A) 10.376
B) 5.19
C) 7.289
D) 2.144
Correct Answer: B
Solution :
For bcc lattice, number of atoms per unit cell =2 Now \[d=\frac{n\times M}{{{a}^{3}}\times {{N}_{o}}}=\frac{2\times 100}{{{(4\times {{10}^{-8}}cm)}^{3}}\times 6.02\times {{10}^{23}}}\] \[=200/38.528=5.19\text{g/cc}\]
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