Question

what is the contribution of an atom per unit cell if the atom is on the face of the cube ​

Answer 1

Explanation:

${ {3}^{2} }^{3}  \times  {(2 \times  {3}^{5}) }^{ - 2}  \times  {18}^{2 }  \\  \\  =  {3}^{6}  \times  \frac{1}{4 \times  {3}^{10} }  \times  {18}^{2}  \\  \\  =  \frac{ {18}^{2} }{4 \times  {3}^{4} }  \\  \\  =  \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}}  \: ^{ \cancel{6 }} \:  ^{ \cancel{2}} \:  ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} }  \\  \\  = { \huge{ \red{ \boxed{1}}}}$

Answer 2

Explanation:

, a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as 12 atom per unit cell, giving 6×12=3 Au atoms per unit cell.

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