A metal (M), crystallizes in $\text{fcc}$ lattice with edge length of $4.242 \text{ \AA}$. What is the radius of $\text{M}$ atom (in $\text{\AA}$)?
Show Hint
Remember the geometric relations for cubic unit cells:
- $\text{fcc}$: Atoms touch along the face diagonal: $4r = a\sqrt{2}$.
- $\text{bcc}$: Atoms touch along the body diagonal: $4r = a\sqrt{3}$.
- $\text{sc}$: Atoms touch along the edge: $2r = a$.
Step 1: State the relationship between the atomic radius and edge length in an $\text{fcc$ lattice.}
In a face-centered cubic ($\text{fcc}$) lattice, the atoms touch along the face diagonal.
The length of the face diagonal is $4r$, where $r$ is the atomic radius.
The face diagonal is also related to the edge length $a$ by the Pythagorean theorem: $d_{\text{face}}^2 = a^2 + a^2 = 2a^2$, so $d_{\text{face}} = a\sqrt{2}$.
Equating the two expressions for the face diagonal:
\[
4r = a\sqrt{2}.
\]
Step 2: Derive the formula for the atomic radius $r$.
The atomic radius $r$ is:
\[
r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}.
\]
Step 3: Substitute the given edge length and calculate the radius.
The edge length is $a = 4.242 \text{ \AA}$. We will use the approximation $\sqrt{2} \approx 1.414$ and observe that $4.242$ is very close to $3\sqrt{2} \approx 3(1.414) = 4.242$.
\[
r = \frac{4.242 \text{ \AA}}{2\sqrt{2}}.
\]
Using the approximation $4.242 \approx 3\sqrt{2}$:
\[
r \approx \frac{3\sqrt{2}}{2\sqrt{2}} \text{ \AA} = \frac{3}{2} \text{ \AA} = 1.5 \text{ \AA}.
\]
Using the direct calculation:
\[
r = \frac{4.242}{2 \times 1.414} \text{ \AA} = \frac{4.242}{2.828} \text{ \AA} \approx 1.500 \text{ \AA}.
\]
The radius of the $\text{M}$ atom is $1.5 \text{ \AA}$.
