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# X atoms are present at corners in a cube and Y atoms are at face centres of a cube. Then the number of the X and Y atoms present in the cube is:(A) $8,8$(B) $6,8$(C) $8,6$(D) $6,6$

Last updated date: 07th Sep 2024
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Hint:Each atom at the corner is shared by eight adjacent unit cells and each atom at the centre of faces is shared between two adjacent unit cells. There are eight corner atoms and six face-centred atoms in a simple cubic unit cell.

Now in the question, it is given that X atoms are present in the corners and Y atoms at the face-centred.
We know that among the categories of unit cells, face-centred unit cells contain one particle present at the centre of each face besides the atoms that are at its corners.
So from the question, we can understand that the packing is fcc where X atoms are at corners and Y atoms at face centres of the cube.
Therefore a face-centred unit cell contains atoms at all the corners and the centre of all faces of the cube.
In the question, it is given that the X atoms are present at the corners in a cube. We know that there are eight corners in a cube and atoms will be present there. So eight atoms at the corners.
Since X atoms are present at corners of the cube, the number of X atoms is eight.
We also know that there are six faces for a cube and the Y atoms will be present at each face of the cube.
Since six faces are there, the number of Y atoms present in the faces of the cube is six
Thus the answer is an option (C) $8,6$ where the number of X atoms is eight and the number of Y atoms is six.

Note:
We can calculate the number of X atoms in corners and the number of Y atoms at face centres simply by knowing how many corners and how many faces are there in a cube. There are eight corners and six faces for a cube. Thus the number of X atoms present at the corners is eight and the number of Y atoms present at faces is six.