Question

Metal M of radius 50nm is crystallized fcc type and made of cubic crystal such that the face of unit cells is aligned with the face of cubic crystal. If the total number of metal atoms of M at all faces of cubic crystal is $6X{{10}^{30}}$ , then the area of one face of cubic crystal is $AX{{10}^{16}}{{m}^{2}}.$find the value of A.

Answer 1

A face centered cubic (FCC) unit cell contains atoms at all the corners and at the centre of all the faces of the cube. Each atom in a face -centered cubic located at the face-centre is shared between two adjacent unit cells and only $\dfrac{1}{2}$ of each atom belongs to a unit cell.

Complete step by step answer:
Let us consider one face of unit cell,

No of atoms on one face= $4(corners)X\dfrac{1}{8}+1(facecentre)X\dfrac{1}{2}(facecentreshare)=\dfrac{1}{2}+\dfrac{1}{2}=1$ per face
No of metal atoms of M at all metal faces of cubic crystal = $6X{{10}^{30}}$
Therefore, number of atoms on once face =$6X{{10}^{30}}X\dfrac{1}{6}={{10}^{30}} atoms$
Hence number of unit cells at one face of crystal = ${{10}^{30}}$ atoms
So the number of unit cells at the edge of the crystal = $\sqrt{{{10}^{30}}}={{10}^{15}}$
Metal M radius (r) = 50nm

Now, edge length of the unit cell = $\dfrac{4}{\sqrt{2}}Xr=\dfrac{4}{\sqrt{2}}X50nm$
Edge length of cubic crystal = $\dfrac{4}{\sqrt{2}}X50X{{10}^{15}}nm$

So, area of face crystal = $\begin{aligned} & ={{(edge\text{ }length\text{ }of\text{ }cubic\text{ }crystal)}^{2}}={{(\dfrac{4}{\sqrt{2}}X50X{{10}^{15}})}^{2}}n{{m}^{2}}=\dfrac{16}{2}X50X50X{{10}^{30}}X{{10}^{-18}}{{m}^{2}} \\\ & =2X{{10}^{16}}{{m}^{2}} \\\ \end{aligned}$
Therefore, $AX{{10}^{16}}{{m}^{2}}=2X{{10}^{16}}{{m}^{2}}$
Hence , the value of A = 2.

Note: We know that any crystal lattice is made up of very large unit cells and every lattice point as occupied by one constituent particle (atom, molecule or ion). We must consider three types of cubic unit cells and for simplicity assume that the constituent particle is an atom.