### The initial diagram of the three planes

The three different crystallographic planes shown are for a unit cell of a hypothetical metal. I have completed this question based on the assumption that each of the planes is rectangular. Therefore, the angles between each of edges of the planes must equal 90o.

Each of the plane dimensions must completely be enclosed by the boundaries of the unit cell of the metal. Even though there are infinite parallel planes having Miller indices equivalent to those of the given planes, these are not enclosed by the unit cell of our metal.Thus, these equivalent planes do not need to be considered to determine the geometry, crystal system and crystal structure of our unit cell. Although we are given the edge lengths of the three crystallographic planes, this gives no information about the orientation of those edges. Therefore, since each plane is rectangular, there are two possible orientations for each plane.

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The correct orientation must allow the dimensions of all three planes to be correlated with one another. An initial diagram of the three planes of the unit cell is shown below:We can see from the above figure that one edge of the (001) plane must be equal in length to one edge of the (101) plane. There is only one possible orientation of the two planes that can produce this condition – the alignment of the 0. 40 nm edge of the (001) plane to the 0.

40 nm edge of the (101) plane. The (001) plane and (101) planes have been orientated so that their 0. 40 nm edges are coincident in the following diagram of the unit cell: The diagonals of each of these planes can be calculated using the pythagorean theorem. Regardless of the orientation of any of the planes, their diagonals will always be of the same length.The diagonal of the (001) plane is of particular interest, because it is aligned with one of the edges of the (110) plane.

This implies that the length of the diagonal of the (001) plane must be equal to the length of one of the edges of the (110) plane. The length of the diagonal of a given plane can be calculated by the pythagorean theorem in the following manner: where lengthplane_diagonal is the length of the diagonal of a given plane, lengthplane_edge_1 is the length of one of the edges of that plane, and lengthplane_edge_2 is the length of the other edge of that plane. Rearranging this formula,For the (001) plane, length(001)_plane_edge_1 = 0.

30 nm, and length(001)_plane_edge_2 = 0. 40 nm. We can therefore apply the above formula to calculate the length of the diagonal of the (001) plane. with significant figures applied The diagonal of the (001) plane is coincident with one of the edges of the (110) plane, so their lengths must be equal. Therefore, the edge of the (110) plane that is coincident with the diagonal of the (001) plane must have a length of 0.

50 nm. There is only one possible orientation of the (110) plane that will allow this situation. This orientation has been established in the following diagram of the unit cell.I have used the symbols ? , ? , and ? , as defined on page 38 of Callister, to denote the three interaxial angles.

Symbols ? 1 and ? 2 have also been used to denote two angles of interest. As a final verification that the above diagram of the unit cell is correct, we can compare the diagonals of the (110) and (101) planes. It can be seen from the above diagram that these diagonals should be coincident in our unit cell. In order for the two diagonals to be coincident, they must have the same length. We can verify that the diagonals possess the same length by applying the pythagorean theorem to determine the lengths of the diagonals.with significant figures applied with significant figures applied Using the appropriate number of significant figures, the diagonals of the (110) plane and (101) plane indeed have the same length. We must now determine the interaxial angles of the unit cell. This can be accomplished by considering the intersections between the given planes.

Miller indices are defined in the cartesian coordinate system. The x-axis, y-axis and z-axis of the cartesian coordinate system are orthogonal by definition. The (001) plane is defined to be parallel to the x-y plane. Therefore, as shown in the above diagram of the unit cell, the 0.30 nm edge of the (001) plane must lie parallel to the x-axis, and the 0. 40 nm edge of the (001) plane must lie parallel to the y-axis. Therefore, by shifting the 0.

40 nm edge, we can see that it equivalently lies along the y-axis. By shifting the 0. 30 nm edge, we can see that it equivalently lies along the x-axis. The interaxial angle between the x-axis and the y-axis, which is denoted by the symbol ? , must be defined by the angle between the 0.

30 nm edge of the (001) plane and the 0. 40 nm edge of the (001) plane. Since I have assumed that the three given planes are rectangular, ? must therefore be 90o.We can also determine the interaxial angle between the y-axis and the z-axis, which is denoted by the symbol ? , by considering the (101) and (110) planes of the unit cell.

From the above diagram of the unit cell, we can see that the 0. 35 nm edge of the (110) plane lies parallel to the z-axis. Therefore, by shifting this edge, we can see that it equivalently lies along the z-axis. From the above diagram of the unit cell, we can see that the 0. 40 nm edge of the (101) plane lies parallel to the y-axis.

Therefore, by shifting this edge, we can see that it equivalently lies along the y-axis.Since the z-axis and y-axis in the cartesian coordinate system are orthogonal by definition, ? must therefore be 90o. We can also calculate the interaxial angle between the x-axis and the z-axis, which is denoted by the symbol ? , 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 ? represents the angle between side2 and side3. This formula can be rearranged to solve for ?. The angles ? 1, ? 2 and ? , 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 ? = 90o, ? = 90o and ? = 90o.These values for ? , ? , ? , ? 1 and ?2 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. 022×1023 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.

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