A certain metal crystallizes in a bcc lattice | vedclass

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

Question

(B) In a $bcc$ lattice, the nearest neighbours are at a distance of $d_1 = \frac{\sqrt{3} a}{2}$, where $a$ is the edge length of the unit cell.Given

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(B) In a $bcc$ lattice, the nearest neighbours are at a distance of $d_1 = \frac{\sqrt{3} a}{2}$, where $a$ is the edge length of the unit cell.Given $d_1 = 433 	ext{ pm}$, we have $\frac{\sqrt{3} a}{2} = 433 	ext{ pm}$.Therefore, the edge length $a = \frac{2 	imes 433}{\sqrt{3}} = \frac{866}{1.732} \approx 500 	ext{ pm}$.The next nearest neighbours in a $bcc$ lattice are located at the corners of the unit cell along the edge, so the distance $d_2 = a$.Thus, the distance between the next nearest neighbours is $500 	ext{ pm}$.

$A$ certain metal crystallizes in a $bcc$ lattice. The distance between the nearest neighbours is $433 \text{ pm}$. The distance between the next nearest neighbours is :- .............. $\text{pm}$

Similar Questions

Which of the following contains rock salt structure?

$A$ compound $M_pX_q$ has a cubic close packing $(ccp)$ arrangement of $X$. Its unit cell structure is shown below. The empirical formula of the compound is

In a face centred cubic lattice,atom $A$ occupies the corner positions and atom $B$ occupies the face centre positions. If one atom of $B$ is missing from one of the face centred points,the formula of the compound is:

Platinum crystallizes in a face-centered cubic $(FCC)$ crystal with a unit cell edge length $a$. The distance between nearest neighbour atoms is

Explain the calculation of the number of atoms in a primitive cubic unit cell.

Vedclass Test Series

Exam Paper Generator

Online Exam Module