Nuclear Fission - Calculations of Energy

  • Thread starter Lissajoux
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Homework Statement



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?

Homework 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.

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.
 

Answers and Replies

  • #2
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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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