Question

For a BCC structure, if a = 351 pm, find r. Lithium forms a BCC structure having an edge length of a unit cell 351 pm, then find the atomic radius of lithium.

Answer 1

Relationship between the lattice constant (a) and atomic radius (r) for a BCC structure:

For a body-centered cubic (BCC) structure, the relationship between the lattice constant \( a \) and the atomic radius \( r \) is given by:

\[ a = \frac{4 \sqrt{2} r}{3} \]

Step 1: Solve for r:
To find the atomic radius \( r \), we can rearrange the formula:

\[ r = \frac{3a}{4 \sqrt{2}} \]

Step 2: Substitute the given value of \( a \):
The given lattice constant \( a \) for lithium is 351 pm (or 3.51 Å). Substituting this value into the equation, we get:

\[ r = \frac{3 \times 3.51 \, \text{Å}}{4 \sqrt{2}} \]

Step 3: Simplify the expression:
Now, simplify the expression:

\[ r \approx \frac{10.53}{4 \sqrt{2}} \, \text{Å} \]

Next, calculate \( 4 \sqrt{2} \) which equals approximately 5.656, so we get:

\[ r \approx \frac{10.53}{5.656} \, \text{Å} \approx 1.86 \, \text{Å} \]

Final Answer:
Therefore, the atomic radius of lithium in its BCC structure is approximately 1.53 Å.