Practice Problems Set II

Transcription

1 P1. For the HCP crystal structure, (a) show that the ideal c/a ratio is 1.633; (b) show that the atomic packing factor for HCP is (a) A sketch of one-third of an HCP unit cell is shown below. Consider the tetrahedron labeled as JKLM, which is reconstructed as 1

2 (b) The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus

3 P. Assuming hard-sphere model. Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of 1.0 g/cm3, and an atomic weight of g/mol. 3

4 P3. (Assuming hard-sphere model.) Titanium has an HCP crystal structure and a density of 4.51 g/cm 3. (a) What is the volume of its unit cell in cubic meters? (b) If the ratio is 1.58, compute the values of c and a. (c) Denote the radius of Titanium as r, do we have r > a or r < a or r = a? (a) 4

5 (b) (c) r < a P4. (Assuming hard-sphere model.) Below are listed the atomic weight, density, and atomic radius for three hypothetical metals. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in figure below. A BCC B Simple cubic C BCC P5. (Assuming hard-sphere model.) Methane (CH4) has a tetrahedral structure similar to that of SiO (Figure below), with a carbon atom of radius cm at the center and hydrogen atoms of radius cm at four of the eight corners. Calculate the size of the tetrahedral cube for methane. 5

6 3a r C r H 8 8 3a ( ) P6 (Assuming hard-sphere model.) Titanium has an HCP unit cell for which the ratio of the lattice parameters c/a is If the radius of the Ti atom is nm and atomic weight is A Ti = g/mol, (a) determine the unit cell volume, and (b) calculate the density of Ti and compare it with the literature value, 4.51 g/cm 3. REMARK: please note that the Callister solution for this problem IS WRONG. So please DO NOT FOLLOW Callister solution manual (in case you have one). One important thing to note is that c/a value is smaller than the ideal one, As a result, the six atoms on the bottom layer WILL NOT touch each other (also the three atoms in the middle layer WILL NOT touch each other). BUT for any atom in the middle layer, it will still touch three atoms in the bottom or top layer. You can imagine this scenario when you compress the HCP along the c-axis. As a result (denoting the radius of Ti as R), a R! Instead, we have a R Also follows we have From the above we have And the unit cell volume is From which we can have V C 3 c a 3 3 3ac a.044 R R ? 0 cm /unit cell g / cm The same as the value in literature. 6

7 P7. Zinc has an HCP crystal structure, a c/a ratio of 1.856, and a density of 7.13 g/cm 3. Compute the atomic radius for Zn. REMARK: please note that like P6, Callister s solution is wrong in its reasoning, though the final answer remains unchanged. One important thing to note is that c/a value is larger than the ideal one, As a result, the six atoms on the bottom layer WILL touch each other (also the three atoms in the middle layer WILL touch each other). BUT for any atom in the middle layer, it WILL NOT touch any atoms in the bottom or top layer. You can imagine this scenario when you elongate the HCP along the c-axis. As a result (denoting the radius of Ti as R), The relation a R Also follows we have remains valid. And we have V 3 3 ac C = 6 R c 3 R nm P8. Within a cubic unit cell, sketch the following directions: (a) [1 10], (e) [1 1 1], (d) (b) [1 1], (f) (c) [01 ], (g) [1 ], [133], (h) [103]. [1 3 ], The directions asked for are indicated in the cubic unit cells shown below. 7

8 P9. Determine the indices for the directions shown in the following cubic unit cell: Solution : Direction A is a [4 30] direction; Direction B is a [3 ] direction; Direction C is a [133] direction; Direction D is a [136] direction. 8

9 P10. What are the indices for the two planes drawn in the sketch below? Solution : Plane 1 is a (00) plane. (Please note this is not reduced to (010). Please see the footnote in Callister, P64 for more information.) Plane is a ( 1) plane. P11. Determine the Miller indices for the planes shown in the following unit cell: Solution Plane A: it is a (3) plane; For plane B it is a (101) plane. P1. Convert the (010) and (101) planes into the four-index Miller Bravais scheme for hexagonal unit cells. Solution 9

10 For (010), h = 0, k = 1, and l = 0, the value of i is equal to i (h k) (0 1) 1 (011 0) Therefore, the (010) plane becomes. Now for the (101) plane, h = 1, k = 0, and l = 1, this leads to i (h k) [1 0] 1 such that (101) becomes (10 1 1). P13. Determine the indices for the planes shown in the hexagonal unit cells below: Solution (a) This plane passes through the origin of the coordinate axis system; therefore, we translate this plane one unit distance along the x axis, per the sketch shown below: 10

11 This is a (10 1 0) plane. (b) For this plane, intersections with the a 1, a, and z axes are a, a, and c. In terms of a and c these intersections are 1, 1, and 1. Using i ( h k), this is a (1 101) plane. 11