Determine the repeat distance, linear density, and packing fraction for FCC nickel, which has a lattice parameter of 0.35167 nm, in the[100],[110] , and [111] directions. Which of these direction is close packed?

Silver Ag has the fcc crystal structure as shown in the figure below. Determine its atomic packing factor APF? Given: The radius of Germanium: r(Ag)= 0.144 nm.  Select one: a. 0.14 b. 0.34 c. 0.68 d. 0.12 e. 0.4 f. 0.51 g. 0.74 h. 0.21

Nickel and aluminum both have FCC crystal structures. Which of the following statement iscorrect about packing factor of their unit cell?Atomic mass of Ni> Atomic mass of AlAtomic radius of Ni< Atomic radius of Ala. The unit cell of nickel has a higher packing factor than the unit cell of aluminumb. The unit cell of nickel has a lower packing factor than the unit cell of aluminumc. The unit cell of nickel has the same packing factor as the unit cell of aluminumd. Provided information is not enough to make a judgment

a) Draw the [11 0], [-1 1 0], [ 0-1-1] directions for FCC and prove they are crystallographically equivalent. The family of directions is shown with these brackets. So we can say these directions are part of the family of directions. b) Draw the (1 o 0), (-1 0 0), (0 -1 0), and (00-1) planes for FCC and prove they are crystallographically equivalent. The family of directions is shown with {} these brackets. So we can say these planes are part of the {10 0} family of planes. h me Note that I cannot put a bar as the minus sign.

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Vanadium (V) has a BCC crystal structure. The atomic radius is R = 0.132 nm and the atomic mass is M = 50.94 g/mole. What is the density of Vanadium in g/mm3? Given: Avogadro’s Number NA = 0.6023 × 1024 (atoms/mole) Select one: a. 1.5 b. 0.021 c. 0.011 d. 0.0087 e. 0.00597

The spacing of adjacent atoms in a NaCl crystal is 0.282 nm, and the masses of the atoms are 3.82*10-26 kg (Na) and 5.89*10-26 kg (CI). Use this information to calculate the density of NaCl. p= kg/m³

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Nickel has the fcc crystal structure and a lattice parameter of a = 0.3517 nm. Using this information, determine the atomic radius of Ni in angstroms.