Question #dfe68

1 Answer
Nov 23, 2015

Answer:

#"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.

https://penyayangbercahaya.wordpress.com/2010/03/09/packing-efficiency/

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.