A) \[\left( \frac{2-\sqrt{3}}{\sqrt{3}} \right)R\]
B) \[\left( \sqrt{\frac{5}{3}}-1 \right)R\]
C) \[\left( \sqrt{\frac{4}{3}}-1 \right)R\]
D) \[(\sqrt{2}-1)R\]
Correct Answer: B
Solution :
| In a body centred lattice this position where the atom of maximum size can be fitted should not be lying on the edge centre as the distance between surface of two atoms which are at corners of a cube is very small so will be off the edge so will be at distance of \[\left( \frac{a}{4} \right)\] from edge center on\[\bot \] bisector of edge of cube. |
| Hence we get |
| \[R+r=\sqrt{\frac{{{a}^{2}}}{u}+\frac{{{a}^{2}}}{16}}=\sqrt{\frac{5{{a}^{2}}}{16}}\] |
| \[=\sqrt{\frac{5}{16}\times \frac{16{{R}^{2}}}{5}}\] |
| So \[r=\left( \sqrt{\frac{5}{3}}-1 \right)R\] |
| \[[4R=\sqrt{3}\,a]\] |
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