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Question: An element A crystalizes in a Fcc structure. \(200g\) of this element has \(4.12 \times {10^{24}}\) ...

An element A crystalizes in a Fcc structure. 200g200g of this element has 4.12×10244.12 \times {10^{24}} atoms. The density of A is 7.2 gcm−37.2\,gc{m^{ - 3}} . Calculate the edge length of the unit cell.

A

26.97×10−24cm26.97 \times {10^{ - 24}}cm

B

299.9 pm299.9\,pm

C

5×10−12 cm5 \times {10^{ - 12}}\,cm

D

2.99 cm2.99\,cm

Answer

299.9 pm

Explanation

Solution

  1. Determine atomic mass (M):
     Total atoms = 4.12×10²⁴ in 200 g
     One atom weighs = 200 g/(4.12×10²⁴) 
     Thus, atomic weight M = (200/4.12×10²⁴)×(6.022×10²³) ≈ 29.23 g/mol

  2. Relate density (ρ), atomic mass (M) & unit cell volume (a³):
     For FCC, number of atoms per cell Z = 4
     When a is in pm, unit cell volume [in cm³] = a³×10^(–30)
     Density formula:
      ρ = (Z×M) / (a³ × Nₐ × 10^(–30))
     Solve for a³:
      a³ = (Z×M×10^(30)) / (ρ×Nₐ)

  3. Substitute values:
     a³ = (4×29.23×10^(30)) / (7.2×6.022×10²³)
       = (116.92×10^(30)) / (43.3584×10²³)
       ≈ 26.97×10⁶ (pm³)

  4. Find edge length:
     a = (26.97×10⁶)^(1/3) ≈ 299.9 pm