Question

A crystal has $BCC$ structure and its lattice constant is $3.6$ $\stackrel{\circ }{A}$. What is the atomic radius?

• A. $3.6$ $\stackrel{\circ }{A}$
• B. $1.8$$\stackrel{\circ }{A}$
• C. $1.27$$\stackrel{\circ }{A}$
• D. $1.56$$\stackrel{\circ }{A}$

Answer 1

Answer: (D)

$1.56$$\stackrel{\circ }{A}$

Atomic radius for $BCC$ structure,
$r=\frac{a\sqrt{3}}{4}=\frac{3.6\sqrt{3}}{4}=1.56\stackrel{\circ }{A}$