b. What is the state of argon at 85 K? Is it a solid, a liquid, or a gas, and why? The peaks indicate some nonrandom arrangement of the atoms is present, which means the material is not a gas. If the peaks were discrete, the distance to the neighbors would be exact, and the material would be a solid. Since the function is continuous and the peaks have nonzero width, the material must be a liquid at 85 K. c. What is your estimate for the diameter of an argon atom? The distance from the r = 0 to the beginning of the first peak in the g(r) plot is the average distance from the center of one atom to the center of the nearest neighbor atom. 2 R ≈ 3.2 Å diameter = 2 R ≈ 3.2 Å This diameter is somewhat larger than expected. Argon is a noble gas with electronic configuration [Na] 3s23p6. If the material is a liquid, however, the distance from the orgin of g(r) to the first peak may not be the distance from the center of one hard sphere touching an adjacent hard sphere. d. What do the peaks on the g(r) shown above represent? The peaks represent large fluctuations from the average density of the material. The large fluctuations at short distances from the origin occur because the closest neighbors reside at regular intervals from the atom at the origin. Areas of empty space result in troughs in g(r), and areas where a nearest neighbor resides is a peak in N/V. The peaks show the first through the fourth nearest neighbors from the origin. e. Why does the pair-correlation function flatten beyond 15-20 Å? g(r) flattens because distinct neighbors are not present at long range in the liquid. The number of neighbors per volume in the shell at (r,r+dr) is equal to the density. 2Question 2. Nematic liquid crystals a. What characterizes a nematic liquid crystal? How is it different from a cholesteric liquid crystal, or from a smectic one? Discuss these differences in terms of orientation and translation order parameters, both long-range and short-range. The translation (Σ) and orientation (S) order parameters are given by ⎟⎠⎞⎜⎝⎛=Σazπ2cos 21cos32−=θS Nematic phases show strong translational order in short range. They show strong orientational order in long range due to a unique axis along which the mesogen orient themselves. The nematic phases have some, but weaker orientational order at short range. Cholesteric phases are similar to nematic phases, but in cholesteric phases, the mesogen orientation twists along the unique axis. Cholesteric show orientational order at short range but little orientational order at long range. Cholestric phases show little translational order. Smectic phases have long range translational and orientational order. b. The orientation order parameter S for a neumatic liquid crystal is (given above), where θ is the angle between a mesogen and the average preferred orientation n, and the angular brackets represent an average over all the mesogens in the sample. Show what the order parameter S will be if all the mesogens are oriented perfectly in the direction , and what will it be if they are randomly oriented (derive explicitly your result). ˆnˆ mesogens orient perfectly in Æ θ = 0 nˆ cos2 θ = 1 \ S = 1 mesogens randomly oriented Æ θ = 0 …90 ° cos θ = +1 …0 cos2 θ = +1 …0 average of cos2 θ = <cos2 θ> = 0.5 \ S = 0 4b. Explain how the Laue conditions arise for constructive interference of a plane wave incident on a monoatomic crystal that has one atom at each point of the Bravais lattice 321),,( anamalnmlrrr++= 6d. Why do we use monochromatic x-rays in a Debye-Scherrer experiment? To measure diffracted rays from a sample, one must either look at the sample from a range of views but with a single wavelength or one must look from one direction and monitor a variety of wavelengths in order to determine characteristic d spacings. The Debye-Scherrer experiment uses the first approach. A camera circles a sample of crystalline powder irradiated with x-rays. The powder contains many orientations of the crystals, and diffracted beams occur in many directions. Using Bragg’s Law nλ = 2 d sinθ, many angles θ are known, λ is constant, and thus d can be solved for.
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