question_answer

If a be the edge length of the unit cell and r be the radius of an atom then for fee arrangement, the correct relation is

A)  \[4a=\sqrt{3}\,r\]                           

B)  \[4r=\sqrt{3}\,a\]                           

C)  \[4r=\sqrt{2}\,a\]                           

D)  \[4r=\frac{a}{\sqrt{2}}\]

Correct Answer: C

Solution :

Given, the edge length of the unit cell = a and, radius of the sphere = r Consider \[\Delta \,ABC,\] \[A{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\]          \[={{a}^{2}}+{{a}^{2}}\]          \[=2{{a}^{2}}\] \[\therefore \]  \[AC=\sqrt{2}\cdot a\] But, from the figure, AC = 4r \[\therefore \]  \[4r=\sqrt{2}\cdot a\]