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Calcium has face-centred cubic lattice and radius of calcium atom is 195.6 picometre. Determine the number of Ca atoms present on surfaces of a mm3 block of calcium metal assuming that atoms in the closest packing calcium metal assuming that atoms are in the closest packing. |
Answer» Solution : In FCC, the relation is  `4r=sqrt(2a) impliesa=2sqrt(2r)=553.24xx10^(-9) mm` `implies` Area of a face pf unit cell `=a^(3)=3.06xx10^(-13)mm^(2)` `implies` SURFACE area of metal BLOCK =6 `mm^(2)` `implies` Total number of FACES of unit CELLS present on surface `=(6)/(3.06xx10^(-13))=1.96xx10^(13)` Each face contributes two two Ca atoms on surface as `implies` Total number of Ca atoms present on surface `=2xx1.96xx10^(13)=3.92xx10^(13)` |
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