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A metal M(at.wt = 40.0) depending on temperature crystallises in f.c.c and b.c.c structure whose unit cell length are 3.5 and $3.0 A^{\large\circ}$ respectively. The ratio of its densities in f.c.c and b.c.c structures is

$(a)\;1.259\qquad(b)\;2\qquad(c)\;\large\frac{8}{\sqrt6}\qquad(d)\;\large\frac{16}{3\sqrt6}$

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$\rho_1 = \large\frac{4\times at.wt}{N_A \times a_1^3}$
$\rho_2 = \large\frac{2\times at.wt}{N_A \times a_2^3}$
$\large\frac{\rho_1}{\rho_2} = \large\frac{4}{2}\times \large\frac{a_2^3}{a_1^3}$
$=\large\frac{4}{2}\times\large\frac{(3.0\times10^{-8})^3}{(3.5\times10^{-8})^3}$
=1.259
Hence answer is (a)
answered Feb 25, 2014 by sharmaaparna1
 

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