The mass of an atom present in a unit cell is | vedclass

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(B) The density of a unit cell is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar

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Face centred unit cell

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(B) The density of a unit cell is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $a^3$ is the volume of the unit cell $(V)$.Rearranging the formula,we get $ho 	imes V = \frac{Z 	imes M}{N_A}$.We know that the mass of one atom is $m = \frac{M}{N_A} = 4.4 	imes 10^{-23} \ g$.Given $ho 	imes V = 1.792 	imes 10^{-22} \ g$.Substituting the values: $1.792 	imes 10^{-22} = Z 	imes (4.4 	imes 10^{-23})$.$Z = \frac{1.792 	imes 10^{-22}}{4.4 	imes 10^{-23}} = \frac{17.92}{4.4} \approx 4.07$.Since $Z \approx 4$,the unit cell is a Face centred unit cell $(FCC)$.

The mass of an atom present in a unit cell is $4.4 \times 10^{-23} \ g$ and the product of density and volume of the unit cell is $1.792 \times 10^{-22} \ g$. What is the type of cubic unit cell?

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