Question

Molybdenum has an atomic mass $96gmo{l^{ - 1}}$ with density $10.3gc{m^{ - 1}}$. The edge length of the unit cell is $314pm$. Determine its lattice structure.

Answer 1

Hint: We already know that As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell and by knowing the dimensions of a unit cell, we can calculate the volume, density, structure of lattice and other parameters.

Complete Step by step answer: As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.
So, by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, the mass and number of atoms in the unit cell and the density of a unit cell as well.
Let the edge length of a cubic crystal of an element or compound be $a\;cm$.
And we are given with edge length as $314pm$or in terms of centimetre we can write it as $3.14 \times {10^{ - 10}}cm$.
Density of the unit cell which is $10.3gc{m^{ - 1}}$.
Atomic mass of a unit cell of Molybdenum is given as $96gmo{l^{ - 1}}$.
And we know that Avogadro’s number is equal to $6.022 \times {10^{23}}$
Now using the formula for calculation of unit cell dimension:
$d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$, where ‘d’ is the density of unit cell, ‘Z’ is the number of atoms in one unit cell, ‘M’ is the atomic mass and ‘a’ is the edge length of unit cell.
After putting all the given values in the formula we get:
$Z = \dfrac{{d \times {N_A} \times {a^3}}}{M}$
$\Rightarrow Z = \dfrac{{10.3 \times {{(3.14)}^3} \times {{10}^{ - 30}} \times 6.022 \times {{10}^{23}}}}{{96}}$
$\Rightarrow Z = 2$.
We know that a body centred cubic structure of a unit cell possess $8$ lattice point at $8$ corners and each one of these is shared by $8$ cells and another lattice point is completely inside the unit cell or present at the centre of the body thus the number of points or atoms per unit cell of bcc is $8 \times \dfrac{1}{8} + 1 = 2$ therefore the value of $Z = 2$. And we are also aware that for body centred cubic unit cells the number of atoms per unit cell is $2$.

Therefore, the structure of crystal lattice is BCC.

Note: Similarly, the structure of crystal lattice of face centred cubic unit cell can be calculated which contains $8$ points at corners and $6$lattice points at face centres of each unit cell. Thus the total number of atoms per unit cell in fcc comes out to be $4$. So, remember for bcc $Z = 2$ and for fcc $Z = 4$.