The unit cell of copper corresponds to a face centered cube of edge length $3.596 \circ A$ with one copper atom at each lattice point. The calculated density of copper in $k g / m^{3}$ is . [Molar mass of Cu: $63.54 g :$ , Avogadro number $= 6.022 \times 10^{23}]$
(JEE Main 2021)

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Solution

Copper crystallises in $f c c$ unit cell with edge length, $a = 3.596 \circ A$ $Density (ρ) = \frac{Z_{e f f} \times M}{N_{A} \times a^{3}}$ Here, $a is the edge length$ $Z_{e f f} is the effective number of particles in a unit cell$ $M is the molar mass N_{A} is the avogadro number$ For a fcc unit cell, 8 atoms are present at corners and 6 atoms at face centres : $Number of atoms in unit cell (Z_{e f f}) = (8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 4$ $ρ = \frac{4 \times M}{N_{A} (a)^{3}}$ $= \frac{4 \times 63.54 \times 10^{- 3}}{6.022 \times 10^{23} (3.596 \times 10^{- 10})^{3}} k g / m^{3}$ $≃ 9076.2 k g m^{- 3}$

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k g / m^{3}

^{- 3}

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