Answer :
Answer:
(a) a = 5.08x10⁻⁸ cm
(b) r = 179.6 pm
Explanation:
(a) The lattice parameter "a" can be calculated using the following equation:
[tex] \rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}} [/tex]
where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and [tex]N_{A}[/tex]: is the Avogadro constant = 6.023x10²³ atoms/mol.
Hence the lattice parameter is:
[tex] a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3} [/tex]
[tex] a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm [/tex]
(b) We know that the lattice parameter of a FCC structure is:
[tex] a = \frac{4r}{\sqrt{2}} [/tex]
where r: is the atomic radius of Th
Hence, the atomic radius of Th is:
[tex]r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm[/tex]
I hope it helps you!
