Question

Gallium has an orthorhombic structure with a = 0.45258 nm, b = 0.45186 nm, and c = 0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 , and the atomic weight is 69.72 g/mol. Determine the number of atoms in each unit cell and the packing factor in the unit cell.

Answer 1

Answer:

Number of atoms will be "8 atoms/cell". The further explanation is given below.

Explanation:

The given values are:

a = 0.45258 nm

b = 0.45186 nm

c = 0.76570 nm

Atomic radius = 0.1218 nm

Density = 5.904 g/cm³

Atomic weight = 69.72 g/mol

Now,

The volume of unit cells will be:

⇒  $V=a_{0}b_{0}c_{0}$

On putting the estimated values, we get

⇒      $=0.45258\times 0.45186\times 0.76570$

⇒      $=0.1566 \ nm^3$

⇒      $=1.56\times 10^{-22} \ cm^3$

(a)...

From equation of density we get,

⇒  $Density=\frac{(x \ atoms/cell)(Atomic weight)}{(Volume)(6.02\times 10^{23} \ atoms/mol)}$

⇒  $5.904=\frac{x\times 69.72}{(1.566\times 10^{-22})(6.02\times 10^{23})}$

⇒        $x=8 \ atoms/cell$

(b)...

From equation of PF we get,

⇒  $PF=\frac{(3 \ atoms/cell)(\frac{411}{3} )(0.1218)}{0.1566}$

⇒         $= 0.387$

Answer 2

The number of atoms in each unit cell and the packing factor in the unit cell are;

a) Number of atoms in each unit cell = 8 atoms

b) Atomic packing factor = 0.3876

We are given;

a = 0.45258 nm ≈ 0.45258 × 10^(-7) cm

b = 0.45186 nm = 0.45186 × 10^(-7) cm

c = 0.76570 nm = 0.7657 × 10^(-7) cm

Atomic radius; r = 0.1218 nm = 0.1218 × 10^(-7)

density; ρ = 5.904 g/cm³

Atomic weight = 69.72 g/mol

Avogadro's number = 6.02 × 10²³ atom/mol

a) The number of atoms per cell is calculated by clearing it from the equation;

Number of atoms/cell = (ρ × cell volume × Avogadro number)/atomic weight

Where;

cell volume = a × b × c

Cell volume = 0.45258 × 10^(-7) × 0.45186 × 10^(-7) × 0.7657 × 10^(-7)

Cell volume = 1.5659 × 10¯²² cm³

Thus;

Number of atoms per cell = (5.904 × 1.5659 × 10¯²² × 6.02 × 10²³)/69.72

Number of atoms per cell = 7.9 atoms

Number of atoms per cell = 8 atoms

b Formula for the packing factor is calculated from the formula:

FE = (number of atoms per cell x atom volume)/cell volume

Formula for atom volume is;

Atom volume = 4πr³/3

Atom volume = (4π × 0.1218 × 10^(-7))³/3

Atom volume = 7.5875 × 10¯²⁴ cm³

Thus;

FE = (8 x 7.5875 x 10¯²⁴)/(1.5659¯²²)

FE = 0.3876

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