Question
How will derive the formula of density of a unit cell?
✓
Answer
$1.$ Using the edge length of a unit cell, we can calculate the density $(p )$ of the crystal by considering a cubic unit cell as follows.
$\text { Density of the unit cell } \rho=\frac{\text { Mass of the unit cell }}{\text { Volume of the unit cell }}$
Mass of the unit cell $=\left\{\begin{array}{c}\text { Total number of atoms belong } \\ \text { to that unit cell }\end{array}\right\} \times$ mass of one atom
Mass of one atom $=\frac{\text { Molar mass }\left( g mol ^{-1}\right)}{\text { Avagardo number }\left( mol ^{-1}\right)}$
$ =\frac{ M }{ N _{ A }}$
Substitute the value $(3)$ in $(2)$
Mass of the unit cell $= n \times \frac{M}{N_A}$
For a cubic unit cell, all the edge lengths are equal. i.e., $a=b=c$
Volume of the unit cell $= a \times a \times a = a ^3$
$\therefore$ Density of the unit cell $=\rho=\frac{n M}{a^3 N_A}$
$\text { Density of the unit cell } \rho=\frac{\text { Mass of the unit cell }}{\text { Volume of the unit cell }}$
Mass of the unit cell $=\left\{\begin{array}{c}\text { Total number of atoms belong } \\ \text { to that unit cell }\end{array}\right\} \times$ mass of one atom
Mass of one atom $=\frac{\text { Molar mass }\left( g mol ^{-1}\right)}{\text { Avagardo number }\left( mol ^{-1}\right)}$
$ =\frac{ M }{ N _{ A }}$
Substitute the value $(3)$ in $(2)$
Mass of the unit cell $= n \times \frac{M}{N_A}$
For a cubic unit cell, all the edge lengths are equal. i.e., $a=b=c$
Volume of the unit cell $= a \times a \times a = a ^3$
$\therefore$ Density of the unit cell $=\rho=\frac{n M}{a^3 N_A}$
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