Energy released when 238U is divided into two 119Pd.

[Ulrik Nordin]

Hi! I have a question about nuclear fission. Here is the problem:

238_U (B/A = 7.6 MeV/n) is divided into two 119_Pd ( B/A = 8.5 MeV/n). How much energy will be released?

I was thinking since the 119_Pd has larger B/A (binding energy per nucleus), energy is needed for this reaction to happen. But in my notes, the teacher has made this calculation 238*(8.5-7.6) = 214 MeV and states that this is released from the reaction. I do not undertand.

Best regards
ulrik

 

[jtbell]

Binding energy is actually negative. That is, you must subtract it from the sum of the mass-energies of the individual nucleons, in order to get the total mass-energy of a nucleus. So if you think in terms of "average mass-energy per nucleon", this is smaller for ¹¹⁹Pd than for ²³⁸U.

 

[phyzguy]

Although binding energy is defined in such a way that the numbers come out positive, it is actually a negative energy, since it represents how much energy is released when the nucleus is assembled from its constituents. so we can write:

E(U238) = E(238 separate nuclei) - 238 * B/A(U238); E(Pd119) = E(119 separate nuclei) - 119 * B/A(Pd119)

So if you subtract the final energy from the initial energy, you get:

E(238 separate nucleu) - 238 * 7.6 - 2*(E(119 separate nuclei) - 119 * 8.5) = 238*(8.5 - 7.6)