Answer:
[tex]r=215pm[/tex]
[tex]N_{Mn}=20[/tex]
Explanation:
From the question we are told that:
Edge length of the unit cell [tex]l=608pm[/tex]
a)
Generally the equation for The relationship between edge length and radius is mathematically given by
[tex]4r=\sqrt{2a}[/tex]
Therefore
[tex]4r=\sqrt{2*608}[/tex]
[tex]r=\frac{\sqrt{2*608}}{4}[/tex]
[tex]r=215pm[/tex]
b)
From the question we are told that:
Density [tex]\rho=7.297[/tex]
Edge length of [tex]l=630.0 pm=>630*10^-{10}[/tex]
Therefore Volume is given as
[tex]V=l^3[/tex]
[tex]V=630*10^-{10}^3[/tex]
[tex]V=2.50047*10^{−22}[/tex]
Generally the equation for Mass is mathematically given by
[tex]m=Volume*density[/tex]
[tex]m=V*\rho[/tex]
[tex]m=2.50047*10^{−22}*7.297[/tex]
[tex]m=1.83*10^{-21}g[/tex]
Therefore Molarity is given as
[tex]n=\frac{M}{Molar M}[/tex]
[tex]n=\frac{1.83*10^{-21}g}{55}[/tex]
[tex]n=3.32*10^{-23}[/tex]
Finally The atoms in a unit cell is
[tex]N_{Mn}=Moles*Avogadro\ constant[/tex]
[tex]N_{Mn}=3.32*10^{-23}*6.023*10^{23}[/tex]
[tex]N_{Mn}=20[/tex]