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Force required to break the atomic bond

  1. Dec 31, 2011 #1
    I have a two questions.

    1. What is the force (not energy) required to break the H-O atomic bond?
    2. What is the minimum additional distance required to separate the H from the O to consider the bond broken?

    From this table I see that the energy of the bond is 459000J/mol and the radius of the bond is 9.6 x 10-11 m.
    What I first tried (I know it is mostly likely wrong though) was take the energy per mole number and divide it by the radius to get the force per mole. Then I divided that by Avogadro's number to get the force per bond.

    Thinking that was incorrect I decided to try the concept of
    Force to break = tensile strength x cross sectional area.

    That is to say, I compared it to breaking by pulling apart a thin iron bar. I know Young's modulus comes into play here so that's why I guessed I need to know how much further I must separate the H atom from the O atom to consider the bond being broken.

    I know this calculation may not be possible or practical but all I need is a lower limit (estimate). Meaning, I want to know X in the inequality Force required is > or = X. To be honest, my main purpose for this is to determine how strong the comic character Silver Surfer was when he claimed to amp his strength to atomic strength. He is 225lb at a height of 6' 4" so I figured he has approximate density of water, which is why I chose the H-O bond instead.
  2. jcsd
  3. Dec 31, 2011 #2


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    Gold Member

    For a rough calculation I would use Coulonmb's law

    and use the seperation as that between a sodium ion and a chlorine ion in an ionic bond.
    Coulomb's law requires point charges which the sodium and chlorine ions are definitely not when they approach their ionic molecular distance so you will have some error.

    For H-O , the bond is covalent.

    For a metallic bond, as you mentioned, you have the tensile strength, cross sectional area and atomic radius. You should be able to calculate some bond force there also,

    Compare this to the ionic and let us know what you get.
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