Calculating Cu Atom Radius in FCC Lattice
The problem asks for the radius of a Copper (Cu) atom which crystallizes in a Face-Centered Cubic (fcc) lattice. We are given the unit cell edge length, a = 361 pm.
Understanding FCC Lattice Geometry
In an fcc structure, atoms are arranged such that they touch each other along the face diagonal of the unit cell. The relationship between the atomic radius (r) and the unit cell edge length (a) in an fcc lattice is derived from this geometry.
The length of the face diagonal is $a\sqrt{2}$. Since the atoms touch along this diagonal, the diagonal length is equal to four times the atomic radius ($4r$).
Therefore, the relationship is:
$a\sqrt{2} = 4r$Solving for Atomic Radius
We can rearrange the formula to solve for the atomic radius, r:
$r = \frac{a\sqrt{2}}{4}$Alternatively, this can be written as:
$r = \frac{a}{2\sqrt{2}}$Applying Given Values
Substitute the given unit cell edge length, a = 361 pm, into the formula:
$r = \frac{361 \text{ pm}}{2\sqrt{2}}$Calculate the value:
$r = \frac{361 \text{ pm}}{2 \times 1.4142...}$ $r = \frac{361 \text{ pm}}{2.8284...}$ $r \approx 127.64 \text{ pm}$Conclusion
The calculated radius of the Cu atom is approximately 127.64 pm. Comparing this value with the given options, 127 pm is the closest value.
Important Questions from Solid State
The packing efficiency (in %) of spheres for a body-centered cubic (bcc) lattice is approximately
In NaCl crystal, the radius ratio is :
Minimum interplanar spacing required for Bragg’s diffraction is:
What does 'θ' represent in Bragg's Law?
Which of the following is molecular solid?
