KVPY Sample Paper KVPY Stream-SX Model Paper-26

Lithium forms a \[\operatorname{b}.c.c.\] lattice. If the lattice constant is \[3.50\times {{10}^{-10}}m\] and the experimental density is \[5.30\times {{10}^{2}} kg {{m}^{-3}}\], the percentage occupancy of Li metal is (Li = 7).

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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