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Single crystal XRD

Single crystal XRD. Energy as a function of wavelength. X-rays – High energy, highly penetrative electromagnetic radiation Energy E = h n = hc / λ Because n = c/ λ is the frequency c is the speed of light in a vacuum λ is the wavelength h is Planck’s constant

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Single crystal XRD

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  1. Single crystal XRD

  2. Energy as a function of wavelength • X-rays – High energy, highly penetrative electromagnetic radiation • Energy E = hn = hc/λ • Because n = c/λ • is the frequency • c is the speed of light in a vacuum • λ is the wavelength • h is Planck’s constant • λ(X-rays) = 0.02-100Å (avg. ~1 Å) • λ(visible light) = 4000-7200Å As λgets smaller, energy E gets bigger.

  3. Powder X-ray Diffraction • When the geometry of the incident X-rays impinging the sample satisfies the Bragg Equation, constructive interference occurs and a peak in intensity occurs. The powder improves the chance that all possible planes are sampled. The sample rotates through angles θ while the detector counts photons.

  4. Kb would make confusing double peaks • The Copper anode makes two strong peaks, Ka and Kb. The Kb would cause confusing double peaks, so we filter it out with a Nickel filter.

  5. Calculations • Recall last time we had a homework problem • A strong peak is recorded at a 2q = 28.29 degrees. Calculate d using Bragg’s Law. Assume a l = Ka for copper of 1.54 Å. and use n=1 • Hint: 2q is given, you need q. a. 1.625 angstroms b. 3.15 angstroms We used Braggs Equation nλ = 2d sin θ

  6. Instrument Plots • Here is a typical plot with a strong reflection at • 2q = 28.29. Lets look at the other reflections

  7. The next stronger peak • There is another reflection at • 2q = 47.12. Calculate d. nλ = 2d sin θ

  8. The next stronger peak • There is another reflection at • 2q = 55.74. Calculate d. nλ = 2d sin θ

  9. Identification • A mineral with d - spacings of 3.15, 1.93, 1.65 is Fluorite, CaF2 After we have the d-spacings, we want to calculate the dimensions of the unit cell, and decide which planes are giving strong reflections. This will allow us to check our model building ideas.

  10. Calculating Unit Cell Size • A formula exists for calculating the dimensions of the unit cell for each crystal system. • For Isometric crystals such as Fluorite or Halite, the formula is simple • a2 = d2 (h2 + k2 + l2) where the (hkl) are the Miller indices • Let’s try it with Halite, where the XRD reports d = 2.8, 1.98, 1.62, 1.404

  11. Halite NaCl d d2x (h2 + k2 + l2) for the zones zones[100] [110] [111] [200] [210] [211] sum of squares 1 2 3 4 5 6 2.8 7.847.84 15.08 23.52 31.36 39.20 47.04 1.98 3.923.927.84 11.76 15.68 19.6 23.52 1.62 2.622.62 5.24 7.86 10.48 13.1 15.72 1.40 1.971.97 3.96 5.94 7.86 9.9 11.88 Discussion: which is a2 ? Which zones made which reflections?

  12. Homework and Lab • Now complete the last question on HW8-9, then do the lab using your data printouts from your unknown.

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