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2 April, 14:05

Assume the radius of an atom, which can be represented as a hard sphere, is r  1.95 Å. The atom is placed in a (a) simple cubic, (b) fcc, (c) bcc, and (d) diamond lattice. As-suming that nearest atoms are touching each other, what is the lattice constant of eachlattice

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  1. 2 April, 15:33
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    Answer: Simple cubic=0.39nm

    Face centred cubic

    =0.55nm

    Body centered cubic

    =0.45nm

    Diamond lattice = 0.9nm

    Step-by-step explanation: The lattice constant (a)

    for SC=2*r

    Fcc=4*r/√2

    Bcc = 4*r/√3

    Diamond lattice=8*r/√3

    Here,

    r is the atomic radius measured in nm

    r = 1.95Å * 1nm/10Å

    =0.195nm

    Now let's calculate (a)

    SC = 2*r = 2*0.195 nm=0.39nm

    Fcc = 4*r/√2 = 4*0.195nm/√2

    = 0.55nm

    Bcc = 4*r/√3 = 4*0.195nm/√3

    = 0.45nm

    Diamond lattice = 8*r/√3

    =8*0.195nm/√3

    = 0.9nm
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