Question

A metal crystallizes in fcc lattice. The edge length of the unit cell is 200 pm. What is the radius (in m) of the metal atom?

• A. \(\sqrt[3]{0.353} \times 10^{-10}\)
• B. \(\sqrt[3]{0.512} \times 10^{-10}\)
• C. \(\sqrt{0.353} \times 10^{-10}\)
• D. \(\sqrt[3]{0.253} \times 10^{-10}\)

Hint:

Remember the relationship between edge length and radius for different types of cubic unit cells (simple cubic, bcc, fcc).

Answer 1

The Correct Option is A

1. Relationship between edge length (a) and radius (r) in an fcc lattice:

In an fcc lattice, the atoms are located at the corners and the centers of the faces.
The atoms touch along the face diagonal.
The face diagonal is equal to \(4r\), where \(r\) is the radius of the atom.
The face diagonal is also equal to \(a\sqrt{2}\), where \(a\) is the edge length.
Therefore, \[ 4r = a\sqrt{2} \] \[ r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}} \]

2. Convert the edge length to meters:

\( a = 200 \) pm \( = 200 \times 10^{-12} \) m

3. Calculate the radius (r):

\[ r = \frac{200 \times 10^{-12} \, \text{m}}{2\sqrt{2}} \] \[ r = \frac{100 \times 10^{-12} \, \text{m}}{\sqrt{2}} \] \[ r = \frac{100}{\sqrt{2}} \times 10^{-12} \, \text{m} \] \[ r = \frac{100 \times \sqrt{2}}{2} \times 10^{-12} \, \text{m} \] \[ r = 50\sqrt{2} \times 10^{-12} \, \text{m} \] \[ r = 50 \times 1.414 \times 10^{-12} \, \text{m} \] \[ r = 70.7 \times 10^{-12} \, \text{m} \] \[ r = 7.07 \times 10^{-11} \, \text{m} \]

4. Check the given options:

Let's cube the radius: \[ (7.07 \times 10^{-11})^3 = 353.4 \times 10^{-33} \] \[ \sqrt[3]{353.4 \times 10^{-33}} = \sqrt[3]{0.3534 \times 10^{-30}} = \sqrt[3]{0.353} \times 10^{-10} \]

Final Answer:
\( \sqrt[3]{0.353} \times 10^{-10} \) m.