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Question: Niobium is found to crystallize with bcc structure and found to have the density of 8.55 \(g/c{{m}^{...

Niobium is found to crystallize with bcc structure and found to have the density of 8.55 g/cm3g/c{{m}^{3}}. Determine the atomic radius of niobium if its atomic mass is 93 u.

Explanation

Solution

You can use the formula ρ=ZMa3NA\rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}}, where ρ\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, NA{{N}_{A}} is the Avogadro’s number. Another formula will be r=a34r=\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:
ρ=ZMa3NA\rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}}
Where ρ\rho is the density of niobium and the given density of niobium is 8.55 g/cm3g/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.
NA{{N}_{A}} is the Avogadro’s number and its value is 6.022 x 10236.022\text{ x 1}{{\text{0}}^{23}} .
By putting all the values mentioned above, we can calculate the edge length of the unit cell as:
ρ=ZMa3NA a3=ZMρNA  \rho =\dfrac{ZM}{{{a}^{3}}{{N}_{A}}} \\\ {{a}^{3}}=\dfrac{ZM}{\rho {{N}_{A}}} \\\
a3=2 x 938.55 x 6.022 x 1023=3.61 x 10−23cm3{{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 x 10−8cma=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=a34r=\dfrac{a\sqrt{3}}{4}
r=3.3 x 10−8 x 34r=\dfrac{3.3\text{ x 1}{{\text{0}}^{-8}}\text{ x }\sqrt{3}}{4}
r=1.43 x 10−8cmr=1.43\text{ x 1}{{\text{0}}^{-8}}cm
Hence, the radius will be 1.43 x 10−8cm1.43\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=a2r=\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=a22r=\dfrac{a}{2\sqrt{2}}