Question

The edge length of the unit cell of a bcc lattice of a metal is $350\text{pm}$.
What is the radius of the atom?
($\sqrt{3}=1.7$)
[1 Mark]

• A. $148.75\stackrel{o}{A}$
• B. $110.24\stackrel{o}{A}$
• C. $94.38A\stackrel{o}{A}$
• D. $208.17\stackrel{o}{A}$

Answer 1

The correct option is A $148.75\stackrel{o}{A}$

In a bcc structure, the atoms toch each other along the body diagonal.

let the radius of the atom $=r$
Given, edge length of unit cell = $a$
From the diagram,
$A{B}^2=B{C}^2+A{C}^2$
$B{C}^2=B{D}^2+C{D}^2=2a^2$
$\therefore A{B}^2=2a^2+a^2=3a^2$
$\therefore AB=\sqrt{3}a$
But $AB=4r$
$\therefore 4r=\sqrt{3}a\Rightarrow r=\sqrt{3}a/4$
$\Rightarrow r=\sqrt{3}\times \frac{350}{4}\text{pm}$
$=\sqrt{3}\times \frac{350}{4}\times {10}^{-12}\text{m}$
$=\sqrt{3}\times \frac{350}{4}\times {10}^{-10}\text{m}=148.75\times {10}^{-10}\text{cm}$