A) 0.0887
B) 1.0887
C) 1.546
D) 1.544
Correct Answer: B
Solution :
[b] In bcc structure, |
body diagonal \[=4r(Fe)=\sqrt{3}a\] |
\[\therefore \] \[a=\frac{4}{\sqrt{3}}r(Fe)\] |
\[Z=2\] atoms per unit cell |
\[\therefore \] |
\[d(\alpha -\text{form})=\frac{Zm}{{{N}_{0}}{{a}^{3}}}=\frac{56\times 2}{(6.02\times {{10}^{23}}){{\left( \frac{4}{\sqrt{3}} \right)}^{3}}}\] |
In fee structure, |
face diagonal \[=4r(Fe)=\sqrt{2}{{a}^{1}}\] |
\[\therefore \] \[{{a}^{1}}=2\sqrt{2}R(Fe)\] |
\[Z=4\] atoms per unit cell |
\[\therefore \] \[d(\gamma -form)=\frac{mZ}{{{N}_{0}}{{({{a}^{b}})}^{3}}}\] |
\[=\frac{56\times 4}{(6.02\times {{10}^{23}}){{(2\sqrt{2}r)}^{3}}}\] |
\[\Rightarrow \] Density ratio of \[\gamma -\]form to \[\alpha -\]form |
\[=\frac{2\times {{\left( \frac{4}{\sqrt{3}}r \right)}^{3}}}{{{(2\sqrt{2}r)}^{3}}}\] |
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