question_answer

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

A)  \[4.9\,\overset{\text{o}}{\mathop{\text{A}}}\,\]                           

B)  \[4.3\,\overset{\text{o}}{\mathop{\text{A}}}\,\]           

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

D)         \[3.4\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

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. \[\therefore \]  \[2r=3.7\] \[\Rightarrow \]               \[r=\frac{3.7}{2}=1.85\overset{\text{o}}{\mathop{\text{A}}}\,\] The spacing between unit cells in various directions are called its lattice parameter a. Hence, in bcc lattice                 \[r=\sqrt{\frac{3}{4}}a\](\[a\]is lattice parameter)                                 \[1.85=\frac{\sqrt{3}}{4}a\]             \[\Rightarrow \]   \[a=\frac{1.85\times 4}{\sqrt{3}}=4.3\overset{\text{o}}{\mathop{\text{A}}}\,\]