# Question #c698b

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to use the definition of the *unified atomic mass unit*, **single** nickel atom, then use the characteristics of a *face-centered cubic structure* to find the **total mass** of a unit cell.

Once you know the mass of a cell, use nickel's density to determine the cell's volume.

So, the unified atomic mass unit is defined as

#"1 u" = 1.660539 * 10^(-24)"g"#

The mass of a single nickel atom expressed in *grams* will thus be

#58.7 color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-24)"g")/(1color(red)(cancel(color(black)("u")))) = 9.747 * 10^(-23)"g"#

Now, a **face-centered cubic cell structure** is characterized by a total of

one lattice point in every corner of the cell#-># a total ofeightcorner lattice pointsone lattice point in every face of the cell#-># a total ofsixface lattice points

Each corner lattice point will hold **total number of atoms** of nickel that can be found in a face-centered cubic cell will thus be

#"no. of atoms" = overbrace(1/8 * 8)^(color(blue)("corners")) + overbrace(1/2 * 6)^(color(red)("faces")) = "4 atoms"#

So, if each cell contains a total of **total mass** of a cell will be

#4 color(red)(cancel(color(black)("atoms"))) * (9.747 * 10^(-23)"g")/(1color(red)(cancel(color(black)("atom")))) = 3.899 * 10^(-22)"g"#

Use the given density to find the **volume** of a unit cell

#3.899 * 19^(-22) color(red)(cancel(color(black)("g"))) * "1 cm"^3/(8.9 color(red)(cancel(color(black)("g")))) = 4.381 * 10^(-23)"cm"^3#

As you know, the volume of a cube is given by the formula

#color(blue)(V = l xx l xx l = l^3)" "# , where

This means that you have

#l = root(3)(V)#

In your case, this is equivalent to

#l = root(3)(4.381 * 10^(-23)"cm"^3) = 3.525 * 10^(-7)"cm"#

The length of the unit cell is usually given in *nanometers*, so convert the result to get

#3.525 * 10^(-8) color(red)(cancel(color(black)("cm"))) * (10^7"nm")/(1color(red)(cancel(color(black)("cm")))) = "0.3525 nm"#

Rounded to two sig figs, the number of sig figs you have for nickel's density, the answer will be

#l = color(green)("0.35 nm")#