Question

Need the answer of #3 and #4 A and B

3. a-Polonium crystallizes in a body-centered cubic unit cell. The edge length of its unit cell is 336 pm. a. What is the radius of polonium? b. Determine the density of a-Polonium da listabaid 4. Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm. a. What is the atomic radius of a calcium ion? AP + 6 - 12 >(558.8pm)2 558.pm)2 = (4R) R=(1558.8 pm)² + (558.4pm) ² = 297.6pm e for Ca. b. Calculate the density of calcium in this structure. Number of atoms in FCC = 4 molar mass 40.1 6.658x10232 6.66x1025 6.022-X1023 uraadrost mass of it atomis = 4 x 6.66x1023 .=2,664x1024 volume of cake L² = 15588

Answer 1

solution: - z-a) The relation between edge tength (a) and radius of atom (r) for BCC lattice 3a = 48 8= (3 a, where a= edge length &=radius of atom. and a is given, 336pm. so, &- (3 x 3361m = 145.49 pm 6) To find density of the lattice we use the formula, et= 2x m ². NA where da density 2= Number of atom m=molecular weight of each atom NA- Avogrado's Number. D Effective number of atoms in body centered cubic is 2. So d= 2 x 209. (336x10 -12) 3x 6.023x10 23 d= 1.829 x 107 g/m3 / 18.29 91 m3

4-a) The relation between edge length and radius of atem is face-centered ubic structure is given by 48=129, . 9 given, a=558. 8 Pm. so = 558 8 = 197.56 pm. 212 b) Density of the lattice is given by, d = zx m 23 x Na for face centered cubic structure 2=4. anda in for calcium is 40 g. .. d= 4 x 40 (558+8x10-12)3x 6'023X6023 de 1.522 x 106 glm or 1.522 91 cm3
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