Question

Nickel has an fcc structure with a density of .

(a) Calculate the nearest neighbor distance in crystalline nickel.

(b) What is the atomic radius of nickel?

(c) What is the radius of the largest atom that could fit into the interstices of a nickel lattice, approximating the atoms as spheres?$\mathbf{8}\mathbf{.}\mathbf{90}{\mathbf{\text{g cm}}}^{\mathbf{-}\mathbf{3}}$

Short answer

a) The nearest neighbor distance in crystalline nickel is$11.8\stackrel{°}{A}$.

b) The atomic radius of nickel is $5.9\stackrel{°}{A}$.

c) The radius of the largest atom that could fit into the interstices of a nickel lattice is$1.24\stackrel{°}{A}$

Step-by-step solution

Subpart (a) The nearest neighbor distance in crystalline nickel.

As density;

$\rho =\frac{\text{zM}}{{\text{a}}^{\text{3}}{\text{N}}_{\text{A}}}$

Here,

Given the density is$8.90{\text{g cm}}^{\text{- 3}}$

For fcc; z = 4

$\text{M(Ni)}=58.7g$

Putting the values in equation;

$8.90{\text{g cm}}^{\text{- 3}}=\frac{4\times 58.7g}{{\text{a}}^{\text{3}}{\text{×6.022×10}}^{23}}$

Therefore;

$$\begin{gathered} {\text{a}}^{\text{3}}=\frac{4\times 58.7\text{g}}{{\text{8.90g cm}}^{\text{- 3}}{\text{×6.022×10}}^{23}} \\ {\text{a}}^{\text{3}}=\frac{\text{234.8g}}{{53.610}^{23}{\displaystyle \times }{\text{g cm}}^{\text{- 3}}} \\ {\text{a}}^{\text{3}}=43.8\times {10}^{-24}{\text{cm}}^{\text{3}} \end{gathered}$$

And so;

$\text{a}=3.52\times {10}^{-8}\text{cm}$

Changing the unit;

$1\stackrel{°}{A}={10}^{-8}\text{cm}$

$1\text{cm}={10}^8\stackrel{°}{A}$

$$\begin{gathered} 3.52\times {10}^{-8}\text{cm}=3.52\times {10}^{-8}\times {\text{10}}^8\stackrel{°}{A} \\ 3.52\times {10}^{-8}\text{cm}=3.52\stackrel{°}{A} \end{gathered}$$

In the case of fcc, the nearest distance is calculated as;

$$\begin{gathered} \text{d}=\frac{\text{a}\sqrt{\text{2}}}{2} \\ \text{d}=\frac{3.52\stackrel{°}{A}\sqrt{2}}{2} \\ \text{d}=11.8\stackrel{°}{A} \end{gathered}$$

Subpart (b) The atomic radius of nickel.

The atomic radius of nickel is

$$\begin{gathered} \text{r}=\frac{\text{d}}{\text{2}} \\ =\frac{11.8\stackrel{°}{A}}{2} \\ =5.9\stackrel{°}{A} \end{gathered}$$

Subpart (c) The radius of the largest atom that could fit into the interstices of a nickel lattice.

In the case of fcc, the interstitial site is

Here,

$$\begin{gathered} =2\sqrt{2}r \\ AB=a=2(r+R) \end{gathered}$$

So,

$$\begin{gathered} 2(r+R)=2\sqrt{2}r \\ (r+R)=\sqrt{2}r \\ R=r\left(\sqrt{2}-1\right) \end{gathered}$$

Also, as in fcc

$\text{r =}\frac{a\sqrt{2}}{4}$

Therefore;

$$\begin{gathered} R=r\left(\sqrt{2}-1\right) \\ R=\frac{a\sqrt{2}}{4}\left(\sqrt{2}-1\right) \\ R=\frac{2a-a\sqrt{2}}{4} \\ R=\frac{a\sqrt{2}}{4} \end{gathered}$$

As calculated

$a=3.52\stackrel{°}{A}$

Hence;

$$\begin{gathered} R=\frac{3.52\stackrel{°}{A}\sqrt{2}}{4} \\ R=1.24\stackrel{°}{A} \end{gathered}$$