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The radius of a Niobium atom Niobium has a density of 8.57 and crystallizes with the body-centered cubic unit cell?
Wiki User
∙ 13y ago
in essence bcc crystal is one where there one atom at the center of the cubic structure, with (1/8) atom at each corner.
number of atoms per cell =2
if the side of the cude is a and the radius of atom is r,
4r = (sqrt(3)*a) (using pythogarus theorem)
that is, a = (sqrt(3))/(4r) = 1.732/(4r)
a few other obvious things:
number of atoms/mole = avagadro's number = 6.022*10^23
density = mass/volume. in the case of crystal lattice with side a = crystal mass/a^3
niobium atomic weight = 93
so in 93g of niobium, you'd have 6.022*10^23atoms
so each atom weighs (93)/(6.022*10^23) g
so
density = mass of 2 atoms in the crystal cell/a^3
mass of 2 atoms = 2*Atomic Mass = 2*(93)/(6.022*10^23)
a^3 = (1.732)^3/(4r)^3
denisty=8.57g/cm^3
hence
8.57 = {2*(93)/6.022*10^23)}/{(1.732)^3/(4r)^3)}
solve for r
Wiki User
∙ 13y ago
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