A metal crystallises in a face-centred cubic | vedclass

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

Question

(C) In a face-centred cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.Let the radius of each atom be '$r$'.The length of th

Vedclass · Accepted Answer

$\frac{a}{\sqrt{2}}$

Vedclass · Answer

(C) In a face-centred cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.Let the radius of each atom be '$r$'.The length of the face diagonal is given by $\sqrt{2}a$.Since the atoms along the face diagonal touch each other,the length of the face diagonal is also equal to $4r$.Therefore,$4r = \sqrt{2}a$.The closest approach between two atoms is the distance between their centers,which is $2r$.From $4r = \sqrt{2}a$,we get $2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.

$A$ metal crystallises in a face-centred cubic $(FCC)$ structure. If the edge length of its unit cell is '$a$',the closest approach between two atoms in the metallic crystal will be:

Similar Questions

Which of the following companies started selling Humulin in $1983$?

If $x y \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$, where $m, n \notin N$, then $\frac{d y}{d x}$ is equal to

Where is the Central Rice Research Institute located?

The energy $(E)$,angular momentum $(L)$,and universal gravitational constant $(G)$ are chosen as fundamental quantities. The dimensions of the universal gravitational constant in the dimensional formula of Planck's constant $(h)$ is

Which of the following represents the correct electronic configuration of a neutral $Carbon$ atom in its ground state?

Vedclass Test Series

Exam Paper Generator

Online Exam Module