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Question: The density of copper metal is 8.95 gm\(c{m^{ - 3}}\), if the radius of copper atom is 127.8 pm is t...

The density of copper metal is 8.95 gmcm−3c{m^{ - 3}}, if the radius of copper atom is 127.8 pm is the copper unit cell a simple cubic, a body centered cubic of face centered cubic structure? (At mass of Cu = 63.54 g mol−1mo{l^{ - 1}}andNa=6.02×1023mol−1{N_a} = 6.02 \times {10^{23}}mo{l^{ - 1}})

Explanation

Solution

As we all know that density of a cubic crystal is equal to the mass by unit cell by volume of the unit cell. Based on this equation we will solve this question. Also, the mass of a unit cell is equal to the product of the number of atoms and the mass of each atom present in a unit cell.

Complete step by step answer:
As we know that the density of a crystal of ionic compounds
ρ=Z×Ma3×Na\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_a}}}
Where, Z is the number of atoms per unit cell
Na{N_a}is the Avogadro number which is equal to Na=6.02×1023mol−1{N_a} = 6.02 \times {10^{23}}mo{l^{ - 1}}
M is the mass of one atom
For a simple cubic unit cell, each corner of the unit cell is defined by a lattice point in which an atom, ion, or a molecule can be found in the crystal lattice.
In case of simple cubic unit cell,
A = 2r, Z =1, Mass of Cu = 63.54 g mol−1mo{l^{ - 1}}
Now, we will substitute the values in the formula of density.
Therefore,ρ=1×63.54(2×127.8×10−10)3×6.02×1023\rho = \dfrac{{1 \times 63.54}}{{{{(2 \times 127.8 \times {{10}^{ - 10}})}^3} \times 6.02 \times {{10}^{23}}}} = 6.31g/cm36.31g/c{m^3}
In a BCC unit cell there are atoms at each corner of the cube and an atom at the centre of the structure.
Thus, Z = 2, a = 43r\dfrac{4}{{\sqrt 3 }}r
Therefore, ρ=2×63.54(43×127.8×10−10)3×6.02×1023=8.2g/cm3\rho = \dfrac{{2 \times 63.54}}{{{{(\dfrac{4}{{\sqrt 3 }} \times 127.8 \times {{10}^{ - 10}})}^3} \times 6.02 \times {{10}^{23}}}} = 8.2g/c{m^3}
In a FCC unit cell there are atoms at all the corners of the crystal lattice and at the centre of all the faces of the cube.
Thus, Z = 4
a=42r ρ=4×63.54(42×127.8×10−10)3×6.02×1023=8.92g/cm3  a = \dfrac{4}{{\sqrt 2 }}r \\\ \rho = \dfrac{{4 \times 63.54}}{{{{(\dfrac{4}{{\sqrt 2 }} \times 127.8 \times {{10}^{ - 10}})}^3} \times 6.02 \times {{10}^{23}}}} = 8.92g/c{m^3} \\\
As we know that the density of copper metal is 8.95 g/cm3c{m^3}which is nearest to FCC crystal lattice. Therefore, the copper unit cell is face centered.

Note:
From this question it is evident to us that solid copper metal Cu can be described as the arrangement of copper atoms in a face centered cubic (FCC) configuration. Thus, a copper atom is found at each corner and in the center of each face of a cube.