Vacant space in body centered cubic (BCC) lat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a $BCC$ unit cell,the number of atoms per unit cell is $2$. The packing efficiency is calculated as: $	ext{Packing Efficiency} = \frac{Z 	im

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) For a $BCC$ unit cell,the number of atoms per unit cell is $2$. The packing efficiency is calculated as: $	ext{Packing Efficiency} = \frac{Z 	imes \frac{4}{3} \pi r^3}{a^3} 	imes 100$. For $BCC$,the relation between edge length $a$ and radius $r$ is $a = \frac{4r}{\sqrt{3}}$. Substituting this,$	ext{Packing Efficiency} = \frac{2 	imes \frac{4}{3} \pi r^3}{(\frac{4}{\sqrt{3}} r)^3} 	imes 100 = 68 \%$. Therefore,the vacant space (void space) is $100 \% - 68 \% = 32 \%$.

Vacant space in body centered cubic $(BCC)$ lattice unit cell is about (in $\%$)

Similar Questions

What is the number of spheres in contact with a tetrahedral void?

An ionic compound is formed by elements $X$ and $Y$. Element $Y$ forms a face-centered cubic $(FCC)$ lattice, and element $X$ occupies one-third of the tetrahedral voids. What is the formula of the compound?

The packing efficiency of the two-dimensional square unit cell shown below is (in $\%$)

In a closed packed array of $N$ spheres,the number of tetrahedral holes is

In a metal oxide,$O^{2-}$ ions are arranged in $hcp$ lattice while metal ions occupy $2/3$ of the octahedral voids. The simplest formula of the oxide is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module