A) \[87.78%\]
B) \[99.87%\]
C) \[97.78%\]
D) \[94.12%\]
Correct Answer: C
Solution :
we know that, |
Theoretical density=\[\frac{zM}{NV}=\frac{zM}{N\left( {{a}^{3}} \right)}\] |
for a b. c. c. lattice: z=2 and given that\[a=3.50\times {{10}^{-10}}m\] |
\[\operatorname{M}=7\times {{10}^{-3}}\,\operatorname{kg}/mole\]\[{{\operatorname{d}}_{cal}}=\frac{2\times 7\times \left( {{10}^{-3}} \right)}{6.022\times {{10}^{23}}\times {{\left( 3.50\times {{10}^{-10}} \right)}^{3}}}\]\[=5.42\times {{10}^{2}}\operatorname{kg}\,{{m}^{-3}}\] |
\[\therefore \] Percentage occupancy \[=\frac{\rho \exp }{\rho \operatorname{cal}}\times 100\] |
\[=\frac{5.30\times {{10}^{2}}}{5.42\times {{10}^{2}}}\times 100=97.78%\] |
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