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Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm^-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)(a) 97.7%(b) 95.4%(c) 4.6%(d) 2.3%The question was posed to me in an interview for internship.This interesting question is from Solid State in division Solid State of Chemistry – Class 12

Answer»

Right answer is (d) 2.3%

To explain I would say: Given,

Edge length (a) = 350 pm = 3.5 x 10^-8 cm

Atomic mass (M) = 7 amu

Avogadro’s NUMBER (N0) = 6.02 x 10^23

Density (ρ) = (Z x M)/(a^3 x N0)

= (2 x 7)/((3.5 x 10^-8)^3 x 6.02 x 10^23)

= 0.542 G cm^-3

% of lithium atoms occupied = (Experimental density/Theoretical density) x 100

= 0.53/0.542 x 100

= 97.7%

% of unoccupied lattice SITES = 100 – 97.7

= 2.3%.


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