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Suppose the mass of a single Ag atom is ‘m’. Ag metal crystallizes in fcc lattice with unit cell of length ‘a’. The density of Ag metal in terms of ‘a’ and ‘m’ is ............(A) \(\cfrac{4m}{a^3}\)(B) \(\cfrac{2m}{a^3}\)(C) \(\cfrac{m}{a^3}\)(D) \(\cfrac{m}{4a^3}\) |
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Answer» Correct answer is: (A) \(\cfrac{4m}{a^3}\) (\(\therefore\) fcc type unit cell contains total 4 atoms) Edge length of fcc unit cell = a Volume of fcc unit cell = a3 Density of silver (Ag) =\(\cfrac{Mass\,of\,fcc \,unit \,cell}{Volume\,of\,fcc\, unit\,cell}\) \(\therefore\) Density of silver (Ag) = \(\cfrac{4m}{a^3}\) |
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