Question

Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3. Aluminum crystallizes in a face centered cubic structure.

Answer 1

Let’s do it steps-

1. ..First calculate mass of 1 Al atom -
   26.98 g/mole x 1 mole/6.02 x 10^23 atoms = 4.48 x 10^-23 g/atom
   One unit cell has 4 atoms, since each corner contains 1/8 atom and the 6 faces each have ½ atom.
   4 atoms/unit cell x 4.48 x 10^-23 g/atom = 1.792 x 10^-22 grams/unit cell

2..Volume of unit cell:
1.792 x 10^-22 grams/unit cell x 1 cm^3/2.71 g = 0.6613 x 10^-22 = 6.613 x 10^-23 cm^3

3..Edge length = (6.613 x 10^-23)^1/3 = 4.05 x 10^-8 cm
The face diagonal of a unit cell
(Length edge)^2 + (length edge)^2 = (face diagonal)^2
(4.05 x 10^-8)^2 + (4.05 x 10^-8)^2 = (face diagonal)^2

face diagonal = 5.73 x 10^-8 cm
The face diagonal contains 4 radii.

4..Radius of Al atom = 1.43 x 10^-8
1.43 x 10^-8 cm x 10^12 picometers/meter x 1 meter/100 cm = 143 pm

Hope this will help you. Write me for any query.