Respuesta :
Answer:
Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]
Explanation:
We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )
M = atomic weight = 195.08 gm/mol.
N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]
a= size of side of cube.
Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex] ....1
Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]
Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]
Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]
Putting value of a in equation 1 we get,
[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]
Hence, this is the required solution.
