Question 1:a) For BCC, we have the following diagram:Hence, the relationship between R and a can be deduced asb 2 a 2 a 2 2a 2 c 2 a 2 b 2 a 2 2a 2 3a 2 c 2 3a 4 R a4R3b) We have# of atoms = 24383volume of two atoms = 2 R 3 R 3 4 volume of a unit cell = a R 3 33Hence, the APF (Atomic Packing Factor) for BCC can be determined as follow:APF volume of two atomsvolume of a unit cell88 R3 R3 3 3364 3 4 RR3 3 3 3 68%8Question 2:a) Converting pm to cm, we have:405 pm 1 cm 405 1010 cm 4.05 108 cm10110 pmThe volume of the unit cell is calculated by: 4.05 108cm 6.64 1023 cm33The mass of a tantalum per FCC unit cell is calculated by:6.64 1023 cm3 2.70 g / cm3 1.79 1022 gb) Since its given that Aluminum has FCC crystal structure, the number of atoms is n 4 .Hence, the mass of a tantalum per atom is calculated by:1.79 1022 g 4.48 1023 g4c) The atomic weight of 26.98 g / mol . Hence, the number of moles in four unit cells of Al iscalculated by:4.48 1023 g 1.66 1024 mol26.98 g / molQuestion 3:The atomic mass of Niobium (Nb) is equal to 92.91 g / mol , and if Niobium had a FCC structure,we would have the following calculation for density ...

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