Calculate the number of atoms in 0.3 g of a | vedclass

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(B) For a $bcc$ unit cell,the number of atoms per unit cell is $n = 2$. The mass of one unit cell is given by the product of density $(\rho)$ and volu

Vedclass · Accepted Answer

$2.0 	imes 10^{21}$

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(B) For a $bcc$ unit cell,the number of atoms per unit cell is $n = 2$. The mass of one unit cell is given by the product of density $(ho)$ and volume $(a^3)$,which is $ho 	imes a^3 = 3 	imes 10^{-22} \ g$. The number of unit cells in $0.3 \ g$ of the metal is $\frac{	ext{Total mass}}{	ext{Mass of one unit cell}} = \frac{0.3 \ g}{3 	imes 10^{-22} \ g} = 1.0 	imes 10^{21}$ unit cells. Since each $bcc$ unit cell contains $2$ atoms,the total number of atoms is $2 	imes (1.0 	imes 10^{21}) = 2.0 	imes 10^{21}$ atoms.

Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

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