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Unitcell of theironcrystal hasedgelengthof 288pmanddensityof 7.86 g cm^(-3). Determinte the typeof crystal lattice. Atomicmass ofFe= 56 g mol^(-1) |
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Answer» Densityof crystal = d= 7.86 g `cm^(-1)` A vogadro number= 6.022 `xx10^(-23) mol^(-1)` Atomicmass ofFe= 56 g `mol^(-1)` Typeof crystallattice = ? Massof one Fe atom = `(56)/(6.022 xx 10^(23)) = 9 .3 xx 10^(-23) g` If thereare z atoms in the unitcell , then Massof unitcell=mass ofz atoms = z `xx 9.3 xx 10^(-23) g` Volumeof unitcell `= a^(3)= (2.88 xx 10^(-8) )^(3)` `=23 .88 xx 10^(-24) cm^(3)` Densityof unitcell`d= ("massof unit CELL")/("Volumeof unitcell")` `7.86 = (z xx 9.3 xx 10^(-23))/(23. 88 xx 10^(-24))` `:. z =(7.86 xx 23.88 xx 10^(-24))/(9.3xx 10^(-23))= 2.01 =2` Sincethe numberof atomsin THEUNIT cellis 2,thecrystal latticemust be of bcctype. |
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