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Aluminium metal forms a cubic face centred closed packed crystal structure. Its atomic radius is `125xx10^(-12)`m. (a) Calculate the length of the side of the unit cell. (b) How many unit cells are there in `1.0m^(3)` of aluminium? |
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Answer» (a) For a face centred cubic lattice (fcc). Radius (r) `=(a)/(2sqrt2)` `a=rxx2sqrt2=125xx10^(-12)xx2sqrt2m` `=125xx2xx1.414xx10^(-12)=354xx10^(-12)m` (b) Volume of unit cell `(a)^(3)=(354xx10^(-12))^(3)m^(3)=4.436xx10^(-29)m^(3)` No. of unit cells in `1.0m^(3)` of Al `=((1.0m^(3)))/((4.436xx10^(-29)m^(3)))=2.25xx10^(28)` |
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