Question

The edge length of a solid possessing cubic unit cell is \( 2\sqrt{2}\,r \) (structure I), based on hard sphere model, which upon subjecting to a phase transition, a new cubic structure (structure II) having an edge length of \( \frac{4r}{\sqrt{3}} \) is obtained, where \( r \) is the radius of the hard sphere. Which of the following statements is true?

• A. Density of the structure II is lower than structure I
• B. Density of structure II is higher than structure I
• C. The pore volume in structure I is 1.2 times higher than that of structure II
• D. The pore volume of both the structures are equal
• E. The octahedral voids in structure I is transformed into tetrahedral voids in structure II

Hint:

Memorize key relations: \( 2\sqrt{2}r \rightarrow \text{FCC} \), \( \frac{4r}{\sqrt{3}} \rightarrow \text{BCC} \). These appear frequently in exam questions.

Answer 1

The Correct Option is A

Concept: In cubic unit cells: - For FCC: \( a = 2\sqrt{2}r \) - For BCC: \( a = \frac{4r}{\sqrt{3}} \) Packing efficiency: - FCC \( = 74\% \) - BCC \( \approx 68\% \)

Step 1: {Identify structure I.}
Given: \[ a = 2\sqrt{2}r \] This corresponds to: \[ \text{FCC structure} \]

Step 2: {Identify structure II.}
Given: \[ a = \frac{4r}{\sqrt{3}} \] This corresponds to: \[ \text{BCC structure} \]

Step 3: {Compare number of atoms per unit cell.}
\[ \text{FCC: } 4 \text{ atoms/unit cell} \] \[ \text{BCC: } 2 \text{ atoms/unit cell} \]

Step 4: {Compare packing efficiency.}
\[ \text{FCC} = 74\% \quad (\text{more efficient}) \] \[ \text{BCC} \approx 68\% \]

Step 5: {Conclusion.}
On phase transition: - Structure changes from FCC → BCC - Packing efficiency decreases - Density decreases - Volume increases

Step 6: {Final statement.}
\[ \boxed{\text{Structure I is FCC and Structure II is BCC (density decreases)}} \]