Question

Using X ray diffraction, it is found that nickel \[\left( {{\text{mass }} = 59{\text{ g mo}}{{\text{l}}^{ - 1}}} \right)\] crystallizes as ccp. The edge length of the unit cell is \[3.5\mathop {\text{A}}\limits^{\text{o}} \] . if density of \[{\text{Ni}}\] is \[{\text{9}}{\text{.0 g}}/{\text{c}}{{\text{m}}^3}\] , then value of Avogadro number from the data is:
A.\[6.05 \times {10^{23}}\]
B.\[6.11 \times {10^{23}}\]
C.\[6.02 \times {10^{23}}\]
D.\[6.023 \times {10^{23}}\]

Answer 1

Hint:Using the formula of density of unit cell and substituting the known variable will get the value of Avogadro’s constant. The number of atoms in an fcc unit cell is 4. The cube of edge length will give us the volume.

Formula used: \[{\text{d}} = \dfrac{{{\text{n}} \times {\text{M}}}}{{{{\text{N}}_a} \times {\text{V}}}}\]
Here d is the density of crystal, n is the total number of atoms in a unit cell, M is the molar mass and V is the volume of the unit cell. \[{{\text{N}}_a}\] is the Avogadro constant.

Complete step by step answer:
Cubic close packing which is also known as face centred cubic packing is an arrangement of particles of solid in such a manner that one atom is present at each corner of the unit cell and also one atom at each face centre of the unit cell. In a unit cell eight corners are there and atoms are present at 8 corners. The atoms at the corner have a contribution of one eighth. Atom at the face centre has a contribution of half and there are 6 face centres in a cube. According to this the number of atoms in face centred cubic cell will be:
\[\dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 = 4\]
Molar mass of nickel is given to us. The volume can be calculated using the edge length.
\[{\text{volume }} = {{\text{a}}^3}\]
\[3.5\mathop {\text{A}}\limits^{\text{o}} = 3.5 \times {10^{ - 10}}{\text{m}}\]
Substituting the values in the formula we will get:
\[{{\text{N}}_a} = \dfrac{{{\text{n}} \times {\text{M}}}}{{{\text{d}} \times {\text{V}}}}\]
\[ \Rightarrow {{\text{N}}_a} = \dfrac{{4 \times 59}}{{9 \times {{\left( {3.5 \times {{10}^{ - 10}}{\text{m}}} \right)}^3}}}\]
Solving the above we will get:
\[{{\text{N}}_a} = 6.11 \times {10^{23}}\]

The correct option is B.

Note:
The unit cell is the building block of a crystal lattice. The crystal lattice follows the same property as a unit cell. There are various types of unit cell such as face centred unit cell, body centred unit cell, end centred unit cell and edge centred unit cell.