Ask Question
18 July, 09:33

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

+3
Answers (1)
  1. 18 July, 09:41
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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, ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers