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Homework Help: Nuclear Fission - Calculations of Energy

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Example nuclear fission:

    [tex]{}_{92}^{235}U+{}_{0}^{1}n \rightarrow {}_{38}^{90}Sr+{}_{54}^{163}Xe+10{}_{0}^{1}n+Q[/tex]

    How much energy does such fission release?

    2. Relevant equations

    I'm given the values of [itex]M({}^{235}U), m_{n}, M({}^{90}Sr), M({}^{136}Xe)[/itex] but I've not listed them here.

    3. The attempt at a solution

    I thought that Q(energy) = energy in final state - energy in initial state. I.e. that:

    [tex]Q={}_{38}^{90}Sr+{}_{54}^{163}Xe+9{}_{0}^{1}n -{}_{92}^{235}U[/tex]

    But this gives me a negative answer, of around [itex]-0.15u[/itex]. I thought that it should be positive, unless this is just from how I've set out the equation and can just say it's positive. Done a few things similar to this, but still rather confused with getting the hang of it.
  2. jcsd
  3. Apr 24, 2010 #2
    To get it straight just remember that you are conserving energy, so [tex]E_i = E_f[/tex]. Using this you get:

    [tex]U+n = Sr+Xe+10n+Q[/tex]

    [tex]Q = (U+n)-(Sr+Xe+10n)[/tex]

    So, Q = (initial masses) - (final masses). Now it will be positive, meaning energy was released during this reaction. It all depends on where you put the Q in the reaction formula. It would be the other way around if Q was on the initial side (meaning it takes energy to perform the reaction).
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