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Mass Defect Calculation

  1. Oct 2, 2015 #1
    Hi, this may be a very trivial question but,

    When calculating energy, is it BE(products) - BE(reactants) or is it the other way round?
    I have been solving questions and the solutions were products - reactants for some and reactants - products for some.

    Also for mass defect , is it mass ( products ) - mass ( rxts ) or again is it the other way round ?
    Does it change for fusion/fission ?

    Edit: I think I got it, if energy is released then mass defect = (mass reactants) - (mass products) , if it is energy absorbed it is the other way round.
    For energy it is BE(Products)-BE(reactants) for energy released and the other way round for energy absorbed.
    Am I right?
     
    Last edited: Oct 2, 2015
  2. jcsd
  3. Oct 2, 2015 #2

    mfb

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    Calculating energy of what? There are two reasonable definitions, those two have equal magnitude but opposite sign.

    It is a property of the nucleus, independent of its origin.

    Sure, this is just energy conservation.
     
  4. Oct 3, 2015 #3
    i think since both fission and fusion reactions release energy so mass defect = mas of reactants- mas of product
    so,i think it doesn't change
     
  5. Oct 3, 2015 #4

    mfb

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    It depends on the reaction.
     
  6. Oct 3, 2015 #5
    oh really, depends on what reaction exactly?

    both fission and fusion releases energy right? isn't that the mass of reactants will always be higher than the mass of products?
    maybe my physics syllabus is too basic, would you mind explaining it to me?
     
  7. Oct 3, 2015 #6

    mfb

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    Fusion of hydrogen to helium releases energy, fission of helium to hydrogen needs energy. The second part is a direct consequence of energy conservation.
    Fission of uranium releases energy, fusion of the fission products to uranium needs energy.

    As a rough guideline, fusing nuclei lighter than iron and fission of nuclei heavier than iron releases energy, while the opposite directions need energy. This is not always true, but it is a good approximation.
     
  8. Oct 3, 2015 #7

    jtbell

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  9. Oct 6, 2015 #8
    Sorry for the late reply. Thanks alot for your help again!
     
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