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Diameter of the iron atom

  1. Jun 9, 2014 #1
    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: $$\frac{0.071}{84,436,877}=8.4\times 10^{-8}$$
    The right answer is d=2.8E-10
     
  2. jcsd
  3. Jun 9, 2014 #2

    ehild

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    It is correct so far.

    What do you mean?????

    In what units???? Your answer is wrong anyway.

    ehild
     
  4. Jun 9, 2014 #3
    Don't drop the units. Keep the units throughout your calculation and you will see your mistake.
     
  5. Jun 10, 2014 #4
    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?
     
  6. Jun 10, 2014 #5

    ehild

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    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
  7. Jun 15, 2014 #6
    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
     
  8. Jun 15, 2014 #7

    haruspex

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    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
  9. Jun 15, 2014 #8

    tms

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    What am I missing? Wikipedia give the radius as 126 pm ##= 1.26 \times 10^{-10}## m.
     
  10. Jun 15, 2014 #9

    haruspex

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    Please see my correction to earlier post.
     
  11. Jun 15, 2014 #10
    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?
     
  12. Jun 15, 2014 #11

    ehild

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  13. Jun 17, 2014 #12

    DrClaude

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    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}##.
     
  14. Jun 17, 2014 #13

    ehild

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

    ehild
     
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