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If a is the length of the side of a cube...

If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:

A

`(2)/(sqrt(3))a`

B

`(4)/(sqrt(3))r`

C

`(sqrt(3))/(4)a`

D

`(sqrt(3))/(2)a`

Text Solution

Verified by Experts

The correct Answer is:
4
The distance between the body centered atom and one corner atom in the cube is actually the nearest neighbour distance the distance between centres of touching spheres.
In body centered cubic unit cell ,the atom at the centre is in touch with the other two atoms diagonally arranged at the corners,if a is the lenght of the side of a cube , the length of a body diagonal is equal to `sqrt(3)a` .it is also equal to `4r`,where `r` is the radius of the sphere (atom) as all the three sphere (atoms) along the diagonal touch each other
Therefore
`sqrt(3)a=4r`
`r=(sqrt(3))/(4)a`
The nearest neighbour distance `(d)` is `2r`.
Therefore
`d=2r=2((sqrt(3))/(4)a)=sqrt(3)/(2)a` Note that nearest neighbouring distance `d` is half of the body diagonal.
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Knowledge Check

  • If a is the length of the side a cubic unit cell the distance between the body -centred atoms and one correct atoms in the cube will be

    A
    `(2)/(sqrt(3)) a`
    B
    `sqrt(3)/(2) a`
    C
    `(4)/(sqrt(3)) a`
    D
    `sqrt(3)/(4) a`

  • The distance between the body-centred atom and one corner atom in sodium is (a = 424 pm)

    A
    7.35 Å
    B
    1.5 Å
    C
    2.12 Å
    D
    3.67 Å

  • The distance between the body centred atom and a corner atom in sodium (a = 4.225 Å) is

    A
    `3.66 Å`
    B
    `3.17 Å`
    C
    `2.99 Å`
    D
    `2.54 Å`

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