Question

An atom crystallizes in fcc crystal lattice and hence a density of $10gc{m^{ - 3}}$ with unit cell edge length of $100pm$ . Calculate number of atoms present in $1g$ of crystal. $\left( {4 \times {{10}^{23}}atoms} \right)$ .

Answer 1

Hint: This question gives the knowledge about the fcc crystal lattice. Fcc crystal lattice is called as face-centered cubic crystal lattice. It contains six atoms at each face of the cube and eight atoms at the corners of the cube.

Formula used:
The formula used to determine the number of atoms is as follows:
$\smallint = \dfrac{{zM}}{{{a^3}{N_A}}}$
Where $\smallint $ is density, $z$ is an effective number of atoms in a unit cell, $M$ is molar mass, ${N_A}$ is Avogadro’s number and $a$ is the edge of the cube.

Complete step by step answer:
Fcc crystal lattice contains six atoms at each face of the cube and eight atoms at the corners of the cube. The effective number of atoms in a fcc crystal unit cell is always $4$. The coordination number of FCC lattice is $12$ .
The formula to determine the number of atoms is considered as follows:
$ \Rightarrow \smallint = \dfrac{{zM}}{{{a^3}{N_A}}}$
Rearrange the formula to determine molar mass as follows:
$ \Rightarrow M = \smallint \dfrac{{{a^3}{N_A}}}{z}$
Consider this as equation $1$.
Now, to determine the number of atoms as follows:
$ \Rightarrow n = \dfrac{m}{M} \times {N_A}$
Consider this as equation $2$.
Where $m$ is given mass, $n$ is the number of atoms and ${N_A}$ is Avogadro’s number.
Substitute the value of $M$ in equation $2$ as follows:
$ \Rightarrow n = \dfrac{m}{{\smallint \dfrac{{{a^3}{N_A}}}{z}}} \times {N_A}$
On simplifying, we get
$ \Rightarrow n = \dfrac{{m \times z}}{{\smallint \times {a^3}}}$
Substitute $m$ as $1$ , $z$ as $4$ , $\smallint $ as $10$ and $a$ as ${10^{ - 8}}$ in the above formula to determine the no. of atoms.
$ \Rightarrow n = \dfrac{{1 \times 4}}{{10 \times {{\left( {{{10}^{ - 8}}} \right)}^3}}}$
On simplifying, we get
$ \Rightarrow n = \dfrac{4}{{10 \times {{10}^{ - 24}}}}$
On further simplifying, we get
$ \Rightarrow n = \dfrac{4}{{{{10}^{ - 23}}}}$

Therefore, the number of atoms are $4 \times {10^{23}}$ atoms.

Note: Always remember that the effective number of atoms in a face centered cubic crystal unit cell is always $4$ . FCC crystal lattice contains six atoms at each face of the cube and eight atoms at the corners of the cube.