Respuesta :
Answer:
Explanation:
For face centered cubic structure , no of atoms in a cell will be 4 .
so n = 4
density = n x atomic mass / a³ x N
N is avogrado no
12.4 = n x 103 / (a³ x 6.02 x 10²³)
a³ = 4 x 103 / (12.4 x 6.02 x 10²³ )
= 5.52 x 10⁻²³
= 55.2 x 10⁻²⁴
a = 3.8 x 10⁻⁸ cm
if atomic radius is r
a = 2√2 x r
r = a / 2√2
= 3.8 x 10⁻⁸ /2√2 cm
= 1.34 x 10⁻⁸ cm
The radius of the Rhodium FCC crystal has been [tex]\rm \bold{1.34\;\times\;10^-^8\;cm}[/tex].
The face centered cubic crystal lattice has been the arrangement of the atoms in the cubic structure. The number of atom in a face centered crystal lattice has been 4.
The density of the face centered lattice has been given as:
[tex]d=\dfrac{n\;\times\;\text{atomic number}}{a^3\;\times\;N}[/tex]
Where, the number of atoms in FCC crystal, n=4
The atomic number of Rd, =103
The edge length of the crystal, a
The Avogadro constant, [tex]N=6.023\;\times\;10^2^3[/tex]
The density of the lattice, [tex]d=12.4\;\rm g/cm^3[/tex]
Substituting the values for calculating edge length, a:
[tex]12.4=\dfrac{4\;\times\;103}{a^3\;\times\;6.023\;\times\;10^2^3}\;\text {cm}\\a^3=55.2\;\times\;10^-^2^4\;\text{cm}\\a=3.8\;\times\;10^-^8\;\rm cm[/tex]
The edge length of the FCC structure has bee [tex]\rm \bold{3.8\;\times\;10^-^8\;cm}[/tex].
The radius of the FCC crystal has been given as:
[tex]a=2\2\sqrt{2}r[/tex]
Where, the edge length of the crystal, [tex]a= {3.8\;\times\;10^-^8\;\rm cm}[/tex]
The radius (r) of the crystal has been given as:
[tex]3.8\;\times\;10^-^8=2\sqrt{2}\;\times\;r\\r=\dfrac{3.8\;\times\;10^-^8}{2\sqrt{2} } \\r=1.34\;\times\;10^-^8\;\rm cm[/tex]
The radius of the Rhodium FCC crystal has been [tex]\rm \bold{1.34\;\times\;10^-^8\;cm}[/tex].
For more information about the radius of FCC, refer to the link:
https://brainly.com/question/14934549
