Answer:
FCC.
Explanation:
Hello,
In this case, since the density is defined as:
[tex]\rho =\frac{n*M}{Vc*N_A}[/tex]
Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:
[tex]Vc_{BCC}=(\frac{4r}{\sqrt{3} } )^3=(\frac{4*0.1345x10^{-7}cm}{\sqrt{3} } )^3=2.997x10^{-23}cm^3\\\\Vc_{FCC}=(2\sqrt{2}r)^{3} =(2\sqrt{2} *0.1345x10^{-7}cm)^3=5.506x10^{-23}cm^3[/tex]
Hence, we compute the density for each crystal structure:
[tex]\rho _{BCC}=\frac{n_{BCC}*M}{Vc_{BCC}*N_A}=\frac{2*102.9g/mol}{2.337x10^{-23}cm^3*6.022x10^{23}/mol} =14.62g/cm^3\\\\\rho _{FCC}=\frac{n_{FCC}*M}{Vc_{FCC}*N_A}=\frac{4*102.9g/mol}{5.506x10^{-23}cm^3*6.022x10^{23}/mol} =12.41g/cm^3[/tex]
Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.
Regards.
