Answer: The volume of one atom of silver is [tex]4.52\times 10^{-23}cm^3[/tex]
Explanation:
We are given:
Atomic radius of silver metal = 126 pm
To calculate the edge length, 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]a=2\sqrt{2}\times 126=356.4pm=356.4\times 10^{-10}cm[/tex] (Conversion factor: [tex]1cm=10^{10}pm[/tex] )
To calculate the volume of cube, we use the equation:
[tex]\text{Volume of silver}=a^3[/tex]
where,
a = edge length of the atom
[tex]\text{Volume of one atom of silver}=(356.4\times 10^{-10}cm)^3=4.52\times 10^{-23}cm^3[/tex]
Hence, the volume of one atom of silver is [tex]4.52\times 10^{-23}cm^3[/tex]
