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(B) In a Schottky defect, an equal number of cations and anions are missing from the lattice, which reduces the effective number of atoms per unit cel

Vedclass · Accepted Answer

$2.1$

Vedclass · Answer

(B) In a Schottky defect, an equal number of cations and anions are missing from the lattice, which reduces the effective number of atoms per unit cell $(Z_{eff})$. For an $FCC$ lattice, the number of atoms per unit cell is $Z = 4$. With $0.1 \%$ Schottky defects, the new effective number of atoms is $Z_{new} = 4 - (4 	imes 0.001) = 3.996 \approx 4$. However, using the provided calculation logic: $Z_{new} = 4 - 0.004 = 3.996$. Given the atomic mass of Potassium $(K)$ is $39 \ g \ mol^{-1}$. The edge length $a = 0.5 \ nm = 5 	imes 10^{-8} \ cm$. The density formula is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. $ho = \frac{3.996 	imes 39}{6.022 	imes 10^{23} 	imes (5 	imes 10^{-8})^3} = \frac{155.844}{6.022 	imes 10^{23} 	imes 125 	imes 10^{-24}} = \frac{155.844}{75.275} \approx 2.07 \ g \ cm^{-3}$. Rounding to the nearest value, we get $2.1 \ g \ cm^{-3}$. Thus, Option $B$ is correct.

Potassium crystallizes in an $FCC$ lattice with a unit cell edge length of $0.5 \ nm$. The approximate density (in $g \ cm^{-3}$), if it contains $0.1 \%$ Schottky defects, is:

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