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The relation between atomic radius and e...

The relation between atomic radius and edge length 'a' of a body centred cubic unit cell :

A

r=a/2

B

`r=sqrt(a//2)`

C

`r=sqrt3/4a`

D

`r="3a"/2`

Text Solution

Verified by Experts

The correct Answer is:
C
Distance between nearest neighbours, `d=(AD)/2`
In right angled `triangle ABC, AC^2=AB^2+BC^2`
`AC^2=a^2+a^2` or `AC=sqrt2a`
Now in right angled `triangleADC`,
`AD^2=AC^2+DC^2`
`AD^2=(sqrt2a)^2 + a^2 = 3a^2 rArr AD=sqrt3a`
`therefore d=(sqrt3a)/2`
Radius , `r=d/2=sqrt3/4` a
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