Question

Aluminium has FCC structure. Its density is 2700 Kg/m3. Calculate its unit cell dimension. Given: Atomic Weight of Aluminium =26.98, Avogadro's No. = 6.023X 10^26 atoms/ k mol.
​

Answer 1

Explanation:

For FCC, number of unit cells= 4

Density= 2.7g/cm

3

d=

a

3

×N

A

Z×M

⇒2.7=

a

3

×6.023×10

23

4×27

⇒a

3

=

2.7×6.023×10

23

4×27

⇒a=4.05×10

−8

cm

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Answer 2

Answer:

The dimension of the unit cell is 4.04×10⁻¹⁰m.

Explanation:

First lets see the relation between unit cell dimension and density,

$d=\frac{Z\times M}{a^3\times N_A}$        (1)

Where,

d=density of the unit cell

Z=number of the unit cell

M=atomic weight of the element

a³=volume of the unit cell

NA=avogadro number=6.023×10²⁶   (given)

From the question we have,

d=2700kg/m³

Z=4              (∴ For FCC the number of the unit cell is 4)

M=26.98     (given)

By substituting the value of d, Z, M, and NA in equation (1) we get;

$2700=\frac{4\times 26.98}{a^3\times 6.023\times 10^2^6}$

$a^3=\frac{4\times 26.98}{2700\times 6.023\times 10^2^6}=\frac{107.92}{162.621\times 10^2^8}=0.663\times 10^-^2^8$

$a=\sqrt[3]{0.663\times 10^-^2^8}=4.04\times 10^-^1^0m$

Hence, the dimension of the unit cell is 4.04×10⁻¹⁰m.