Question

Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

Answer 1

Hey there!

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Answer :

(i) In simple cubic only corners are occupied by points. Since each point on a corner is shared by eight unit cells, therefore, a point at the corner of a unit cell contributes only one-eighth $( \frac{1}{8})$ to the cell. There are 8 corners each occupied by one point.

Even though a total of $\frac{7}{8}$ part is there with neighbours, one neighbour cell only has $\frac{1}{8}$ part.

(ii) In bcc, complete atom is present at body centre.

Number of corner atoms = 8

Number of unit cells sharing atoms at the corner = 8

∴ Contribution to the unit cell = 8 × $\frac{1}{8}$ = 1

Number of atoms at the centre of the cube = 1

Contribution to the unit cell = 1 (as it is not shared)

Hence, the atom in the centre of the cube is not shared by other unit cell. In other words, zeroth (0) portion of an atom shared at body centre of cubic unit cell.