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Distance between atoms based on density

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 * 10^-26 kg. If you simplify and treat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?

    2. Relevant equations



    3. The attempt at a solution

    I managed to come up with (a) on my own, which was:

    1.17789 * 10^-29 m

    However, for (b), I'm confused as to exactly how it got the answer:

    0.282 nm = 0.282 * 10^-9 m

    At first I treated each atom as a cube and got some answers that were way off.. I took the hint and it mentioned that you should treat each atom as a sphere (WTF? It says cube in the problem, but whatever), so using the formula for that, I got

    V = (4/3)*pi*r^3 = 1.17789 * 10^-29 m

    Solving for r^3 I had

    r^3 = 0.281200696 * 10^-29

    Now solving for r in this case was problematic because depending on where you place the decimal point, you get a wildly different answer for the cubed root of this figure, in your scientific calculator. But the figure looks similar to the actual answer, except that the magnitude was cube-rooted or something..

    I'm still pretty stumped... Could someone explain what I'm doing wrong?

    UPDATE: Never mind, I found out that 1) the problem was stated correctly, the hint was a red herring, and 2) I can use Mathematica to find the proper cube-root of numbers in scientific notation.
     
    Last edited: Sep 12, 2009
  2. jcsd
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