The volume of a simple cubic unit cell is x t | vedclass

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

Question

(C) In a simple cubic unit cell,the number of particles per unit cell $(Z)$ is $1$. The volume of the unit cell $(V_{cell})$ is related to the volume

Vedclass · Accepted Answer

$4.0$

Vedclass · Answer

(C) In a simple cubic unit cell,the number of particles per unit cell $(Z)$ is $1$. The volume of the unit cell $(V_{cell})$ is related to the volume occupied by a particle $(V_{particle})$ by the relation: $V_{cell} = Z 	imes V_{particle}$. Given $Z = 1$ and $V_{particle} = 2.1 	imes 10^{-23} \ cm^3$. Therefore,$V_{cell} = 1 	imes 2.1 	imes 10^{-23} \ cm^3 = 2.1 	imes 10^{-23} \ cm^3$. Comparing this with $x 	imes 10^{-23} \ cm^3$,we get $x = 2.1$. Note: The provided options do not contain the correct value $2.1$. Based on the standard definition of a simple cubic unit cell where the edge length $a = 2r$,the volume is $a^3 = (2r)^3 = 8r^3$. If the volume occupied by a particle (sphere) is $\frac{4}{3}\pi r^3 = 2.1 	imes 10^{-23} \ cm^3$,then $r^3 = \frac{2.1 	imes 10^{-23} 	imes 3}{4 	imes 3.14} \approx 0.5016 	imes 10^{-23} \ cm^3$. Thus,$V_{cell} = 8 	imes 0.5016 	imes 10^{-23} \approx 4.01 	imes 10^{-23} \ cm^3$. Therefore,$x \approx 4.0$.

The volume of a simple cubic unit cell is $x \times 10^{-23} \ cm^3$. Calculate the value of $x$ if the volume occupied by a particle in it is $2.1 \times 10^{-23} \ cm^3$.

Similar Questions

An element crystallizes in an $fcc$ unit cell structure with an edge length of $200 \, pm$. If $200 \, g$ of the element contains $24 \times 10^{23}$ atoms, what is its density in $g/cm^3$?

In the Bragg's equation for diffraction of $X$-rays, $n$ represents:

Metal $M$ crystallises in $fcc$ lattice. If the unit cell has an edge length of $4.077 \ \mathring{A}$ and the density is $10.5 \ g \ cm^{-3}$,then the atomic weight of the metal is:

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

$A$ metal with an atomic radius of $141.4 \, pm$ crystallises in the face-centred cubic structure. The volume of the unit cell in $pm^3$ is $.... . \times 10^7$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module