Respuesta :
Answer: The radius of palladium atom is 137.5 pm
Explanation:
To calculate the edge length of the crystal lattice, we use the equation to calculate the density of metal, which is:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density = [tex]12.0g/cm^3[/tex]
Z = number of atom in unit cell = 4 (FCC)
M = atomic mass of crystal = 106.4 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = ?
Putting values in above equation, we get:
[tex]12.0g/cm^3=\frac{4\times 106.4}{6.022\times 10^{23}\times a^3}\\\\a^3=\frac{4\times 106.4}{6.022\times 10^{23}\times 12.0}=5.89\times 10^{-23}cm^3\\\\a=\sqrt[3]{5.89\times 10^{-23}}=3.89\times 10^{-8}cm[/tex]
Converting this into picometers, we use the conversion factor:
[tex]1cm=10^{10}pm[/tex]
So, [tex]3.89\times 10^{-8}cm\times \frac{10^{10}pm}{1cm}=389pm[/tex]
To calculate the radius of atom, we use the relation between the radius and edge length for FCC lattice:
[tex]a=2\sqrt{2}R[/tex]
Putting values in above equation, we get:
[tex]398=2\sqrt{2}\times R\\\\R=\frac{389}{2\sqrt{2}}=137.5pm[/tex]
Hence, the radius of palladium atom is 137.5 pm
