An elemental crystal has a density of 8570 $Kg/m^3$ packing fraction 0.68. If the nearest neighbor distance is 2.86 $A^0$. - Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: May 2013

Given:-

APF = 0.68

ρ = 8570 $Kg/m^3$

d = 2.86 $A^0$ = $2.86 × 10^{-10}$ m = 2 × (Radius of an atom) = 2r

i.e. r = 1.43 × 10-10 m

To Find:-

Mass of one Atom (mA)

Solution:-

Given that, APF = 0.68, the crystal structure is of BCC type.

For a BCC structure,

$$r = \frac{\sqrt{3}}{4}a$$

$$1.43 × 10^{-10} = \frac{\sqrt{3}}{4}a$$

$a = 3.302 × 10^{-10}m$

Therefore, Volume of a Single Unit Cell, $V_{Unit} = a^3 = (3.302 × 10^{-10})^3$

∴ $V_{Unit} = 36.102 × 10^{-30}m^3$

Now, Mass of one Unit Cell, $m_{Unit} = ρ × V_{Unit} = 8570 × 36.102 × 10^{-30}$

Therefore, $m_{Unit} =3.087 × 10^{-25}Kg$

Number of Atoms per Unit Cell in a BCC structure is, n = 2

Hence, Mass of a unit cell is the mass of 2 atoms.

$$m_A = \frac{Mass \ \ of \ \ unit \ \ cell}{n} = \frac{3.087 × 10^{-25}}{2}$$

$m_A = 1.543 × 10^{-25} Kg$