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AB crystallizes in a body-centered cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ions in the lattice is:

Option: 1

335 pm


Option: 2

250 pm


Option: 3

200 pm


Option: 4

300 pm


Answers (1)

best_answer

Given, 

AB crystallizes in a body-centred cubic lattice. It means A will be at the body centre and B will be at the corner of the lattice or vice versa.

So, The distance between two oppositely charged ions in the lattice will be half of the body diagonal.

For body-centred cubic (BCC):

Lattice points: at corners and body centre of the unit cell.

In a bcc lattice, it is clear that the atom at the centre will be in touch with the other two atoms diagonally arranged

The length of the body diagonal c is equal to \mathrm{\sqrt{3}a}

So, the distance between the body centre and the corner atom will be half of the body diagonal c.

\mathrm{\textup{so, distance will be }= \frac{c}{2}=\frac{\sqrt{3}}{2}a}

\mathrm{\textup{AB distance }=\frac{\sqrt{3}}{2}\times387 \ pm}

\mathrm{\textup{AB distance }=335 \ pm}

Hence, option number (1) is correct.

Posted by

Divya Prakash Singh

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