Looking for classes? Ksquare Career Institute, Bengaluru →
If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius in fcc crystal is -
A
2.92 Å
B
1.27 Å
C
1.81 Å
D
2.10 Å
Detailed Solution
In a face centred cubic (fcc) crystal the atoms touch each other along the face diagonal.
Face diagonal $= \sqrt{2}\,a = 4r$, where $a$ is the lattice parameter.
$r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}}$
$r = \dfrac{\sqrt{2} \times 3.6}{4}$
$r = 0.9 \times 1.41$
$r = 1.27$ Å
Face diagonal $= \sqrt{2}\,a = 4r$, where $a$ is the lattice parameter.
$r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}}$
$r = \dfrac{\sqrt{2} \times 3.6}{4}$
$r = 0.9 \times 1.41$
$r = 1.27$ Å