Vanadium crystallizes with a body-centered cubic unit cell. The volume of the unit cell is \(0.0278 \mathrm{~nm}^{3}\). What is its atomic radius?

Short Answer

Expert verified
Answer: The atomic radius of vanadium in the BCC unit cell is approximately 0.131 nm.

Step by step solution

01

Find the lattice constant (a) of the BCC unit cell

The volume of a cubic unit cell is given by \(V = a^3\), where 'a' is the lattice constant. We can solve for 'a', given the volume: $$ a = \sqrt[3]{V} $$ Substitute the given volume, \(V = 0.0278 \mathrm{~nm}^{3}\): $$ a = \sqrt[3]{0.0278} $$ Calculate the lattice constant: $$ a \approx 0.303 \mathrm{~nm} $$
02

Determine the relation between atomic radius and lattice constant for BCC unit cell

In a BCC unit cell, atoms are located at the corners and the center of the cube. If we consider a body diagonal of the unit cell, there are two atomic radii (corner and center atoms) and one lattice constant (edge) on a straight line. We obtain a relation between the atomic radius and the lattice constant using the Pythagorean theorem for a right angle triangle formed using half of the body diagonal, half of the edge (lattice constant), and one atomic radius as its sides: $$ \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 = (2r)^2 $$ Simplifying, we get: $$ \frac{a^2}{2} = 4r^2 $$
03

Calculate the atomic radius using the lattice constant and relation derived in Step 2

Using the derived relation and the calculated lattice constant, we can find the atomic radius 'r': $$ 4r^2 = \frac{a^2}{2} $$ Plug in the lattice constant value, \(a \approx 0.303 \mathrm{~nm}\): $$ 4r^2 = \frac{(0.303)^2}{2} $$ Solve for 'r': $$ r = \sqrt{\frac{(0.303)^2}{8}} $$ Compute the atomic radius: $$ r \approx 0.131 \mathrm{~nm} $$ The atomic radius of vanadium in the BCC unit cell is approximately \(0.131 \mathrm{~nm}\).

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