Question

For both FCC and BCC crystal structures, there are two different types of interstitial sites, one larger and one smaller. For FCC, the larger site is octahedral and is located at the center of each edge of the unit cell. For BCC, the larger site is tetrahedral and is found at 0, 1/2, 1/4 positions - that is, centered on (100) planes, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges. For both FCC and BCC, compute the maximum radius of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.

Answer 1

For FCC, the radius of impurity=0.41 R .

For BCC, the radius of impurity= 0-291 × R

Step-by-step explanation:

a) Given: radius of host atom= R

For FCC crystal structure

The first attached image shows the crystal structure.

From the crystal structure, we can say b -2R=2r

Make r the subject of the formula:

r =
`(b-2R)/(2)`

The lattice parameter can be expressed as b=2R ×√2

Then, solve for r by inputing the value of b;

r =
`(b-2R)/(2)`

r =
`(2R√(2) - 2R )/(2)`

r = 0.41R

b) For BCC crystal structure,

The second attached image shows the crystal structure.

Using Pthygoras theorem for the right hand trangle in the second image:

Hyp² = opp² + adj²

where: hyp² = (R + r)²

opp² =
`((b)/(4) )^(2)`

adj² =
`((b)/(2) )^(2)`

therefore: (R + r)² =
`((b)/(4) )^(2)` +
`((b)/(2) )^(2)`

The lattice parameter is expressed here as b =
`(4R)/(√(3) )`

Substitute the value of b

Therefore;

(R + r)² =
`((4R)/(√(3) ) /4)^(2) +( (4R)/(√(3) )/2)^(2)`

R² + r² + 2Rr =
`((4R)/(4√(2) ) )^(2) +((4R)/(2√(2) ))^(2)`

48R² + 48r² +96Rr = (16R² × 4) + (16R² × 1)

48R² + 48r² +96Rr = 64R² + 16R²

0 = 32R² - 48r² - 96Rr

Divide the 16

0 = 2R² - 3r² - 6Rr

3r² - 2R² -6Rr = 0

Divide by 3

r² + 2Rr -0.66R² = 0

Therefore,

r = 0.291 × R

Answer 2

Step-by-step explanation:

The diameter of an atom that will just fit into this site (2r) is just the difference between that unit cell edge length (a) and the radii of the two host atoms that are located on either side of the site (R); that is

2r = a – 2R

However, for FCC a is related to R according to Equation 3.1 as

a=2R*sqrt(2)

Therefore, solving for r from the above equation gives

r = (a-2R)/2 = [(2R*sqrt(2)) – 2R]/2 = 0.41R

The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this (100) face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write

(a/2)^2 + (a/4)^2 = (R+r)^2

However, from Equation 3.3,

a = 4R/sqrt (3)

Therefore, making this substitution, the above equation takes the form

(4R/2*sqrt(3))^2 + (4R/4*sqrt(3))^2 = R^2 + 2Rr + r^2

After rearrangement the following quadratic equation results:

r^2 + 2Rr – 0.667 R^2 = 0

And upon solving for r:

r = -(2R) ± sqrt{(2R)^2 – 4*1*-0.667R^2} / 2

= -2R ± 2.582R / 2

And, finally

r(+) = -2R + 2.582R/2 = 0.291R

r(-) = -2R – 2.582 R / 2 = -2.291R

As only the r(+) root is possible, therefore, r = 0.291R.

Thus, for a host atom of radius R, the size of an interstitial site for FCC is approximately 1.4 times that for BCC.

cheers, i hope this helps....