Question

Formula of unit cell density is
A. $\dfrac{{ZM}}{{{a^3}{N_0}}}$
B. $\dfrac{{Z{N_0}}}{{{a^3}M}}$
C. $\dfrac{{{N_0}{a^3}}}{{MZ}}$
D. $\dfrac{Z}{{M{N_0}}}$

Answer 1

The density of a unit cell is the mass of its atoms per unit volume of it. The volume of a unit cell can be calculated with the help of unit cell dimensions. With the help of volume and the mass of the unit cell, the density of any unit cell can be calculated very easily.

Complete step by step answer: Let the edge length of a unit cell be ‘a’, the density of the unit cell be ‘d’, the molar mass of the substance be ‘M’, the no of atoms per unit cell be ‘Z’, the mass of one atom be ‘m’ and ${N_0}$be the Avogadro’s number.
The volume of given unit cell, $V = {a^3}$
The mass of given unit cell $= No{\text{ }}of{\text{ }}atoms{\text{ }}in{\text{ }}unit{\text{ }}cell \times mass{\text{ }}of{\text{ }}one{\text{ }}atom = Z \times m$
The mass of one atom can also be given as:
$m = \dfrac{M}{{{N_0}}}$ …… (i)
The density of unit cell can be given as:
$Density = \dfrac{{Mass{\text{ }}of{\text{ }}unit{\text{ }}cell{\text{ }}}}{{Volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}$
$\Rightarrow d = \dfrac{{Z \times m}}{{{a^3}}}$
Putting the value of (i) in it, we get:
$\Rightarrow d = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$
So, the correct answer for the density of unit cell is option (A) $\dfrac{{ZM}}{{{a^3}{N_0}}}$

Note:
-The density of any unit cell is the same as the density of the substance of the unit cell.
-The density of the substance can also be determined by other different methods.
-Out of the parameters involved in the above formula (Z,d,M and a), if any three are known to us, then we can easily calculate the fourth unknown parameter by using the formula.
-${N_0}$ is Avogadro's number. So do not confuse it with the number of atoms while solving problems based on density and volume of unit cells.