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) \[1.85=\frac{\sqrt{3}}{4}a\] \[\Rightarrow \] \[a=\frac{1.85\times 4}{\sqrt{3}}=4.3\overset{\text{o}}{\mathop{\text{A}}}\,\]You need to login to perform this action.
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