A metal X crystallises in a face-centred cubi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a face-centred cubic $(FCC)$ unit cell, the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2}a$, where $a$ is th

Vedclass · Accepted Answer

$609.6$

Vedclass · Answer

(D) In a face-centred cubic $(FCC)$ unit cell, the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2}a$, where $a$ is the edge length.The distance between the centers of two nearest neighbor atoms (shortest separation of nuclei) is equal to half of the face diagonal, which is $\frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Given $a = 862 \, pm$.Shortest separation $= \frac{862}{1.414} \approx 609.6 \, pm$.

$A$ metal $X$ crystallises in a face-centred cubic arrangement with the edge length $862 \, pm$. What is the shortest separation of any two nuclei of the atom? ........... $\, pm$

Similar Questions

Potassium crystallizes with a

In a face-centered cubic lattice,a unit cell is shared equally by how many unit cells?

If the radius ratio for an ionic solid is $0.5248$ and the radius of the cation is $0.95 \ \mathring{A}$,what is the radius of the anion?

What is the total number of unit cells shared by each corner particle of a $bcc$ unit cell?

The space lattice of $CaF_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module