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Posted by on Friday, November 15, 2013 at 3:21pm.

Determine for solid manganese (Mn) the following:

What is the atomic density (atoms/m) along the [110] direction of the Mn crystal?

What are the number of nearest neighbors around an atom in the Mn crystal?

Calculate the interplanar spacing (m) between {220} planes in the Mn crystal.

  • Chemistry - , Saturday, November 16, 2013 at 7:47am

    For molibdenium

    We see from the PT that Mo is a BCC material, so the face diagonal [011] has 2⋅(12) atoms over a distance of 2√a.

    The value of the lattice parameter a can be found in the same manner as the previous problem (noting that the BCC structure has only 2 atoms per unit cell rather than 4). Thus,

    1 atom2√a=12√(2VmolarNAv)1/3=2.24×107

  • Chemistry - , Saturday, November 16, 2013 at 12:56pm


  • Chemistry - , Wednesday, August 27, 2014 at 6:13am

    gg is right

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