Question

Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of the host atom, and compute the radius r of an impurity atom that will just fit into an BCC tetrahedral site in terms of the atomic radius R.

Answer 1

Answer:

FCC: r = 0.414R

BCC: r = 0.291R

Explanation:

For an FCC unit cell, the interstitial site is located at the middle of the edge. An atom that can occupy the interstitial site will have a diameter of 2*r. And we know that:

2*r = a - 2*R        equation (1.0)

a = $2*\sqrt{2}*R$

Therefore, substituting the expression for 'a' in equation (1.0)

2*r =  $2*\sqrt{2}*R$ - 2*R

r = R*($2\sqrt{2} - 2$)/2 = 0.414R

For a BCC unit cell, there is a right-angle triangle formed by 3 arrows. Using the triangle, we have:

$\frac{a^{2} }{2} +\frac{a^{2} }{4} = (R+r)^{2}$        equation (2.0)

a = $\frac{4R}{\sqrt{3} }$

replacing the expression for a in equation (2.0), we have:

$\frac{4R^{2} }{2\sqrt{3} } + \frac{4R^{2} }{4\sqrt{3} } = R^{2} + 2Rr + r^{2}$

Further simplification and rearrangement, the expression above is simplified to:

$r^{2} + 2Rr - 0.667R^{2} = 0$

Solving the above quadratic equation, we have:

$r = \frac{-2R - 2.582R}{2}or\frac{-2R + 2.582R}{2}$

r = - 2.291R or 0.291 R

Since the value of r can only be positive, the correct answer is r = 0.291R