Search any question & find its solution
Question:
Answered & Verified by Expert
If the lattice parameter for a crystalline structure is $3.6 Å$, then the atomic radius in fcc crystal is
Options:
Solution:
1656 Upvotes
Verified Answer
The correct answer is:
$1.27 Å$
Atomic radius is the distance between two nearest neighbours. For fcc crystal Atomic radius $=\frac{1}{2 \sqrt{2}} \times$ lattice parameter
or $r=\frac{a}{2 \sqrt{2}}$
or $r=\frac{3.6}{2 \sqrt{2}} Å=1.27 Å$
or $r=\frac{a}{2 \sqrt{2}}$
or $r=\frac{3.6}{2 \sqrt{2}} Å=1.27 Å$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.