A) 1.259
B) 2
C) \[\frac{8}{\sqrt{6}}\]
D) \[\frac{16}{3\sqrt{6}}\]
Correct Answer: A
Solution :
For FCC \[\operatorname{density} = {{d}_{1}}\], \[=\,\,\,\frac{4\times {{z}_{1}}}{{{N}_{A}}\times a_{1}^{3}}\] For BCC \[density={{d}_{2}}\] \[=\,\,\,\frac{2\times {{z}_{1}}}{{{N}_{A}}\times a_{2}^{3}}\] \[\frac{{{d}_{1}}}{{{d}_{2}}}=\frac{4}{2}\times \frac{a_{2}^{3}}{a_{1}^{3}}\,\,=\,\,\frac{{{(3\times {{10}^{-\,8}})}^{3}}}{{{(3.5\times {{10}^{-8}})}^{3}}}\] \[=\,\,\,1.259\]
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