The edge length of a $bcc$ unit cell is $4 \times 10^{-10} \ m$. What is the radius of the atom occupying the body center of the cubic unit cell?
- A$\sqrt{3} \times 10^{-10} \ m$
- B$4 \times 10^{-10} \ m$
- C$0.414 \times 4 \times 10^{-10} \ m$
- DNone of these
Explore More
Similar Questions
Calculate the number of atoms present in a unit cell of an element having molar mass $190 \ g \ mol^{-1}$ and density $20 \ g \ cm^{-3}$. Given that $[a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}]$.
Silver crystallises in $fcc$ structure. If the edge length of the unit cell is $316.5 \text{ pm}$, what is the radius of the silver atom (in $\text{pm}$)?
Calculate the number of unit cells in $0.4 \ g$ of metal if the product of density and volume of the unit cell is $1.2 \times 10^{-22} \ g$.
Suppose the mass of a single $Ag$-atom is $m$. $Ag$ metal crystallises in $fcc$ lattice with unit cell of length $a$. The density of $Ag$ metal in terms of $a$ and $m$ is
MediumWBJEE 2015
View SolutionIf the edge length of a cubic unit cell in a $bcc$ structure is $400 \, pm$, then the atomic radius of the metal will be ........... $pm$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo