Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel and (b) the octahedral interstitial site in BCC lithium.


(a) For the tetrahedral site in FCC nickel $(a_0 = \pu{3.5167 Å}):$

$$r_\ce{Ni} = \frac{\sqrt{2}(\pu{3.5167 Å})}{4} = \pu{1.243 Å}$$

$r/r_\ce{Ni} = 0.225$ for a tetrahedral site. Therefore:

$$r = (\pu{1.243 Å})(0.225) = \pu{0.2797 Å}$$

(b) For the octahedral site in BCC lithium $(a_0 = \pu{3.5089 Å}):$

$$r_\ce{Li} = \frac{\sqrt{3}(\pu{3.5089 Å})}{4} = \pu{1.519 Å}$$

$r/r_\ce{Ni} = 0.414$ for an octahedral site. Therefore:

$$r = (\pu{1.519 Å})(0.414) = \pu{0.629 Å}$$

In the BCC crystal lattice the radius ratio for the octahedral site, as I learnt it is $~0.155,$ which is derived taking into account the distance between two body centered atoms.

However, in the above answer the radius ratio used is $~0.414,$ which as I've learnt is the radius ratio for the octahedral interstitial site in a FCC crystal lattice.

Is the answer given in the text a mistake, or have I understood something wrong?

  • 1
    $\begingroup$ Please typeset your question: others who might search for this material will be impeded by the image. $\endgroup$ Jul 10, 2021 at 13:13
  • $\begingroup$ It’s a mistake, m.youtube.com/watch?v=rRGJ8zoGYEI&feature=youtu.be $\endgroup$
    – Karsten
    Jul 10, 2021 at 13:35
  • 1
    $\begingroup$ Shouldn’t that be ‘maximum’ radius that will just fit? The minimum radius is 0. $\endgroup$
    – Jon Custer
    Jul 10, 2021 at 15:14

1 Answer 1


Actualluy it was a mistake. The impurities usually fall into the larger interstitial sites of crystal structures. According to this problem solving chapter, the interstitial sites in FCC and BCC crystal structures are described as follows:

For both FCC and BCC crystal structures, there are two different types of interstitial sites. Of these two interstitial sites in each crystal structures, the one site is always larger than the other and is normally occupied by impurity atoms.

Karsten Theis has given excellent video to show which one of the two interstatial sites is the larger than other, with calculations to show that (in here, the radius of the host atom is $R$ while that of the guest or thw impurity atom is $r$). Thus, the octahedral interstitial site with $\frac{r}{R}$ ratio of $0.414$ in FCC is the larger one between two sites, while that in the BCC is the tetrahedral interstitial site with $\frac{r}{R}$ ratio of $0.291$.

Note that the $\frac{r}{R}$ ratio of the tetrahedral interstitial site of FCC is the value of $0.225$. The $\frac{r}{R}$ ratio of the octahedral interstitial site of BCC is equal to the value of $0.154$.

According to the answer given in the solution manual, it has been using both interstitial sites of only FCC crustal structure for the calculations.

The different view of two larger sites in bothe crustal units are shown in following image:

FCC and BCC interstitial sites

  • $\begingroup$ A few typos: 1st line Actually, and in bold cyrstal. $\endgroup$
    – Rishi
    Jul 11, 2021 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.