- 3 A Gold Au Crystal Forms An Fcc Crystal With The Lattice Constant 4 1 10 M And The Atomic Weight Is 197 The E 1 (217.86 KiB) Viewed 27 times
(3) A gold (Au) crystal forms an FCC crystal with the lattice constant 4.1 × 10-¹⁰ m and the atomic weight is 197. The e
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(3) A gold (Au) crystal forms an FCC crystal with the lattice constant 4.1 × 10-¹⁰ m and the atomic weight is 197. The e
(3) A gold (Au) crystal forms an FCC crystal with the lattice constant 4.1 × 10-¹⁰ m and the atomic weight is 197. The elastic stiffness constants of Au are C₁1 = 1.9 × 10¹2 dyn/cm², C₁2 = 1.6 × 10¹2 dyn/cm² and C44 = 0.42 × 1012 dyn/cm². Q1: Does the Au crystal have an optical mode phonon dispersion? Q2: Calculate the longitudinal sound velocity c. Use the unit of m/s. Q3: Calculate the Debye angular frequency w in the Debye model. Use the unit of rad/s. Q4: Calculate the maximum angular frequency from the w-k dispersion relation. Use the theoretical w of a monatomic 1D crystal model and the unit of rad/s. Assume that interatomic spring constant a is equal to E-Ro, here Eis the Young's modulus and Ro is the interatomic distance at the equilibrium. Q5: Experimental Debye angular frequency is about 2.3 × 1013 rad/s. Show whether all phonon angular frequencies are expected to work at the room temperature (300 K).