Question

Metal element 'M' of radius 50nm is crystallized in FCC format and made into cubical crystals such that face of unit cells are aligned with face of cubical crystal.If total number of metal atoms of 'M' at face of cubical crystal is 6 $\times$ ${10}^{30}$, then area of one face of cubical crystal is A $\times$ ${10}^{16}$ $m^2$, the value of 'A' is:

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Answer 1

Consider the unit cell face

So, atoms of Face will be = 4 $\times$ $\frac{1}{8}$ + $\frac{1}{2}$ $\times$ 1 = 1 atoms no. of atoms at one face of crystal = $\frac{1}{6}$ $\times$ ${10}^{30}$

= ${10}^{30}$ atoms. So, no. of units cells of edge of crystal = $\sqrt{{10}^{30}}$ = ${10}^{15}$

Now edge length of one unit cell = $\frac{4}{\sqrt{2}}$ $\times$ 50 $\times$ ${10}^{15}$nm

So, area of face of crystal = ${(\frac{4}{\sqrt{2}}\times 50\times {10}^{15})}^2$ n$m^2$

= $\frac{16}{2}$ $\times$ 25 $\times$ ${10}^2$ $\times$ ${10}^{30}$ = 2 $\times$ ${10}^{34}$ n$m^2$

= 2 $\times$ ${10}^{34}$ $\times$ ${10}^{-18}$$m^2$ = $2\times {10}^{16}$$m^2$.