Given, Density = 10 g cm-3
mass = 1 g
Edge length of unit cell = 100 pm
Volume = \(\frac{mass}{density}\) = \(\frac{1g}{10g\,cm^{-3}}\)
= 0.1 cm3
Volume of unit cell = a3
= (100 x 10-10 cm)3
= 1 x 10-24cm3
Number of unit cell in 1 g of crystal,
= \(\frac{Total\,volume}{Volume\,of\,unit\,cell}\)
= \(\frac{0.1cm^3}{1\times10^{-24}{cm^3}}\)
The given unit cell is of FCC type. Therefore. it contains 4 atoms.
0.1 x 1024 unit cells will contain 4 x 0.1 x 1024 = 4 x 1023 atoms
