Respuesta :
Answer:
The radius of tantalum (Ta) atom is [tex]R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm[/tex]
Explanation:
From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through
[tex]a=\frac{4R}{\sqrt{3} }[/tex]
So the volume of the unit cell [tex]V_{c}[/tex] is
[tex]V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}[/tex]
We can compute the theoretical density ρ through the following relationship
[tex]\rho=\frac{nA}{V_{c}N_{a}}[/tex]
where
n = number of atoms associated with each unit cell
A = atomic weight
[tex]V_{c}[/tex] = volume of the unit cell
[tex]N_{a}[/tex] = Avogadro’s number ([tex]6.023 \times 10^{23}[/tex] atoms/mol)
From the information given:
A = 180.9 g/mol
ρ = 16.6 g/cm^3
Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.
We can use the theoretical density ρ to find the radio R as follows:
[tex]\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}[/tex]
Solving for R
[tex]\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}[/tex]
Substitution for the various parameters into above equation yields
[tex]R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm[/tex]
