# Diameter of the iron atom

## Homework Statement

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

## 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: $$\frac{0.071}{84,436,877}=8.4\times 10^{-8}$$

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ehild
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## Homework Statement

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

## The Attempt at a Solution

The volume of 55.847 grams is: $$\frac{55.847}{7.87}=7.1[cm^3]$$
It is correct so far.

I assume the NA number of atoms is arranged in a cube with an edge consisting of $$\sqrt[3]{6.02\times 10^{23}}\cong 84,436,877$$
What do you mean?????

The diameter: $$\frac{0.071}{84,436,877}=8.4\times 10^{-8}$$

ehild

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

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?

ehild
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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?
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

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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

haruspex
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Gold Member
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
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.

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

haruspex
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Gold Member
What am I missing? Wikipedia give the radius as 126 pm ##= 1.26 \times 10^{-10}## m.
Please see my correction to earlier post.

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
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?

ehild
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DrClaude
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}##.

ehild
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What if the expected solution was simpler?
It is quite possible, but wrong. The spheres do not fill the available volume.

ehild