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Question
With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
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Solution
In ccp, each cube consists of eight cubic components, number of atoms per unit cell in ccp is
Where Nc = No. of atoms at corner
Nf = No. of atoms at face centre
= Nc × Nf × contribution
= `8 xx 1/8 + 6 xx 1/2` = 4
Position of octahedral voids = Edge centre and body centre
Number of actahedral voids per unit cell in cubic close packing = `N_e xx 1/4 + N_b xx 1`
= `12 xx 1/4 + 1 xx 1` = 4
⇒ Number of octahedral voids = 4.
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