CHE 350 Fall 2014 NAME_Solutions__________________________ Exam 1 (100 points) 2(a) The conventional unit cell for the HCP crystal structure is shown below. What is the associated Bravais lattice? On the diagram, draw the basis vectors (a1, a2, a3) for that Bravais lattice. 7 points

4 points for correct basis vectors The associated Bravais lattice for this structure is SIMPLE HEXAGONAL 3 points for correct Bravais lattice 2(b) On the diagram above, sketch and label each plane indicated below. Miller indices are given in terms of the Bravais lattice vectors. 6 points (100), (001), and (002) See above diagram 2 points for each plane 2(c) Which of the set(s) of planes mentioned in (b) do you expect will give rise to diffraction peak(s)? Explain your reasoning. 7 points (100) = Will appear as a diffraction peak

1 point for correct (002) = Will appear as a diffraction peak 1 point for correct (001) = Will NOT appear as a diffraction peak 1 point for correct

Explanation: Looking at the (001) plane, there is another parallel plane of atoms that lies one half-d001 spacing away from this plane the (002) plane. If light diffracts off of the (001) plane, with a wavelength equal to d001, then any light that diffracts off of the intermediate (002) plane will destructively interfere with the (001) scattering light because it will be completely out-of-phase with the scattered light from (001). Neither the (002) nor the (100)

a3

a1

a2

( 0 0 1 )

( 0 0 2 ) ( 1 0 0 )

CHE 350 Fall 2014 NAME_Solutions__________________________ Exam 1 (100 points) planes have intermediate atom planes that lie dhkl away, and so these planes will produce diffraction peaks. 4 points for correct explanation. Partial credit awarded for mentioning destructive interference, or lattices that are d-spacing away from the planes in question. 2(d) How would the diffraction pattern differ between HCP and a simple hexagonal crystal structure? Explain. 5 points The simple hexagonal structure does not have any atoms on the (002) plane. Thus, the destructive interference that eliminates the diffraction peak for the (001) plane in the Hexagonal Close-Packed structure in parts a)-c) will not occur, and this plane (among others) would show up. In general, the simple hexagonal diffraction pattern will have more peaks than the hexagonal close-packed structure. Also, the peak locations will change because the d-spacings between planes in SH should be different than in HCP. For instance, the (001) peak in SH will have a smaller d001-spacing, and thus the diffraction peak will be shifted to a higher diffraction angle. 5 points for saying the SH diffraction pattern will have more peaks present than the HCP pattern. Partial credit awarded for mentioning the 2-atom basis of HCP, the lack of destructive interference in SH, or discussing changes in d-spacing.

CHE 350 Fall 2014 NAME_Solutions__________________________ Exam 1 (100 points) 4(a) CsCl has a coordination number of 8 (i.e. each ion has 8 nearest neighbors) and has cubic symmetry. Below, draw the atoms in a CsCl unit cell. 5 points

5 points for correct structure 4(b) How many of each type of atom is found within the unit cell of CsCl, as drawn in (b)? 4 points Corner atoms (Cl in this case): 8 x (1/8) atom = 1 Cl atom 2 points Central atom (Cs in this case): 1 x (1) atom = 1 Cs atom 2 points 4(c) Find a fully-simplified expression for the ionic packing fraction in CsCl, where the radii of the ions are related by rCs+/rCl- = x. The ionic packing fraction is the fraction of space occupied by the ions, analogous to the atomic packing fraction for non-ionic crystals. (Note that x = 0.939, but your answer should be expressed in terms of x and no other parameters.) 11 points

Ionic Packing Fraction = Volume of Atoms in Unit Cell/Volume of Unit Cell First, we calculate the volume of the unit cell. We need to figure out the close-packing direction to relate the lattice constant, a, to the radii of the ions. We do this by drawing a body-diagonal through the cube (see picture at right):

Chlorine

Cesium

a

CHE 350 Fall 2014 NAME_Solutions__________________________ Exam 1 (100 points) We know that along this line, , a cesium atom contacts to chlorine atoms. Thus, the length of is: = + 3 points for identifying a close-packing direction in the 3D unit cell, even if geometry was off. To relate the lattice constant, a, to the body-diagonal line, , we do some trigonometry (recalling that x=rCs/rCl):

a2 + a2 = 2

= 2a 2 + a2 = 2

= 3a = 2(rCs + rCl )a = 23 (rCs + rCl ) =

23 rCl (1+ x)

2 points for correct expression for a Now that we know the lattice constant, a, in terms of the ion sizes, we can calculate the total volume of the unit cell:

Vcell = a3 =23 rCl 1+ x( )

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=83 3 rCl

3 1+ x( )3

2 points for correctly calculating Vcell from a Next we calculate the total volume of atoms in the unit cell. There is 1 cesium atom and 1 chlorine atom in the unit cell, so we can calculate the total volume occupied by atoms:

Vatom =43 rCl

3 +43 rCs

3 =43 (rCl

3 + rCs3) =43 rCl

3(1+ x3) 2 points for correctly calculating Vatom from the atomic radii Finally, we compute the ionic packing fraction:

I.P.F. = VatomVcell=

43 1+ x

3( )rCl383 3 1+ x( )

3 rCl3= 3 1+ x3( )2 1+ x( )3

2 points for correct I.P.F., following from previously calculated Vatom and Vcell. For each of the above parts, partial credit if math error was made.

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