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What is the volume of an iron atom?

  1. Sep 22, 2004 #1
    Plz help ASAP!!!

    Iron has a mass of 7.87g/cm^3 of volume, and the mass of an iron atom is 9.27x10^(-26)kg. If the atoms are spherical and tightly packed,
    a) a) what is the volume of an iron atom? --> this one I do know how to do it, the answer is 1.18x10^29 m^3
    b) what is the distance between the centers of adjacent atoms?
    This one, my answer is different than the answer in the book
    This was how I did it:
    b) [(1.18 x 10^29 m^3) x 3/4] / pi = r^3
    r= 3.04 x 10^9 m
    => the distance between the centers of adjacen atoms is 6.08 x 10^9 meters
    ****but the answer of the book is 0.282 nm
    Would someone plz show me how to do part b of the problem? Thanks so much
     
  2. jcsd
  3. Sep 22, 2004 #2
    Looks good to me... Are you sure the book contains the right answer?
     
  4. Sep 22, 2004 #3
    For a), I assume you meant [itex]1.18\times10^{-29}[/itex]. All of the other exponents should also be negative, I believe.

    I get [itex]6.05\times10^{-10}[/itex] m for b. I think there's an error in the book. You used the right method.
     
  5. Sep 22, 2004 #4
    got it!!!

    I figured it out!!
    Sirus was right, the radius is 1.18X10^-29. The you jst multiply that by 3/4pi and then take the third root. Finally, you multiply it by 2 and get 2.82X10^-10, which is 282nm.
     
  6. Sep 22, 2004 #5
    Right, I was putting pi in the numerator in my calculator (hytuoc, I suspect this was also your error). Book was right.
     
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