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Question
Predict the type of cubic lattice of a solid element having edge length of 400 pm and density is 6.25 g/ml.
(Atomic mass of element = 60)
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Solution
Given:
Edge length a = 400 pm = 400 × 10−12 m = 400 × 10−12 cm = 4 × 10−8 cm
a3 = (4 × 10−8 cm)3
= 64 × 10−24 cm3
ρ = 6.25 g/mL = 6.25 g.cm−3
M = 60
To find:
n = ?
Formula:
ρ = `("M"xx"n")/("a"^3xx"N"_"A")`
6.25 = `(60xx"n")/(64xx10^-24xx6.022xx10^23)`
6.25 = `(60xx"n")/(6.4xx6.022)`
n = `(6.25xx6.4xx6.022)/60`
n = `(6.25xx6.4)/10`
n = 4
Hence, the type of cubic lattice is FCC, as 4 particles are present per unit cell.
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