An element crystallizes in a bcc lattice. The | vedclass

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

Question

(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \f

Vedclass · Accepted Answer

$2.16 	imes 10^{-22}$

Vedclass · Answer

(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \frac{4r}{\sqrt{3}}$.Given $r = 2.598 \ \mathring{A} = 2.598 	imes 10^{-8} \ cm$.Substituting the value of $r$: $a = \frac{4 	imes 2.598 	imes 10^{-8}}{\sqrt{3}} = \frac{4 	imes 2.598 	imes 10^{-8}}{1.732} = 4 	imes 1.5 	imes 10^{-8} = 6 	imes 10^{-8} \ cm$.The volume of the unit cell is $V = a^3 = (6 	imes 10^{-8} \ cm)^3 = 216 	imes 10^{-24} \ cm^3 = 2.16 	imes 10^{-22} \ cm^3$.

An element crystallizes in a $bcc$ lattice. The atomic radius of the element is $2.598 \ \mathring{A}$. What is the volume (in $cm^3$) of one unit cell?

Similar Questions

$A$ metal crystallizes in a $BCC$ structure. The edge length of its unit cell is $3.04 \ \mathring{A}$. What is the volume of its unit cell in $cm^3$?

Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

Silver metal crystallizes in a $ccp$ lattice. If the edge length of the unit cell is $407 \ pm$, the radius of the silver atom is ............. $pm$.

Calculate the molar mass of an element having density $5.6 \ g \ cm^{-3}$ that forms a $bcc$ structure. $\left[a^3 \times N_{A}=75 \ cm^3 \ mol^{-1}\right]$

Silver forms a $CCP$ structure and its density is $10.5 \ g/cm^3$. What is the edge length of the unit cell? (Molar mass of silver is $107.9 \ g/mol$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module