Question

An element having bcc geometry has atomic mass 50 u. Calculate the density of the unit cell if its edge length is 290 pm.

Answer 1

Hint: The unit cell containing one constituent particle at the centre of the cell, besides those at the vertices of the unit cell is called the body-centred unit cell. We shall substitute the values in the formula given.

Formula: Density = $\dfrac{{{\text{z M}}}}{{{{\text{a}}^{\text{3}}}{\text{ }}{{\text{N}}_{\text{A}}}}}$
where z is the number of the constituents in the unit cell, M is the molecular weight, and ${{\text{N}}_{\text{A}}}$ is the Avogadro’s constant.

Complete stepwise Solution:
The density of a unit cell can be defined as the mass of the unit cell divided by the volume of the unit cell. Now, the mass of the unit cell is equal to the number of moles multiplied by the molecular weight of it. Therefore, mathematically the mass is equal to,
Mass (m) = \[\dfrac{{{\text{number of moles}}}}{{{\text{molecular weight of the substance}}}}\]=$\dfrac{{\text{n}}}{{\text{M}}}$
The volume of the unit cell considering them to be perfectly cubic in nature,
Volume (V) = ${{\text{a}}^{\text{3}}}$, where ‘a’ is the length of the edge of the cubic cell.
According to the question, atomic mass = 50 u, a = 290 pm, ${{\text{N}}_{\text{A}}}$= $6.022 \times {10^{23}}$ atoms of the element.
Hence the density = \[\dfrac{{{\text{2}} \times {\text{50}}}}{{{{\left( {290 \times {{10}^{{\text{ - 10}}}}} \right)}^{\text{3}}}{\text{ }}\left( {6.023 \times {{10}^{{\text{23}}}}} \right)}}\] = \[\dfrac{{10{\text{0}}}}{{{{\left( {290} \right)}^{\text{3}}}\left( {6.023} \right){{10}^{{\text{ - 7}}}}}}\]

So, the final answer is $6.81{\text{g/ c}}{{\text{m}}^3}$.

Notes: The density of the unit cell is the same as the density of the substance. The density of the substance can be determined from other methods as well. The knowledge of any of the five parameters in the equation for density can give the knowledge of the unknown quantity.