A) 1.67
B) 1.26
C) 6.23
D) 1.04
Correct Answer: B
Solution :
[b] FCC has 4 atoms in a unit cell |
BCC has 2 atoms in a unit cell |
\[d=\frac{z\times M}{{{N}_{0}}\times {{a}^{3}}}\] |
\[\frac{{{d}_{FCC}}}{{{d}_{BCC}}}=\frac{4}{2}\frac{{{(3.0)}^{3}}}{{{(3.5)}^{3}}}=1.26\] |
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