1.

Aluminium crystallies in a cubic close packed structure. Radius of the atom in the metal is 125 pm. (i) What is the length of the side of the unit cell ? (ii) How many unit cells are there in 1 cm^(3) of aluminium ?

Answer»

Solution :r = 125 pm = `125 xx 10^(-10)` CM
(i) For ccp, edge length of unit cell,
`a = (4R)/(sqrt(2))`
`= (4 xx 125 xx 10^(-10))/(1.414)`
= `353.60 xx 10^(-10)` cm
`= 354 xx 10^(-10)` cm
(ii) NUMBER of unit cells in 1 `cm^(3)`
`= ("Total volume")/("Volume of 1 unit cell" (a^(3)))`
`= (1 cm^(3))/((354 xx 10^(-10))^(3))`
`= 2.254 xx 10^(22)` Unit cells


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