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Aluminium metal forms a cubic close-packed crystal structure. Its atomic radius is `125xx10^(-12)`m. (a) Calculate the length of the side of the unit cell. (b) How many such unit cells are there in `1.00 m^3` of aluminium ? |
Answer» Correct Answer - (a)354 pm (b)`2.25xx10^28` ccp=fcc. For fcc, `r=a/(2sqrt2)` or `a=2sqrt2r=2xx1.414xx125xx10^(-2) m =354xx10^(-12) m`=354 pm Volume of one unit cell =`(354xx10^(-12)m)^3=4.436xx10^(-29) m^3` `therefore` No. of unit cell in 1 `m^3=1/(4.436xx10^(-29))=2.25xx10^28` |
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