Calcium crystallises in a face-centred cubic | vedclass

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(A) The density of a crystal is given by the formula $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For a face-centred cubic $(FCC)$ unit cell,the number of

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$1.55$

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(A) The density of a crystal is given by the formula $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For a face-centred cubic $(FCC)$ unit cell,the number of atoms per unit cell $Z = 4$.The molar mass of Calcium $(Ca)$ is $M = 40 \ g \ mol^{-1}$.The edge length $a = 0.556 \ nm = 0.556 	imes 10^{-7} \ cm$.Frenkel defects involve the migration of ions into interstitial sites,which does not change the mass or the volume of the unit cell; therefore,the theoretical density remains unchanged.Substituting the values: $d = \frac{4 	imes 40}{6.022 	imes 10^{23} 	imes (0.556 	imes 10^{-7})^3}$.$d = \frac{160}{6.022 	imes 10^{23} 	imes 1.719 	imes 10^{-22}} \approx 1.55 \ g \ cm^{-3}$.

List-$I$ (Defect)	List-$II$ (Examples)
$A$. Frenkel defects	$1$. $FeO$
$B$. Schottky defects	$2$. $NaCl$
$C$. Vacancy defects	$3$. $AgCl$
$D$. Metal deficiency defects	$4$. Crystals with vacant lattice sites

Calcium crystallises in a face-centred cubic unit cell with edge length $a = 0.556 \ nm$. Calculate the density. If it contained $0.1 \%$ Frenkel defects,what is the density in $g \ cm^{-3}$?

Similar Questions

An element $X$ (atomic weight = $24$) crystallises in a face-centered cubic $(FCC)$ lattice. The edge length of the unit cell is $400 \, pm$. If this lattice contains $4\%$ Schottky defects,calculate the observed density of this element in $g/cm^3$ $(N_A = 6 \times 10^{23})$.

Why is $FeO_{(s)}$ not formed in stoichiometric composition?

How do defects arise in crystals? State the types of defects.

Match List-$I$ with List-$II$.

List-$I$ (Defect) List-$II$ (Examples)

$A$. Frenkel defects $1$. $FeO$

$B$. Schottky defects $2$. $NaCl$

$C$. Vacancy defects $3$. $AgCl$

$D$. Metal deficiency defects $4$. Crystals with vacant lattice sites

Frenkel defect is caused due to

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