Calculations for Standard Cubical Unit Lattice

Calcium crystallizes in a face centred cubic unit cell with a $\text{=}\text{0}\text{.560}\text{nm}\text{.}$  The density of the metal if it contains 0.1% schottky defects would be:

Copper crystallises in a face-centred cubic lattice and has a density of  $8.930\mathrm{g}{\mathrm{cm}}^{-3}$ at $393 \mathrm{K}.$ The radius of a copper atom is:
[Atomic mass of$\mathrm{Cu}=63.55\mathrm{u},{\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{\mathrm{mol}}^{-1}]$

Iron has a body-centered cubic unit cell of cell edge $286.65 \mathrm{pm}$. The density of iron is $7.87 \mathrm{g} {\mathrm{cm}}^{-3}$. The Avogadro number is

(Atomic mass of iron $=56\hspace{0.17em}\mathrm{g}\hspace{0.17em}{\mathrm{mol}}^{-1}$)

X-rays diffraction studies show that copper crystallizes in an FCC unit cell with cell edge of  $3.6885\times {10}^{-8}\mathrm{cm}.$ In a separate experiment, copper is determined to have a density of  $\text{8}{\text{.92g/cm}}^{\text{3}}$, the atomic mass of copper would be:

In face-centred cubic $\left(\mathrm{FCC}\right)$ and body centred cubic $\left(\mathrm{BCC}\right),$ whose unit cell lengths are $3.5$ and $3.0$ $\mathrm{\AA }$ respectively, a metal crystallises into two cubic phases. What is the ratio of densities of $\mathrm{FCC}$ and $\mathrm{BCC}?$

In face centred cubic (FCC) crystal lattice, edge length is 400 pm. The diameter of greatest sphere which can be fit into the interstitial void without distortion of lattice is

The density of mercury is 13.6 g ml^-1. The approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom is

A substance ${\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}$ crystallizes in a face centred cubic (FCC) lattice in which atoms $\text{A}$ occupy each corner of the cube and atoms $\text{B}$ occupy face centres of the cube composition of the substance ${\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}$ is

The coordination number of a metal crystallizing in a hexagonal close-packed structure is:

$\mathrm{CsBr}$ has bcc structure with edge length $4.3 \mathrm{\AA }$ . The shortest inter ionic distance in between ${\mathrm{Cs}}^{+}$ and ${\mathrm{Br}}^{-}$ is:

In a compound, atoms of an element $\mathrm{Y}$ form $\mathrm{CCP}$ lattice and those of element $\mathrm{X}$ occupies ${\frac{2}{3}}^{\mathrm{rd}}$ of the tetrahedral voids. The formula of the compound can be

Atom $\mathrm{X}$ occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in $\%$) of the resultant solid is closest to

Copper crystallizes in an FCC unit cell with cell edge of  $3.608\times {10}^{-8}\mathrm{cm}.$ The density of copper is  $\mathrm{8}.92 \mathrm{g}/{\mathrm{cm}}^{\mathrm{3}}$, Calculate the atomic mass of copper.

Assume that the atoms in a face-centered cubic unit cell are represented by spheres of radii $\mathrm{r}$. The volume per atom in this unit cell is given by

Elements $\mathrm{X}$ and $\mathrm{Y}$ form a compound. The atoms of element $\mathrm{X}$ arrange in a cubic close packed lattice. The atoms of element $\mathrm{Y}$ occupy half of the tetrahedral voids and all the octahedral voids. The formula of the compound thus formed is [treat $\mathrm{X}$ and $\mathrm{Y}$ atoms as hard spheres]

The CORRECT order of number of atoms per unit cell in simple cubic (SC) unit cell, body centered cubic (BCC) unit cell and face centered cubic (FCC) unit cell is