1.

In a body centered unit lattice of A_(2) type

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The edge length is equal to `(4R)/sqrt(3)`
The edge length is equal to `2R^(2)+2R^(2`
The edge length is equal to `(2d)/sqrt(3)` where, d=nearest neighbouring distances
The square of the edge length is equal to `16R^(2)` where 'R' is radius of atom.

Solution :In a body CENTRED cubic, close packing the sphers lying on the body diagonal touch each other.
Here, AB is the body diagonal
`(AB)^(2)=(FD)^(2)+("edge")^(2)=(BC)^(2)+(AC)^(2)`
`=2("edge")^(2)+("edge")^(2)=3("edge")^(2)`
`(4R)^(2)=3a^(2)=16R^(2)`=(Body Diagonal)`.^(2) RARR 4R=sqrt(3)a`
`rArr a=(4R)/sqrt(3)` Hence, CHOICE (a) is correct.
(c) Also 2R=d=nearest neighbouring distance `rArr`
`a=(2d)/sqrt(3)` Hence, choice (c) is correct.
(b) is wrong because the SPHERES are not touching each other.
(d) Edge, length: Choice (d) is also wrong because
`a^(2)=(16R^(2))/(3)`


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