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

V_{c}: Volume of unit cell

N_{A}: Avogadro’s number (6.023 × 10^{23} atoms/mol)

Copper has FCC unit cell, thus:

n = 4

V_{c} = a^{3}

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

\({V_c} = {\left( {2\sqrt 2 × 1.28 × {{10}^{ - 10}}} \right)^3} = 4.74 × {10^{ - 29}}\) m^{3}

\(\rho = \frac{{nA}}{{{N_A}{V_c}}} = \frac{{4 × 63.5}}{{6.023 × {{10}^{23}} × 4.74 × {{10}^{ - 29}} × 1000}} = 8.89 × {10^3}\frac{{kg}}{{{m^3}}}\;\)