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2 years ago

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BCC lithium has a lattice parameter of 3.5089x10^-8cm and contains one vacancy per 200unit cells.Calcuate (a)the number of vacan

cies percubic centimeter; and(b)the density of Li.

1 answer:

professor190 [17]2 years ago

6 0

Answer:

a) The number of vacancies percubic centimeter is $1.1573\times 10^{20} vacancy/cm^3$ .

b) The density of Lithium is $1.076 g/cm^3$ .

Explanation:

Edge length of the lithium unit cell = a = $3.5089\times 10^{-8} cm$

Volume of the unit cell = V

$V=a^3=(3.5089\times 10^{-8} cm)^3=4.3203\times 10^{-23} cm^3$

The lithium crystal has one vacancy per 200unit cells.

$\frac{1}{200} vacancy/unit cell=0.005$ vacancy/unit cell

The number of vacancies percubic centimeter:

$\frac{0.005}{V}=\frac{0.005 vacancy}{4.3203\times 10^{-23} cm^3}$

$=1.1573\times 10^{20} vacancy/cm^3$

The number of vacancies percubic centimeter is $1.1573\times 10^{20} vacancy/cm^3$ .

b)

Atomic mass of lithium = 7 g/mol

Number of atom in a unit cell = z = 2

Volume of the unit cell = V

$\rho =\frac{z\times M}{N_A\times V}$

$=\frac{4\times 7g/mol}{6.022\times 10^{23} mol^{-1}\times 4.3203\times 10^{-23} cm^3}=1.076 g/cm^3$

The density of Lithium is $1.076 g/cm^3$ .

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