Copper crystallizes in an FCC unit cell with cell edge of 3 .608 ×...

The number of effective atoms in a FCC unit cell, $Z = 4atoms$

$M = \dfrac{d \times {N}_{0} \times {a}^{3}}{Z}$

Where,

$M$ is atomic mass of copper

$d$ is density of copper= $8 . 92g / c{m}^{3},$

${N}_{0}$ is the Avogadro number $= 6 .022 \times {10}^{23}{mol}^{- 1}$
$a$ is the edge length $= 3.608 \times {10}^{- 8}\text{cm}$

Substituting the values in the formula,

$⇒ M = \dfrac{8 .92 g / {cm}^{3} \times 6 .022 \times {10}^{23}atom / mol \times {( 3 .608 \times {10}^{- 8}cm )}^{3}}{4atoms}$

$⇒ M = 63 .1 g / mol$

Atomic mass of copper $= 63 .1 u$