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Unit cell of the iron crystal has edge length of 288 pm and density of 7.86 g `cm^(-3)`. Determinte the type of crystal lattice. Atomic mass of Fe= 56 g `mol^(-1)` |
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Answer» Correct Answer - Type of crystal lattice is bcc. Given : Edge length =a= 288 pm `=2.88 xx 10^(-8)` cm Density of crystal = d= 7.86 g `cm^(-1)` A vogadro number = 6.022 `xx10^(-23) mol^(-1)` Atomic mass of Fe= 56 g `mol^(-1)` Type of crystal lattice = ? Mass of one Fe atom = `(56)/(6.022 xx 10^(23)) = 9 .3 xx 10^(-23) g` If there are z atoms in the unit cell , then Mass of unit cell= mass of z atoms = z `xx 9.3 xx 10^(-23) g` Volume of unit cell `= a^(3) = (2.88 xx 10^(-8) )^(3)` `=23 .88 xx 10^(-24) cm^(3)` Density of unit cell `d= ("mass of unit cell")/("Volume of unit cell")` `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` Since the number of atoms in the unit cell is 2, the crystal lattice must be of bcc type. |
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