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Atomic radii for irridium

  1. Jun 15, 2006 #1
    hey just wondering where I went wrong

    Iridium Crystallizes in a face centred cubic unit call that has an edge length of 3.833A
    The atom in the cantre of the face is in contact with the corner atoms

    calculate the atomic radii of the irridium atom
    I did this
    using pythag, (3.833^2)+(3.833^2)=c^2
    b^2= 29.384
    Then I divided it by 2 as it was the radius and got 2.710
    But I am assuming that this is wrong due to the qu asking for the atomic radii. So I divide it by 192.272 and I still get the wrong answer.
    The answer is meant to be 1.355Angstrom

    So if anyone could tell me where I went wrong I would be much appreciative
  2. jcsd
  3. Jun 15, 2006 #2


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    Gold Member

    Hello Taryn,

    try to revisualise one face. How many atom-radii will fit into the length of the face-diagonal? You are close to the correct answer. :smile:


    Last edited: Jun 15, 2006
  4. Jun 15, 2006 #3
    figured it out, u divide by 2 as the atomic radii is being asked for and there are two atoms. So simple!
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