
Which of the following statements is/are correct about tetrahedral voids $(TVs)$ in an fcc unit cell
A.Number of TVs per atom in an fcc unit cell is $2$.
B.Number of TVs per unit cell is $8$.
C.Number if TVs is twice the number of atoms in the fcc unit cell.
D.Number of TVs is equal to the number of atoms in the fcc unit cell.
Answer
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Hint: At first we will talk about the tetrahedral voids. Then will look into the fcc unit cell. Then we will know in detail about each of the options given. Then we will find the best option given.
Complete Step by step answer: Step1. Tetrahedral voids- In hexagonal close packing and in the cubic close packing each sphere of the second layer touches the three spheres of the first layer. So it leaves a very small space which is called the tetrahedral site. In other words the vacant space between four touching spheres is called a tetrahedral void.
Step2- In the below layer three spheres are touched by a sphere and three are touched in the above sphere. So here there are two tetrahedral sites linked with that one sphere.
Step3. We must know that if there is a tetrahedral site it does not mean that it is tetrahedral in geometry but it means that it is surrounded by four spheres and at the center of it, there lies pieces of regular tetrahedron.
Step4- In FCC, one corner of the cell and face center of the cell form a tetrahedral.
In FCC, along one cubical diagonal, there are two TVs obtained. So in FCC, in total there are 8TVs present.
IN FCC, there are 4 total numbers of atoms present.
We deduce that in $3D$ close packing, there are $2$ tetrahedral voids attached to one atom.
Hence the correct option is C.
Note: The simple cubic unit cell is the most basic repeating unit in a simple cubic structure. Each corner is called a lattice point at which one atom can be found. All the corners of the unit cell have the same particle. Other particles can be present at faces of edges.
Complete Step by step answer: Step1. Tetrahedral voids- In hexagonal close packing and in the cubic close packing each sphere of the second layer touches the three spheres of the first layer. So it leaves a very small space which is called the tetrahedral site. In other words the vacant space between four touching spheres is called a tetrahedral void.
Step2- In the below layer three spheres are touched by a sphere and three are touched in the above sphere. So here there are two tetrahedral sites linked with that one sphere.
Step3. We must know that if there is a tetrahedral site it does not mean that it is tetrahedral in geometry but it means that it is surrounded by four spheres and at the center of it, there lies pieces of regular tetrahedron.
Step4- In FCC, one corner of the cell and face center of the cell form a tetrahedral.
In FCC, along one cubical diagonal, there are two TVs obtained. So in FCC, in total there are 8TVs present.
IN FCC, there are 4 total numbers of atoms present.
We deduce that in $3D$ close packing, there are $2$ tetrahedral voids attached to one atom.
Hence the correct option is C.
Note: The simple cubic unit cell is the most basic repeating unit in a simple cubic structure. Each corner is called a lattice point at which one atom can be found. All the corners of the unit cell have the same particle. Other particles can be present at faces of edges.
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