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Calculate the void space in closest pack...

Calculate the void space in closest packing of `n` spheres of radius 1 unit, `n` spheres of radiys `0.414` units, and `2n` spheres of radius `0.225` units.

Text Solution

Verified by Experts

Since the packing is closest, therefore it is an `fcc`-type situation, `Z_(eff) = 4//"unit cell"`.
For `fc c, (4r)^(2) = 2a^(2)` (since `r = 1` unit)
`:. a = 2sqrt2`
Also, number of unit cell `= (n)/(4)`
Volume of crystal `= ((n)/(4)) xx a^(3) = (n)/(4) xx (2sqrt2)^(3)`
`= 4sqrt2n`
Volume of atoms `= (n)/(4)[(4xx(4)/(3)pi(1)^(3)+4xx(4)/(3)pi(0.414)^(3)),(+4xx2xx(4)/(3)pi(0.225)^(3))]`
`= 4.58n`
`:.` Percentage of space occupied `= (4.58n)/(4sqrt2n)xx100 = 81%`
Hence, void space `= 100 - 81 = 19%`
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