Question

material engineering

1. Metal 'P' exhibits a FCC structure. Determine its atomic radius, in nm. (30p) Atomic weight (g/mol) 90.0 Metal Density (g/cm2) 11.75 P R (nm)= CC unit cell), 20p

Answer 1

The relation is BenA, where given by equation. UNA P = density, no no. of atoms n=4 for fcc structure A = atomic weight, ua volume, NA= Avagnado's number NA= 6.022 x 1023 atoms / mol. no 4 atoms / whit cell for fcc. = 11.75 (Glam?) A = 90 (almol) Therefore, we get

PEDA ملہ ۷ 1 11.75= 4x 90 V x 61022x1023 V= 4 x 90 - 11.75 +6.022 x2023 V = 5.0877 x 10-23 (cm3) Now for tee structure ve a3 where - a = unit length of fcc structe. a= 2/2 R, where Ra radius.

v= a3 VE (2 FER) 5.0877x10-23 = 16 S2 R² R² = 2.2484 x 10-24 (om) R = 1.31 x 108 cm R=1.31x 108 x 107 nm R = 0.131 nm Atomic Radius of metal Pis 0.131 om
Using the density formula for an atom to get the required answer.