Question

The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of thorium.

Answer 1

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:

$\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}$      

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 $N_{A}$: is the Avogadro constant = 6.023x10²³ atoms/mol.

Hence the lattice parameter is:  

$a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}$

$a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm$

(b) We know that the lattice parameter of a FCC structure is:

$a = \frac{4r}{\sqrt{2}}$

where r: is the atomic radius of Th

Hence, the atomic radius of Th is:

$r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm$    

I hope it helps you!    

Answer 2

Explanation:

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