Question

Iron at room temperature has BCC structure and atomic radius of \(1.24 ~\dot A\). The lattice constant of iron will be
1.

\(3.864~\dot A\)

2. \(2.864~\dot A\)
3. \(1.864~\dot A\)
4. \(0.664~\dot A\)

Answer 1

Correct Answer - Option 2 : \(2.864~\dot A\)

Concept:

At room temperature, iron has a BCC structure.

If space lattice is BCC

\(Lattice~constant = {4 \over \sqrt{ 3}}× r\)

where r is the atomic radius of the atom

Calculation:

Given:

 Iron has a BCC-lattice with \(r=1.24 ~\dot A\)

\(Lattice~constant = {4 \over \sqrt{ 2}}× r ={4 \over \sqrt{ 2}}× 1.24 = 2.863~\dot A\)

characteristics	BCC	FCC	HCP
a to r relation	\(a = \frac{{4r}}{{\sqrt 3 }}\)	\(a = 2\sqrt 2 r\)	\(a = 2r\)
The average number of atoms	2	4	6
Co-ordination number	8	12	12
APF	0.68	0.74	0.74
Examples	Na, K, V, Mo, Ta, W	Ca, Ni, Cu, Ag, Pt, Au, Pb, Al	Be, Mg, Zn, Cd, Te