JEE Main & Advanced Sample Paper JEE Main - Mock Test - 35

  • question_answer
    Element X crystallizes in a 12 co-ordiantion fee lattice. On applying high temperature it changes to 8 co-ordination bcc lattice. Find the ratio of the density of the crystal lattice before and after applying high temperature -

    A) \[1:1\]               

    B)        \[3:2\]

    C) \[\sqrt{2}:\sqrt{3}\]        

    D)        \[2{{(\sqrt{2})}^{3}}:{{(\sqrt{3})}^{3}}\]

    Correct Answer: D

    Solution :

    [d] Co. No. is 12 means it is f.c.c. lattice Co. No. is 8 means it is b.c.c. lattice f.e.c.  \[{{\rho }_{1}}=\frac{4\times at.mass}{{{N}_{a}}\times a_{1}^{3}}\] b.c.c. \[{{\rho }_{2}}=\frac{2\times at.mass}{{{N}_{a}}\times a_{2}^{3}}\] \[\frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{4}{2}\times \frac{a_{2}^{3}}{a_{1}^{3}}=\frac{2\times {{\left( \frac{4r}{\sqrt{3}} \right)}^{3}}}{{{(2\sqrt{2}r)}^{3}}}\]             \[=\frac{2\times 4\times 4\times 4}{2\times 2\times 2\times 2\times \sqrt{2}(3\sqrt{3})}\]             \[=\frac{4\sqrt{2}}{3\sqrt{3}}=\frac{2{{(\sqrt{2})}^{3}}}{{{(\sqrt{3})}^{3}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner