Question

Niobium is found to crystallize with bcc structure and found to have the density of 8.55 $g/c{{m}^{3}}$. Determine the atomic radius of niobium if its atomic mass is 93 u.

Answer 1

Hint: You can use the formula $\rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}}$, where $\rho $ is the density of the substance, Z is the number of atoms in the crystal structure, M is the atomic mass, a is the edge of the unit cell in pm, ${{N}_{A}}$ is the Avogadro’s number. Another formula will be $r=\dfrac{a\sqrt{3}}{4}$, where r is the radius of the atom.

Complete answer:
We are given that the niobium crystallizes at BCC, which means body-centered cubic cell. So, there will be a total of 2 atoms in one unit cell, as 1 from the corner of the unit cell and 1 at the center of the body.
We can use the formula:
$\rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}}$
Where $\rho $ is the density of niobium and the given density of niobium is 8.55 $g/c{{m}^{3}}$.
Z is the number of atoms in the crystal structure and its value is 2.
M is the atomic mass and here the given atomic mass is 93u.
a is the edge of the unit cell in pm.
${{N}_{A}}$ is the Avogadro’s number and its value is $6.022\text{ x 1}{{\text{0}}^{23}}$ .
By putting all the values mentioned above, we can calculate the edge length of the unit cell as:
$
   \rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}} \\
  {{a}^{3}}=\dfrac{ZM}{\rho {{N}_{A}}} \\
$
${{a}^{3}}=\dfrac{2\text{ x 93}}{8.55\text{ x 6}\text{.022 x 1}{{\text{0}}^{23}}}=3.61\text{ x 1}{{\text{0}}^{-23}}c{{m}^{3}}$
$a=3.3\text{ x 1}{{\text{0}}^{-8}}cm$
Since we have edge length of the unit cell, we can calculate the radius of the atom in the BCC structure by using the formula:
$r=\dfrac{a\sqrt{3}}{4}$
$r=\dfrac{3.3\text{ x 1}{{\text{0}}^{-8}}\text{ x }\sqrt{3}}{4}$
$r=1.43\text{ x 1}{{\text{0}}^{-8}}cm$
Hence, the radius will be $1.43\text{ x 1}{{\text{0}}^{-8}}cm$.

Note: If the structure of the crystal was given SCC (simple cubic structure), then the relation of radius and edge length will be:
$r=\dfrac{a}{2}$
If the structure of the crystal was given FCC (face-centered cubic structure), then the relation of radius and edge length will be:
$r=\dfrac{a}{2\sqrt{2}}$