question_answer

Sodium has body centered packing. Distance between two nearest atoms is 3.7 \[\overset{o}{\mathop{A}}\,\]. The lattice parameter is :

A)  4.9 \[\overset{o}{\mathop{A}}\,\]

B)  4.3 \[\overset{o}{\mathop{A}}\,\]

C)  3.8 \[\overset{o}{\mathop{A}}\,\]            

D)  3.4 \[\overset{o}{\mathop{A}}\,\]

Correct Answer: B

Solution :

Key Idea: Diameter of an atom is defined as centre-to-centre distance between two atoms. The body, centered cubic structure of sodium is as shown: The distance (2r), where r is radius of an atom, can be defined as the centre-to-centre distance between two atoms packed as   tightly, together as possible. This provides a type of effective radius for the atom and is sometime called the atomic radius. \[\therefore \] \[2\,r=3.7\] \[\Rightarrow \] \[r=\frac{3.7}{2}=1.85\,\overset{o}{\mathop{A}}\,\] The spacing between unit cells in various directions are called its lattice parameter a. Hence, in bcc lattice                 \[r=\frac{\sqrt{3}}{4}\,a\] (a is lattice para meter)                 \[1.85=\frac{\sqrt{3}}{4}\,a\] \[\Rightarrow \] \[a=\frac{1.85\times 4}{\sqrt{3}}=4.3\,\overset{o}{\mathop{A}}\,\]