Question

Please do all questions with working provided.

Background information Al is a metal, silvery in colour, with a resistivity of 2.65 μΩ.cm at room temperature. As depicted in Fig. 1(a), it has the so-called face centred cubic (FCC) structure, in which the fractional coordinates of the atoms within the (conventional cubic, side length of a 4.046 A) unit cell are: (0,0,0), CS,%,0), (14,0%), (0,%,-ב The intermetallic compound AbAu is bright purple in colour, has a melting point of 1060°C, is rather brittle, and has a resistivity of ρ-8 μΩ.cm.! As depicted in Fig. 1(b), the repeating unt is also cubic (with a side length we shall denote a'), and has atoms at fractional coordinates: Au: (0,0,0), Cs, ,0), (%,0,%), (0,%,%) (b Fig. 1: Atomic structure of the repeating units of (a) Al and (b) AlbAu. (Images produced using VESTA: J. Appl. Crystallogr., 44 (2011) 1272.) It can be shown (though you need not do so), that the structure factor for an FCC material is where f denotes the atomic scattering amplitude of the element of which the material is comprised. (c) (3 marks) Simplify Eq. (4) into a prescription for which (hkl) combinations produce forbidden reflections and which do not. Note: it will not suffice to simply write down the prescription - you need to show the reasoning that justifies it. (d) (5 marks) For general (hkl), evaluate the structure factor for AlhAu. Simplify your expression to the point where you can again provide a simple prescription for which (hkl) combinations produce forbidden reflections and which do not. You should assume that the atomic scattering amplitude for Al atoms is different to that for Au atoms. (e) (4 marks) Using crosses to indicate which Miller indices (hkl) correspond to forbidden reflections and ticks to indicate which correspond to allowed reflections (i.e. any reflection that is not forbidden), complete the following table. Note: you should be able to complete this and subsequent parts even if you have not obtained an "optimally" simple solution to part (d) Al Au Example (hk (100) (110) Al (FCC 200) (210) (211) 4 (220) (221) (310) (311) (222 12

Answer 1

廿 The athuchau factor fo FCC cau be coleulated usi mula k,) value sho let K=0() all even. Th 21 type 4 4チ。 = 1-1+1-1 A Vowed Aljouses ル asn bidden

Et values must be sepadtater bn a even," Coos 1 factor j "j A. Au can、 be calculated as 2 srom equaton , folawing extincthon sule can be darved 去,hd ame all odd (re.composed onul ot fAU 2 di rib ?

Example (A,H) (2 11 9 12 リッジ 丿// 22 \ 2 3.32