What is the edge length of the unit cell of a body-centred cubic $(BCC)$ crystal of an element whose atomic radius is $75 \ pm$ (in $pm$)?

$CsBr$ has $bcc$ structure with edge length $4.3 \ \mathring{A}$. The shortest interionic distance in between $Cs^{+}$ and $Br^{-}$ (in $\mathring{A}$) is :

If each edge of a cubic unit cell of an element having atomic mass $120$ and density $6.25 \ g \ cm^{-3}$ measures $400 \ pm$,then the crystal lattice is

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 $7.2 \times 10^{-22} \ g$.

An element crystallises in $bcc$ type having atomic radius $1.33 \times 10^{-8} \ cm$,the edge length of the unit cell will be:

If the edge length of a cubic unit cell in a $bcc$ structure is $400 \, pm$, then the atomic radius of the metal will be ........... $pm$.