Calculate the volume of fcc unit cell in cm^3 | vedclass

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

Question

(B) For an $fcc$ unit cell,the packing efficiency is $74\%$. This means the volume occupied by atoms is $0.74 	imes V_{cell}$. The void volume is the

Vedclass · Accepted Answer

$1.6 	imes 10^{-23}$

Vedclass · Answer

(B) For an $fcc$ unit cell,the packing efficiency is $74\%$. This means the volume occupied by atoms is $0.74 	imes V_{cell}$. The void volume is the remaining volume,which is $100\% - 74\% = 26\%$. Therefore,the void volume is $0.26 	imes V_{cell}$. Given that the void volume is $4.16 	imes 10^{-24} \ cm^3$,we have: $0.26 	imes V_{cell} = 4.16 	imes 10^{-24} \ cm^3$. $V_{cell} = \frac{4.16 	imes 10^{-24}}{0.26} \ cm^3$. $V_{cell} = 16 	imes 10^{-24} \ cm^3 = 1.6 	imes 10^{-23} \ cm^3$. Thus,the correct option is $B$.

Calculate the volume of $fcc$ unit cell in $cm^3$ if void volume of it is $4.16 \times 10^{-24} \ cm^3$.

Similar Questions

What is the packing efficiency of silver metal in its unit cell (in $\%$)?

The ratio of densities if the same element undergoes $FCC$ and $HCP$ close packing is:

The stacking pattern in a body-centered cubic $(bcc)$ lattice is of the type............

In an ionic oxide,the oxide anions form an $hcp$ arrangement,while the metal cations occupy $\frac{2}{3}$ of the octahedral voids. Determine the formula of this oxide.

$A$ compound is formed by elements $X$, $Y$, and $Z$. Atoms of $Z$ (anions) make $FCC$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids. Atoms of $Y$ (cations) occupy $\frac{1}{3}$ rd of the tetrahedral voids. The formula of the compound is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module