Question

Density of nickel (fcc unit cell) is 8.94 g/cc. at 20°c. what is the diameter of the atom ? (atomic weight = 59

Answer 1

Thanks for asking the question.

ANSWER::

Density of a cubic crystal = (M.z)/(N.a³)
where,
M = Molecular weight of substance = 59 (given)
z = Number of atoms per unit cell = 4 (FCC unit cell)(given)
a = Edge length of cubic cell (to find to take out diameter of atom)
N = Avogadro's Number (6.022 x 10²³)

Applying values in above formula...

8.94 = (59x4)/(6.022x10²³x a³)
53.83668 x 10²³ x a³ = 236
 a³ = 4.38 x 10⁻²³
a = (4.38 x 10⁻²³)¹/³ cc per unit cell

Now , we know that a = 2D/√2 for FCC unit cell

D = diameter of atom
(4.38 x 10⁻²³)¹/³ = 2D/√2
D = ((4.38 x 10⁻²³)¹/³ )/√2
D = 2.49 x 10⁻⁸ cm

If we convert it into picometre then ,
D = 2.49 x 10⁻⁸ x 10¹⁰ pm = 249 pm

Hope it helps!
          

Answer 2

Answer:

ρ (Density) = 8.94 g/cm³

Ni = 59

Na = 6.022 × 10²³

Formula:-

              ρ = Z × M / a³ × Na

FCC :- r = √2/4 a

= 8.94 = 4 × 59 / 6.022 × 10²³ × a³

a³ = 4 × 59 / 6 × 8.94 × 10⁻²³

a³ = 4.39 × 10⁻²³ cm³

a = √43.9 × √10⁻²⁴ cm³

= 3.5 × 10⁻⁸ cm

= 0.35 × 10⁻⁷ cm

a = 0.35 nm

Now,

      r = √2 × 0.35 / 4

Answer :- 0.125 nm