# Diameter of the iron atom

1. Jun 9, 2014

### Karol

1. The problem statement, all variables and given/known data
55.847 grams of iron contain NA=6.02E23 atoms. the density is ρ=7.87 [gr/cm3]. What is the diameter of the iron atom

2. Relevant equations

3. The attempt at a solution
The volume of 55.847 grams is: $$\frac{55.847}{7.87}=7.1[cm^3]$$
I assume the NA number of atoms is arranged in a cube with an edge consisting of $$\sqrt[3]{6.02\times 10^{23}}\cong 84,436,877$$
The diameter: $$\frac{0.071}{84,436,877}=8.4\times 10^{-8}$$

2. Jun 9, 2014

### ehild

It is correct so far.

What do you mean?????

ehild

3. Jun 9, 2014

### dauto

Don't drop the units. Keep the units throughout your calculation and you will see your mistake.

4. Jun 10, 2014

### Karol

11

I think of the round atom to be a cube with edge d=the diameter of the atom, then the 7.1[cm3] volume is filled with cubes:
$d^3\cdot N_A=7.1\times 10^{-6}[m^3] \Rightarrow d=2.3\times 10^{-10}$
The answer should be 2.8E-10, how to account for the difference? is my method right? maybe because of the different arrangement of round atoms?

5. Jun 10, 2014

### ehild

Your result would be correct if the iron crystallized in simple cubic lattice. But it is body-centered cubic.
There are two atoms in the elementary cell, a cube. One atom is at each vertex, shared between 8 cells, and the other one is at the centre of the cube. The diameter of the atoms is half the body-diagonal of the cube. http://www.chemprofessor.com/solids_files/image027.jpg [Broken]

ehild

Last edited by a moderator: May 6, 2017
6. Jun 15, 2014

### Karol

gg

The BCC structure. every cube has in it one complete atom + $8\times\frac{1}{8}$ of atoms on the corners, which gives 2 atoms in a cube.
In the diagonal of the cube are 4 radii: $\sqrt{3}a=4r\rightarrow r=\frac{4}{\sqrt{3}}$
$$\left(\frac{4}{\sqrt{3}} r\right)^3\frac{N_A}{2}=7.1E-6\Rightarrow d=2r=2.07E-10$$
And still wrong

7. Jun 15, 2014

### haruspex

Some rounding error? Using your equations and numbers I get 2.5E-6. This agrees with the value at http://en.wikipedia.org/wiki/Iron
Typo: I meant 2.5E-10.

Last edited: Jun 15, 2014
8. Jun 15, 2014

### tms

What am I missing? Wikipedia give the radius as 126 pm $= 1.26 \times 10^{-10}$ m.

9. Jun 15, 2014

### haruspex

Please see my correction to earlier post.

10. Jun 15, 2014

### Kvothe8

Your equation seems correct to me, although I think a mistake was made with the final calculation. Using your equation, I get
$$d = 2.48*10^{-10}[m]$$
Maybe the "solution" you've been given is incorrect?

11. Jun 15, 2014

### ehild

12. Jun 17, 2014

### Staff: Mentor

What if the expected solution was simpler?

Starting as the OP with the total volume
$$\frac{55.847\ \mathrm{g}}{7.87\ \mathrm{g/cm}^3} = 7.10\ \mathrm{cm}^3$$
then, dividing by the number of atoms, we get the volume per atom
$$\frac{7.10\ \mathrm{cm}^3}{6.022 \times 10^{23}} = 1.18 \times 10^{-23}\ \mathrm{cm}^3 = 1.18 \times 10^{-29}\ \mathrm{m}^3$$
Assuming a spherical atom, we have
$$r = \left( \frac{3}{4 \pi} V \right) = \left( \frac{3}{4 \pi} 1.18 \times 10^{-29}\ \mathrm{m}^3 \right)^{1/3} = 1.41 \times 10^{-10}\ \mathrm{m}$$
thus a diameter of $d = 2.8 \times 10^{-10}\ \mathrm{m}$.

13. Jun 17, 2014

### ehild

It is quite possible, but wrong. The spheres do not fill the available volume.

ehild