Fall 2001

Office: Rogers Hall 308

Phone and Voice Mail: (541) 737-7016

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CRYSTAL STRUCTURE

1) Callster 3.4

2) Callister 3.6

3) Calculate the radius of an atom of palladium (Pd). The density of Pd is 12,020 kg/m^3, the atomic weight is 106.42, and the crystal structure is FCC.

4) The unit cell of tin (Sn) has tetragonal symmetry, with lattice parameters a = b = 583 pm, and c = 318 pm. Given that the density is 7,300 kg/m^3, and the atomic radius is 151 pm, what is the atomic packing factor?

5) Show with a drawing or drawings that the coordination number of an FCC crystal is 12.

6) In a simple cubic crystal structure, compute the size of the largest atom that could fit into the interstitial space centered on the body center.

7) On a set of cubic unit cells, draw the following directions:

a) [ 1 1 0 ] | f) [ 0 1 -4 ] |

b) [ 0 2 1 ] | g) [ 0 2 4 ] |

c) [ -1 1 0 ] | h) The close packed direction of BCC. |

d) [ 3 -3 3 ] | i) [ 2 -2 1 ] |

e) [ -1 -1 1 ] | j) The close packed direction of FCC. |

NOTE: PLEASE only put TWO directions at most on each unit cell drawing. Be sure to label your axes!

1) Callster 3.4

The best way to approach this problem is to draw the unit cell, and then draw the diagonal plane through the unit cell as shown on the drawing to the right. This is the (110) plane. Redrawing the (110) plane and showing the position of the atoms on the plane is the next step. The height of the plane is the lattice parameter a, and the width of the plane is the length of a face diagonal, or sqrt(2)*a. Drawing the right triangle made of the long and short edges of the (110) plane and the plane diagonal, and recognizing that the plane diagonal is 4R long, we can use the pythagorean theorem to find the relationship between a and R as a = 4 R / sqrt(3). |

2) Callister 3.6

The atomic packing factor is defined as the volume of the atoms in the unit cell divided by the volume of the unit cell. There are 8 corner atoms, each counting 1/8, plus the entire body center atom in the BCC unit cell, for a total of 2 atoms/unit cell. The APF calculation then becomes:

3) Calculate the radius of an atom of palladium (Pd). The density of Pd is 12,020 kg/m^3, the atomic weight is 106.42, and the crystal structure is FCC.

4) The unit cell of tin (Sn) has tetragonal symmetry, with lattice parameters a = b = 583 pm, and c = 318 pm. Given that the density is 7,300 kg/m^3, and the atomic radius is 151 pm, what is the atomic packing factor?

Since this problem deals with a non-cubic system, we need to use the lattice parameters to find the volume of the unit cell. Using the density, and looking up the atomic weight, we can find the number of atoms in the unit cell:

5) Show with a drawing or drawings that the coordination number of an FCC crystal is 12.

My three drawings show the atoms in the top face of the crystal (001) that touch the face center atoms, followed by the atoms in the middle of the lower unit cell (002), and those in the middle of the upper unit cell (another (002)). 4 + 4 + 4 = 12.

6) In a simple cubic crystal structure, compute the size of the largest atom that could fit into the interstitial space centered on the body center.

Take a slice through the unit cell at the (110) plane, and draw the atom placement and dimensions as shown to the right. The calculaltion is then easy to do, with R = atomic radius of the simple cubic atoms, and r = atomic radius of the interstitial atom: |

The interstitial atom is 73.2% the size of the simple cubic atom.

7) On a set of cubic unit cells, draw the following directions:

a) [ 1 1 0 ] |
f) [ 0 1 -4 ] |

b) [ 0 2 1 ] Reduce it to fit in the unit cell by dividing through by 2: |
g) [ 0 2 4 ] |

c) [ -1 1 0 ] |
h) The close packed direction of BCC is [1 1 1]. |

d) [ 3 -3 3 ] Reduce it to [1 -1 1] to fit in the unit cell. |
i) [ 2 -2 1 ] |

e) [ -1 -1 1 ] |
j) The close packed direction of FCC is a face diagonal, [110]. |

End of File.