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Consider a metal with atomic radius 1 angstrom, which crystallizes in a conventional fcc unit cell. To one decimal place and in angstroms, what is the unit cell edge length of the fcc lattice?

Respuesta :

Answer:

[tex]a=2.8 angstroms[/tex]

Explanation:

In the figure can be seen the Face centered cubic (FCC) structure. It contains 4 halves of an atom and 8 "1/8" of atom.

The side lengh:

[tex]b=4*r[/tex]

Using pithagoras:

[tex]b^2=a^2+a^2[/tex]

[tex](4*r)^2=2*a^2[/tex]

[tex]a=\sqrt{\frac{(4*r)^2}{2}}[/tex]

[tex]a=\sqrt{8}*r[/tex]

[tex]a=\sqrt{8}*1 ang.=2.8 angstroms[/tex]

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