The unit cell of copper corresponds to a face | vedclass

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(C) For an $FCC$ unit cell,the number of atoms per unit cell $(Z)$ is $4$. The formula for density $(d)$ is $d = \frac{Z 	imes M}{N_{A} 	imes a^{3}}

Vedclass · Accepted Answer

$9077$

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(C) For an $FCC$ unit cell,the number of atoms per unit cell $(Z)$ is $4$. The formula for density $(d)$ is $d = \frac{Z 	imes M}{N_{A} 	imes a^{3}}$. Given: $Z = 4$,$M = 63.54 \, g/mol = 0.06354 \, kg/mol$,$N_{A} = 6.022 	imes 10^{23} \, mol^{-1}$,and $a = 3.596 \, \mathring{A} = 3.596 	imes 10^{-10} \, m$. Substituting the values: $d = \frac{4 	imes 0.06354}{6.022 	imes 10^{23} 	imes (3.596 	imes 10^{-10})^{3}} \approx 9076.6 \, kg/m^{3}$. Thus,the density is approximately $9077 \, kg/m^{3}$.

The unit cell of copper corresponds to a face-centered cubic $(FCC)$ lattice with an edge length of $3.596 \, \mathring{A}$. The calculated density of copper in $kg / m^{3}$ is ....... .
[Molar mass of $Cu = 63.54 \, g/mol$; Avogadro Number $= 6.022 \times 10^{23} \, mol^{-1}$]

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The unit cell of copper corresponds to a face-centered cubic $(FCC)$ lattice with an edge length of $3.596 \, \mathring{A}$. The calculated density of copper in $kg / m^{3}$ is ....... . [Molar mass of $Cu = 63.54 \, g/mol$; Avogadro Number $= 6.022 \times 10^{23} \, mol^{-1}$]