Correct Answer - Option 3 : 8.9 × 10
3 kg/m
3
Concept:
Theoretical density of a crystal structure:
\(\rho = \frac{{nA}}{{{N_A}{V_c}}}\)
N: Number of atoms associated with each unit cell
A: Atomic weight
Vc: Volume of unit cell
NA: Avogadro’s number (6.023 × 1023 atoms/mol)
Copper has FCC unit cell, thus:
n = 4
Vc = a3
In FCC:
\(a = 2\sqrt 2 \;r\)
Calculation:
Given, r = 1.28 Å = 1.28 × 10-10 m
\(a = 2\sqrt 2 × 1.6 × {10^{ - 10}}\)
Vc = a3
A = 63.5
\({V_c} = {\left( {2\sqrt 2 × 1.28 × {{10}^{ - 10}}} \right)^3} = 4.74 × {10^{ - 29}}\) m3
\(\rho = \frac{{nA}}{{{N_A}{V_c}}} = \frac{{4 × 63.5}}{{6.023 × {{10}^{23}} × 4.74 × {{10}^{ - 29}} × 1000}} = 8.89 × {10^3}\frac{{kg}}{{{m^3}}}\;\)