Solved: PHYSICS Question for NEET

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If the lattice parameter for a crystalline structure is $3.6\text{ \AA}$ , then the atomic radius in fcc crystal is:

$1.81\text{ \AA}$

$2.10\text{ \AA}$

$2.92\text{ \AA}$

$1.27\text{ \AA}$

Identify the Crystal Structure: The problem specifies a face-centered cubic (fcc) crystal.
Recall the Formula: For an fcc lattice, the atoms touch each other along the face diagonal. The length of the face diagonal is $\sqrt{2}a$ , where $a$ is the lattice parameter (edge length). Since there are three atoms along the face diagonal (one at each corner and one at the face center), the face diagonal equals $4r$ , where $r$ is the atomic radius. Thus, the relationship is $4r = \sqrt{2}a$ .
Calculate Atomic Radius ( $r$ ): Rearranging the formula to solve for $r$ gives $r = \frac{\sqrt{2}a}{4} = \frac{a}{2\sqrt{2}}$ . Substitute the given value $a = 3.6\text{ \AA}$ : $r = \frac{3.6}{2 \times 1.414} = \frac{3.6}{2.828} \approx 1.27\text{ \AA}$
Conclusion: The atomic radius in the fcc crystal is approximately $1.27\text{ \AA}$ .

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