The radius of a barium atom is  r = 2.19  [tex]\times[/tex] 10^-8
Explanation:
The atomic weight of barium is 137.34.
The body-centered cubic structure has two atoms per unit cell.
Therefore, the mass of Ba in a unit cell is calculated as,
             [tex]m =[/tex]  [tex]\frac{2 \times 137.34}{6.023 \times 10^2^3}[/tex]
            [tex]m = 4.56 \times 10^{-22} g[/tex]
       volume = mass / density
              [tex]= \frac{4.56 \times 10^{-22} }{3.50}[/tex]Â
        volume  = [tex]1.30 \times 10^{-22} cm^3[/tex]
The edge length of a cube then is the cube root of[tex]1.30 \times 10^{-22} cm^3[/tex] Â or
           [tex]a = 5.06 \times 10^{-8}[/tex]
The body diagonal is 4 x the radius and equals a  1.732,
Therefore    r = (a [tex]\times[/tex] 1.732) / 4
              = (5.06 [tex]\times[/tex] 10^-8
             [tex]r = 2.19 \times10^{-8}[/tex]
The radius of a barium atom is  [tex]r = 2.19 \times10^{-8}[/tex]cm.
