Question

1. For zine (Zn) with a perfect HCP crystal structure having a c/a ratio of 1.633 and a density of 7.13 g/cm, (18 marks) a) Show that the volume of the Zn unit cell is given by 6v3R2c. (10 marks) b) Compute the atomic radius for Zn in nm. (8 marks)

Answer 1

HCP Volume = base area x height We Can divide the hexagon into 6 dual equilateral trangle Area of equilateral triangle = √3 a2 Вокала з сухо - and a= 2R Base area = 6 x G x 4R² = 6V R2 : Volume = base area x height = 683 R².C

In HCP, no. of atoms per anit Cell - 6. mass NA X Volume volume C = 1.633 -1.633 C 2R c= 2x1.633 R = 7.13 = 6 x 65 6.02 x 1023 x 6v3.x - R²x (= 6x65 6:02 x 1023 x 6Vz -) R² x 2 x 1.633R = 6 x 65 6.02 X 10 23 x 6 V3. 7 R² = 1.968 x 1023 - R² - 19.08 X 1024 =) R (19:08 X1024). R = 2.67 x 10 cm - - 2.67 x 108 nm. lo-7 - 0. 267 nm