A given element A has a face centred unit cell in which A is absent from two of the corners.
The packing fraction is :
A
7π2√2
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B
5π16√2
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C
5π8√2
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D
7π4√2
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E
7π2√2
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F
5π16√2
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G
5π8√2
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H
7π4√2
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Solution
The correct options are B5π16√2 F5π16√2 Contribution for a corner particle is =18 Contribution for a face centre particle is =12
Total no. of particles :
Zeff=(6×18)+(6×12)=154
For a FCC unit cell : Edge length(a)=2√2×r
Here r is the radius of particle A Total volumeV=a3=(2√2×r)3V=16√2×r3 Volume occupied(Vo)=Zeff×43×π×r3Vo=154×43×π×r3 Packing Fraction (P.F)=VoVP.F=154×43×π×r316√2×r3