Consider a face-centered cubic crystal with a host lattice containing atoms of r

Question

Consider a face-centered cubic crystal with a host lattice containing atoms of radius 1.36 Angstroms. One type of interstitial site found in FCC crystals is called an octahedral site, and there are four of these sites per FCC unit cell. 3 of them - total - lie along the 12 edges of the unit cell (within the gap that exists between corner atoms that share a common edge) and 1 is in the exact middle of the cubic unit cell.

If 21 percent of these interstitial sites are filled by impurities, what is the concentration of impurities in units of impurities per cubic centimeter (1/cm3) ?

Note: there are no units for "number of impurities", as this is merely a count.

Explanation / Answer

side of fcc = sqrt8 R

= sqrt8 * 1.36

= 3.8467angstrom

volume of cube = [3.8467e-8 cm]^3

    = 5.692e-23     cm^3

concentration of impurities = 4*0.21/[5.692e-23     cm^3]

= 1.48 *10^22    /cm^3

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.