Question

Vanadium atom has a radius of 121 pm and crystallizes with a BCC unit cell. Determine the number of unit cells present in 1.9 cm3 solid sample of Vanadium. Express answer in scientific notation

Answer 1

Radius=121 pm $= 121 \times 10^{-8}$

BBC No of atoms in unit cell = 2

in BBC

$a=\frac{4}{\sqrt{3}} \times$ (Relation between edge length and atom radius)

$a=\frac{4}{\sqrt{3}}\times 1.21 \times 10^{-8} \\ a=2.794 \times 10^{-8}\;cm$

Volume of one unit cell $= a^3 =(2.794 \times 10^{-8})^3 = 2.181 \times 10^{-23} \;cm^3$

No of unit cells present in $1.9 \;cm^3$

$=\frac{1.9}{Volume \; of \; one \; unit \;cell} \\ = \frac{1.9}{2.181 \times 10^{-23}} \\ = 8.711 \times 10^{22}$

No of unit all present in $1.9\; cm^3 = 8.711 \times 10^{22}$ unit cells.