Question

An elemental crystal has a density of $8570{\text{ kg }}{{\text{m}}^{ - 3}}$. The packing efficiency is 0.68. If the closest distance between neighbouring atoms is $2.86{\text{ {\AA}}}$, the mass of one atom is:
$\left( {1{\text{ amu}} = 1.66 \times {{10}^{ - 27{\text{ }}}}{\text{kg}}} \right)$
A) $186.0{\text{ amu}}$
B) $93.0{\text{ amu}}$
C) $46.5{\text{ amu}}$
D) $43.2{\text{ amu}}$

Answer 1

Hint:We are given that the packing efficiency is 0.68. Thus, we can say that the given element crystallizes in a body centered cubic (BCC) lattice. The ratio of the mass of the unit cell and the volume of the unit cell is known as the density of the unit cell.

Formula Used:
$4r = a\sqrt 3 $
$\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$

Complete step by step solution:
We are given that the closest distance between neighbouring atoms is $2.86{\text{ {\AA}}}$. This distance is known as the diameter which is equal to $2r$.
Thus, $2r = 2.86{\text{ {\AA}}}$
We are given that the packing efficiency is 0.68. Thus, we can say that the given element crystallizes in a body centered cubic (BCC) lattice.
For BCC lattice,
$4r = a\sqrt 3 $
Where $r$ is the radius,
$a$ is the edge length.
Thus,
$a = \dfrac{{4r}}{{\sqrt 3 }}$
$a = \dfrac{{2 \times 2r}}{{\sqrt 3 }}$
$a = \dfrac{{2 \times 2.86{\text{ {\AA}}}}}{{\sqrt 3 }}$
$a = 3.30{\text{ {\AA}}}$
Thus, the edge length of the unit cell is $3.30{\text{ {\AA}}}$. But $1{\text{ {\AA}}} = 1 \times {10^{ - 10}}{\text{ m}}$
Thus, the edge length of the unit cell is ${\text{3}}{\text{.30}} \times {10^{ - 10}}{\text{ m}}$.
We know the equation,
$\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$
Where $\rho $ is the density of the unit cell,
$Z$ is the number of atoms in the unit cell,
$M$ is the atomic mass,
 $a$ is the edge length of the unit cell,
${N_0}$ is Avogadro's number.
Substitute $8570{\text{ kg }}{{\text{m}}^{ - 3}}$ for the density of the unit cell, ${\text{3}}{\text{.30}} \times {10^{ - 10}}{\text{ m}}$ for the edge length of the unit cell, $6.023 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the Avogadro’s number, 2 for the number of atoms in the unit cell. Thus,
$8570{\text{ kg }}{{\text{m}}^{ - 3}} = \dfrac{{2 \times \dfrac{M}{{1000}}}}{{{{\left( {{\text{3}}{\text{.30}} \times {{10}^{ - 10}}{\text{ m}}} \right)}^3} \times 6.023 \times {{10}^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}}}$
$M = \dfrac{{8570{\text{ kg }}{{\text{m}}^{ - 3}} \times 6.023 \times {{10}^{23}}{\text{ mo}}{{\text{l}}^{ - 1}} \times {{\left( {{\text{3}}{\text{.30}} \times {{10}^{ - 10}}{\text{ m}}} \right)}^3} \times 1000}}{2}$
$M = 93.0{\text{ amu}}$
Thus, the mass of one atom is $93.0{\text{ amu}}$.

Thus, the correct option is (B)$93.0{\text{ amu}}$.

Note:The number of atoms in a unit cell is directly proportional to the density and volume of the unit cell and inversely proportional to the atomic mass. The mass of the unit cell is equal to the product of the mass of each atom in the unit cell and the number of atoms in the unit cell. A body centered cubic unit cell has a total of 2 atoms in its unit cell. One atom from the eight corners of the cube and one atom placed in the centre of the cube.