Question

Potassium has a $bcc$ structure with a nearest neighbour distance of $4.52\stackrel{\circ }{A}$. Its atomic weight is $39$. Its density (in $kg{m}^{-3})$ will be:

• A. $454$
• B. $804$
• C. $852$
• D. $908$

Answer 1

The correct option is D $908$

In bcc structure, the nearest neighbor of body-centered atom is the atom present at corner of the atom.

Given, $\frac{a\sqrt{3}}{2}=4.52\stackrel{o}{A}$

Density= $\frac{M\times Z}{N_A\times a^3}$

$\Rightarrow \rho =\frac{39\times 2\times {10}^{-3}kg}{6.022\times {10}^{23}\times {\left(\frac{2\times 4.52\times {10}^{-10}m}{\sqrt{3}}\right)}^3}$

As number of atom in bcc= $2$

$\rho =908kg{m}^{-3}$