Answer:
density = 0.071984 g/cm³
Explanation:
For a simple cubic lattice structure;
There are 8 atoms in the unit cell, with each edge of the cell contributing 1/8th atom to one unit cell.
Thus, in a unit cell, the required number of atoms = 8/8 = 1
The mass of the unit cell can be calculated by using the formula:
[tex]Mass \ of \ the \ unit \ cell = \dfrac{no \ of atoms \ \times Atomic \ weight }{Avogadro \ No}[/tex]
[tex]Mass \ of \ the \ unit \ cell = \dfrac{1 \ \times 5.42 }{6.023\times 10^{23}}[/tex]
[tex]Mass \ of \ the \ unit \ cell = 8.998 \times 10^{24 } \ g[/tex]
Similarly, given the atomic radius = 2.5 angstrom = [tex]2.5 \times 10^{-10} \ meter[/tex]
Thus, volume V = a³
The edge length for the simple cubic radius is:
a = 2 × r
a = 2 × [tex]2.5 \times 10^{-10}[/tex]
a = [tex]5 \times 10^{-10}[/tex] m
a = [tex]5 \times 10^{-8} \ cm[/tex]
Now;
volume V = a³
volume V = [tex](5 \times 10^8)^3[/tex]
volume V = [tex]1.25 \times 10^{-22}[/tex] cm³
Finally; the density = mass/volume
density = [tex]\dfrac{8.998 \times 10^{-24}}{1.25 \times 10^{-22}}[/tex]
density = 0.071984 g/cm³
