AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm . What is the distance between two oppositely charged ions in the lattice?

Question

AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm . What is the distance between two oppositely charged ions in the lattice? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For bcc lattice body diagonal =a√3 The distance between the two oppositely charged ions =(a/2) √3 =(387×1.732/2) =335 pm

Q. $A B$ crystallizes in a body centred cubic lattice with edge length $^{'} a^{'}$ equal to $387 p m$ . What is the distance between two oppositely charged ions in the lattice?

For $b cc$ lattice body diagonal $= a 3$
The distance between the two oppositely charged ions
$= \frac{a}{2} 3$
$= \frac{387 \times 1.732}{2}$
$= 335 p m$

Q. AB crystallizes in a body centred cubic lattice with edge length ′a′ equal to 387pm . What is the distance between two oppositely charged ions in the lattice?

For bcc lattice body diagonal =a3​ The distance between the two oppositely charged ions =2a​3​ =2387×1.732​ =335pm

Q. $A B$ crystallizes in a body centred cubic lattice with edge length $^{'} a^{'}$ equal to $387 p m$ . What is the distance between two oppositely charged ions in the lattice?

For $b cc$ lattice body diagonal $= a 3$
The distance between the two oppositely charged ions
$= \frac{a}{2} 3$
$= \frac{387 \times 1.732}{2}$
$= 335 p m$