Fuel in Nuclear physics Why $^{233}\mathrm{U}$ is an efficient nuclear fuel and  $^{239}\mathrm{Pu}$ cannot be used in thermal reactors??
I couldn't find a good answer for this question in the introductory course I am taking. Could someone explain this? I know that both need to be breeded!
 A: You start by looking as to whether the absorption of a thermal neutron will provide enough energy (binding energy of last neutron) to exceed the activation energy (critical energy) for fission.
Both isotopes satisfy that requirement which is shown in the table below which has been taken from this question where the terms are explained.  

Next one should look to see is the capture cross section for thermal neutrons to produce a fission is different.
For Pu-239 it is 750 barns and for U-233 it is 531 barns so there is nothing in favour of U-233 so far although if one looks further there is another competing cross section which is for capture of a neutron and then the emission of em radiation as explained below.  
After fission both isotopes produce, on average, more than two neutrons per fission which is sufficient to sustain a controlled chain reaction.
The values are approximately 2.9 per fission for Pu-233 and 2.5 per fission for U-233.  
Now look at what happens after a neutron is absorbed and here is possibly the answer to your question?
Once a neutron is absorbed the resulting nucleus can either undergo a fission or emit em radiation.
The emission of em radiation from a nucleus is bad news as far as a fission occurring is concerned because now the nucleus has less energy for a fission to be initiated and the probability of there being a fission is massively reduced.  
After the capture of a thermal neutron by Pu-239 to form Pu-240 73% undergo a fission and 27% emit a gamma ray (called radiative capture of a neutron) whereas for U-233 which forms U-234 the values are 94% for fission and 6% for the emission of em radiation.
So you will see that less of U-233 is "wasted" as compared with Pu-239.
