Question

Question #dfe68

Answer 1

`"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 `1/12"th"` of the mass of a single, unbound carbon-12 atom , is equivalent to approximately

`"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 `1/8"th"` of an atom and the face lattice points will each contain `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

`"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 * 10^(-23)"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

`color(blue)(V = l xx l xx l = l^3)" "`, 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 = root(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

`"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.