If the lattice parameter for a crystalline structure is $3.6 \%$, then the atomic radius in fcc crystal is

1. $1.27 \&$
2. $1.81 Å$
3. $2.10 Å$
4. $2.92 \&$

Solution

$2 R=\frac{a}{\sqrt{2}}$ $\therefore R=\frac{a}{2 \sqrt{2}}$

Asked in: NEET 2008 (Mains)

Practice more Solid State questions on Aicharya