If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:

Options

A. `2sqrt3a`
B. `4/sqrt3a`
C. `sqrt3/4a`
D. `sqrt3/2 a`

Answer

D

Explanation

In body-centred cubic (bcc), oppositely charged ions touch each other along the body diagonal

`therefore` Body diagonal, AE=`2r_(Cs^+)+ 2r_(Cl^-)`
But body diagonal =`sqrt3a`
(From right angled `triangleCDE`, CE=`sqrt(a^2+a^2)=sqrt2a`
From right angled `triangle ACE` , `AE=sqrt(AC^2+CE^2)=sqrt(a^2+2a^2)=sqrt(3a^2)=sqrt3a`)
`therefore 2(r_(Cs^+)+r_(Cl^-))=sqrt3a` or `r_(Cs^+) +r_(Cl^(-))=sqrt3/2a`