The question is: The density of bcc iron is 7900 kg/m3, and its atomic wieght is

Question

The question is: The density of bcc iron is 7900 kg/m3, and its atomic wieght is 56 amu. Using this information, calculate the lattice constant of iron's cubic unit cell and the interatomic spacing (i.e. nearest neighbor distance).

I thought this was mostly unit conversion.

7900 kg/m^3 = 4.757*10^30 amu/m^3
4.757*10^30 amu/m^3 = 8.49552*10^28 atoms of iron / m^3
8.49552*10^28 atoms of iron / m^3 = 4.396057*10^9 atoms of iron / m
4.396057*10^9 atoms of iron / m = 0.439606 atoms of iron / angstrom

Taking the reciprocal of the final answer should get us the space between atoms, which gives me 2.27 angstroms between atoms of iron. I thought this should be the lattice constant, however I'm guessing I've done something wrong since I believe the answer should be 2.86 angstroms.

Any help would be appreciated. Thanks.

Explanation / Answer

Hei,

Yes you are right. The answer is aproximately 2.86 angstroms or better 0,286 nm (nanometer) apologies, i never liked the angstrom unit :D

In order to calculate the lattice constant, the following information are needed.

1. The question has made it clear that iron has a BCC crystal structure and the number of atoms forming a simple

Body-centered cubic crystal is 2. this implies that we have 2 iron atoms.

2. We are given the atomic weight of Fe to be 56 amu. all we need do is convert it to kilogram

      

          i.e mass of Fe = 56 amu * 1.67E-27 kg /amu = 9.35E-26 kg

          (where E denotes 10 to the power)

3. As the name implies body centered cubic, this tells us that the lattice shape is cubic in nature. from basic geometry, we know that the volume of a cube is (L * L x*L)      where L is the length of the cube.

        This length L is the lattice constant for the crystal or the length of the side of the unit cell as denoted in some text book.

      

4. From classical physics,
        density = mass / volume

      for Fe having 2 atoms, the density would be = 2 * mass / volume

                          

and volume = 2 * mass / density                                                        

                                      = 2* 9.35E-26 kg / (7900 (kg/m^3))

                                       2.367E-29 m^3

                and L = lattice constant would be cube root of volume =   0.287 nm

For the nearest neighbour atom at point (x)

           point 'x' is the middle of the BCC cube, taking from geometry , position of a point in the middle of a cube

           is   x = sq root (3)   * length of the cube/ 2

                    = 1.732 * 0.287nm /2

                    = 0.2485E-9

I hope my solutions were explanatory enough.

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