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JEE Main & Advanced Chemistry The Solid State / ठोस प्रावस्था Question Bank Topic Test - Solid State

  • question_answer
    Iron crystallizes in several-modifications. At about \[910{}^\circ C\] 'bcc' form (called \[\alpha -\]form) undergoes transitions, to \[\gamma -\]form with 'fcc' lattice. Assuming that the distance between the nearest neighbors is the same in the two forms at the transition temperature, calculate the ratio of the density of \[\gamma -\]ron to that of \[\alpha -\]iron at the transition temperature.

    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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