alloy, in units of grams per cubic centimeter.
where CMolybdenum is the concentration of molybdenum in our molybdenum-tungsten alloy (in wt%), CTungsten is the concentration of tungsten in the alloy (in wt%), ρMolybdenum is the density of molybdenum in the alloy, and ρTungsten is the density of tungsten in the alloy.
This expression can be substituted into our above formula for NMolybdenum in the following manner:
Additionally, molybdenum and tungsten are the only two components in our desired alloy. Therefore, the total of the concentration of molybdenum in our desired alloy (in wt%) and the concentration of tungsten in our desired alloy (in wt%) must sum to 100. Mathematically, this relationship can be stated as:
Rearranging this relationship to isolate CTungsten yields:
This expression for CTungsten can then be substituted into the above formula for NMolybdenum.
The above equation can be rearranged to solve for CMolybdenum.
According to Appendix B of Van Vlack, the atomic weight of molybdenum is given as 95.94 g/mole. As previously stated, for atoms, Avogadro’s number is equal to 6.022x1023 atoms/mole. The question states that the density of pure molybdenum is 10.22 g/cm3, while the density of pure tungsten is 19.30 g/cm3. Using these values in the above formula, we can solve for CMolybdenum,
with significant figures applied
References:
Callister, W.D. 1997. Material Science and Engineering: An Introduction, 4th Edition. Wiley.
The Cosine Law. 2003. http://whyslopes.com/etc/ComplexNumbers/ch13A.html.
Van Vlack, L.H. 1970. Elements of Materials Science and Engineering. Addison Wesley.