Question

Sodium metal crystallizes as a body-centered cubic lattice with the cell edge $4.29\,{A^ \circ }$ . What is the radius of the sodium atom?
A.$1.857 \times {10^{ - 8}}cm$
B.$2.371 \times {10^{ - 7}}cm$
C.$3.817 \times {10^{ - 8}}cm$
D.$9.312 \times {10^{ - 7}}cm$

Answer 1

We know that a crystal lattice is formed when a small portion of it which has a particular type of arrangement of atoms rotates in three dimensions. Unit cells have eight atoms at the eight corner like as we have eight corners in a room. This is a SCC arrangement for a BCC arrangement; one atom is also present at the body center of the unit cell along with the atoms of eight corners.

Complete step-by-step answer: A unit cell can be of three types mainly, SCC BCC and FCC. In the case of BCC the atoms are at eight corners and one at the body centered on the unit cell. In case of FCC, the eight atoms are at the corner and one each at face centered.
As given in the above example, there is sodium which crystallizes as a body centered cubic lattice with edge length as $4.29\,{A^ \circ }$ we have to see the relation between the edge length and radius of atom. If r is the radius of the sodium atom and if a is the edge length then we can write the relation as,
$r = \dfrac{{\sqrt 3 a}}{4}$
If we put the value of edge length then on solving we get the value of radius of atom of sodium.
$r = \dfrac{{\sqrt 3 \times 4.29 \times {{10}^{ - 8}}cm}}{4}$
$\Rightarrow r = \dfrac{{7.4304 \times {{10}^{ - 8}}cm}}{4}$
$\Rightarrow r = 1.857 \times {10^{ - 8}}cm$
Hence, the radius of sodium atom which crystallize as in body centred crystal is $1.857 \times {10^{ - 8}}cm$

Note: While solving the questions of lattice type note that a small unit cell will give a whole lattice crystal. In case of FCC face centered lattice the unit cell will have eight atoms at eight corners and one at each face in the unit cell. The relation between the edge length and radius of an atom in case of FCC is given as $a = 2\sqrt 2 r$ .