Question

Calculate the radius of a tantalum (ta) atom, given that ta has a bcc crystal structure, a density of 16.6 g/cm3 , and an atomic weight if 180.9 g/mol. (avogadro number, 6.023 ×1023 atoms/mol) (4pts)

Answer 1

It is given that Ta has BCC crystal structure. The density is $16.6 g/cm^{3}$, atomic mass is 180.9 g/mol and Avogadro's number is $6.023\times 10^{23}$ atoms per mol.

The expression for density is as follows:

$\rho =\frac{nA}{V_{c}N_{A}}$

Here, n is number of atoms per unit cell, A is atomic number, $V_{c}$ is volume and $N_{A}$ is Avogadro's number .

Expression for volume is as follows:

$V_{c}=(\frac{4R}{\sqrt{3}})^{3}=\frac{64R^{3}}{3\sqrt{3}}$

For BCC, number of atoms per unit cell is 2, putting the values,

$16.6 g/cm^{3} =\frac{(2 atom)(180.9 g/mol)(3\sqrt{3})}{(64) R^{3}(6.023\times 10^{23}atom /mol)}$

On rearranging,

$R^{3}=\frac{4.87\times 10^{-23} g}{16.6 g/cm^{3}}=2.938\times 10^{-24}cm^{3}$

Thus,

$R=\sqrt[3]{2.938\times 10^{-24}}=1.432\times 10^{-8} cm$

Since, $1 cm=10^{7} nm$

Thus, $1.432\times 10^{-8} cm=0.1432 nm$

Therefore, radius of tantalum atom is 0.1432 nm.