# Question dfe68

Nov 23, 2015

$\text{350 pm}$

#### Explanation:

The idea here is that you need to use the atomic weight of nickel to determine the mass in grams of a single nickel atom, then use the characteristics of a face-centered cubic lattice to determine what the mass of a unit cell is.

Once you know that, use the given density to determine the unit cell's volume.

So, the unified atomic mass unit, $u$, which is defined as the equivalent of $\frac{1}{12} \text{th}$ of the mass of a single, unbound carbon-12 atom , is equivalent to approximately

$\text{1 u" = 1.660539 * 10^(-27)"kg}$

This means that the mass of a single nickel atom, expressed in grams, will be

58.7color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("u")))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = 9.75 * 10^(-23)"g"

Now, a face-centered cubic lattice is characterized by the fact that it contains a total of $14$ lattice points

• a total of eight lattice points that correspond to the corners of the unit cell
• a total of six lattice points that correspond to the faces of the unit cell

Now, the corner lattice points will each contain $\frac{1}{8} \text{th}$ of an atom and the face lattice points will each contain $\frac{1}{2}$ of an atom.

This means that you can estimate the total number of atoms that fit in a face-centered cubic cell by using

$\text{no. of atoms" = overbrace(1/8 xx 8)^(color(red)("8 corners")) + overbrace(1/2 xx 6)^(color(green)("6 faces")) = "4 atoms}$

Use the mass of single nickel atom to estimate the mass of a cubic unit cell**

$9.75 \cdot {10}^{- 23} \text{g"/color(red)(cancel(color(black)("atom"))) * (4color(red)(cancel(color(black)("atoms"))))/"unit cell" = 3.9 * 10^(-22)"g/unit cell}$

Now that you have an idea about what the mass of a unit cell is, you can use nickel's density to determine what its volume would be.

3.9 * 10^(-22)color(red)(cancel(color(black)("g"))) * overbrace( "1 cm"^3/(8.9color(red)(cancel(color(black)("g")))))^(color(blue)("density")) = 4.38 * 10^(-23)"cm"^3

Finally, the volume of a cube is given by the formula

$\textcolor{b l u e}{V = l \times l \times l = {l}^{3}} \text{ }$, where

$l$ - the length of the cube

This means that the length of the unit cell will be equal to

$V = {l}^{3} \implies l = \sqrt[3]{V}$

l = root(3) ( 4.38 * 10^(-23)"cm"^3) = 3.5 * 10^(-8)"cm"

If you want, you can convert this value to picometers by using the fact that

$\text{1 pm" = 10^(-12)"m" = 10^(-10)"cm}$

3.5 * 10^(-8)color(red)(cancel(color(black)("cm"))) * "1 pm"/(10^(-10)color(red)(cancel(color(black)("cm")))) = color(green)("350 pm")#

The answer is rounded to two sig figs, the number of sig figs you have for the density of nickel.