Question

The density of Na crystal with BCC structure having

edge length a = 4.24 Ao

is​

Answer 1

Answer:

Explanation:

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Answer 2

$\color{darkblue}\underline{\underline{\sf Given-}}$

• BCC structure of Na
• Number of atom per unit cell in BCC is (Z) = 2
• Edge lenght (a) = 4.24${\sf A^°}$

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Density of Na

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$\color{darkblue}\underline{\underline{\sf Formula \: Used -}}$

$\color{violet}\underline{\boxed{\sf \rho = \dfrac{ZM}{N_Aa^3}}}$

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${\sf N_A=Avogadro\: number= 6.02×10^{23}}$

a = edge length = ${\sf 4.24A^° \:\:\implies\:\:4.24×10^{-8}cm}$

$\sf {\rho = Density }$

$\implies{\sf \rho =\dfrac{2×23}{6.02×10^{23}×(4.23×10^{-8})^3}}$

$\implies{\sf \rho =\dfrac{46}{6.02×10^{23}×76.2×10^{-24}}}$

$\implies{\sf \rho = \dfrac{46}{458×10^{-1}} }$

$\color{red}\implies{\sf \rho =1.004\:g/cm^3}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Density of Na crystal is $\color{red}{\sf 1.004\:g/cm^3}$.