Aluminium crystallises in a cubic close-packed structure. Its metallicradius is 125 pm.(i) what is the length of the side of the unit cell?(ii) how ma
Answers
Answer:
Explanation:
For the cubic close–packed structure Let a is the edge of the cube and r is the radius of atom
Given that r = 125 pm
a = 2√2 r
Plug the value of r we get
= 2 x 1.414 x125 pm
= 354 pm (approximately)
Volume of one unit cell = side3 = (354 pm)3
1 pm = 10–10 cm
= (354 x 10–10 cm)3
= (3.54 x 10–8 cm)3
= 44.36 x 10–24 cm3
= 4.4 × 10−23 cm3
Total number of unit cells in 1.00 cm3
= total volume / size of each cell
= (1.00cm3)/( 4.4 × 10−23 cm3)
= 2.27 × 1022 unit cell
Answer:-
as we know that, in ccp structure, the formula of Edge length(a) is:-
a = 4r/√2
where, r = radius of metal
a = 4×125/√2 =
a = 353.55 pm