Calculate the total volume of a simple cubic | vedclass

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

Question

(A) $1$. The packing efficiency of a simple cubic unit cell is $52.4\%$, meaning $52.4\%$ of the total volume is occupied by atoms.$2$. The void volum

Vedclass · Accepted Answer

$4.2 	imes 10^{-23}$

Vedclass · Answer

(A) $1$. The packing efficiency of a simple cubic unit cell is $52.4\%$, meaning $52.4\%$ of the total volume is occupied by atoms.$2$. The void volume (empty space) is $100\% - 52.4\% = 47.6\%$ of the total volume.$3$. Let the total volume be $V$. Given that $47.6\% 	ext{ of } V = 2.0 	imes 10^{-23} 	ext{ cm}^3$.$4$. $0.476 	imes V = 2.0 	imes 10^{-23} 	ext{ cm}^3$.$5$. $V = \frac{2.0 	imes 10^{-23}}{0.476} \approx 4.2 	imes 10^{-23} 	ext{ cm}^3$.

Calculate the total volume of a simple cubic unit cell in $\text{cm}^3$ if the void volume is $2.0 \times 10^{-23} \text{cm}^3$.

Similar Questions

What is the percentage efficiency of packing in $BCC$ structure (in $\%$)?

If a compound (oxide) has an $hcp$ arrangement and the cations $M$ occupy two-thirds of the octahedral voids,the formula of the compound is:

The edge of unit cell of $FCC$ $Xe$ crystal is $620 \ pm$. The radius of $Xe$ atom is $..........$ $pm$.

$A$ compound is formed by two elements $X$ and $Y$. Atoms of the element $Y$ (as anions) make $ccp$ and those of the element $X$ (as cations) occupy all the octahedral voids. What is the formula of the compound?

In an $fcc$ crystal lattice,atoms $A$ are at the corners of the cube and atoms $B$ are at the face centers. If one atom is removed from one of the face centers,find the formula of the compound.

Vedclass Test Series

Exam Paper Generator

Online Exam Module