Nuclear Fission - Calculations of Energy

[Lissajoux]

Homework Statement

Example nuclear fission:

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

How much energy does such fission release?

Homework Equations

I'm given the values of M({}^{235}U), m_{n}, M({}^{90}Sr), M({}^{136}Xe) 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:

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

But this gives me a negative answer, of around -0.15u. 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.

 

[nickjer]

To get it straight just remember that you are conserving energy, so E_i = E_f. Using this you get:

U+n = Sr+Xe+10n+Q

Q = (U+n)-(Sr+Xe+10n)

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).