Li forms a body centre cubic lattice. If the lattice constant is 3.50 × 1 0 − 10 m and the experimental density is 5.33 × 1 0 2 kg / m 3 , calculate the percentage occupancy of Li metal. [Li = 7, N A

Question

Li forms a body centre cubic lattice. If the lattice constant is 3.50 × 1 0 − 10 m and the experimental density is 5.33 × 1 0 2 kg / m 3 , calculate the percentage occupancy of Li metal. [Li = 7, N A ​ = 6 × 1 0 23 , 5.33 × ( 3.5 ) 2 = 65.3 ]

TestHub · Accepted Answer

For a BCC lattice, the number of atoms per unit cell is $Z = 2$ .

The theoretical density is given by:

$ρ = \frac{ZM}{a ^{3} N _{A}}$

Substituting the values:

$ρ = \frac{2 \times 7}{( 3.5 \times 1 0 ^{- 10} ) ^{3} \times 6 \times 1 0 ^{23}}$

$ρ = \frac{14}{25.725 \times 1 0 ^{- 7}} \approx 5.44 \times 1 0^{2} k g / m^{3}$

The percentage occupancy is calculated as:

$P erce n t a g eocc u p an cy = \frac{E x p er im e n t a l d e n s i t y}{T h eore t i c a l d e n s i t y} \times 100$

$= \frac{5.33 \times 1 0 ^{2}}{5.44 \times 1 0 ^{2}} \times 100$

$= 98%$

Answer: $98%$

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