A) 1.26
B) 3.25
C) 7.8
D) 5.35
Correct Answer: A
Solution :
| Density \[=\frac{M.Z}{{{N}_{A}}\times {{a}^{3}}}\]Density \[\propto \frac{Z}{{{a}^{3}}}\] |
| In case of fcc, \[Z=4,\]\[a=3.5\overset{{}^\circ }{\mathop{A}}\,\] |
| \[{{d}_{1}}=\frac{4}{{{(3.5)}^{3}}}\] |
| In case of bcc \[Z=2,a=3\overset{{}^\circ }{\mathop{A}}\,\] |
| \[{{d}_{1}}=\frac{2}{{{(3)}^{3}}}\] |
| \[\Rightarrow \]\[\frac{{{d}_{1}}}{{{d}_{2}}}=\frac{{{N}_{1}}}{{{N}_{2}}}\left( \frac{{{a}_{2}}}{{{a}_{1}}} \right)=\frac{4}{2}{{\left( \frac{3}{3.5} \right)}^{3}}=1.26\] |
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