What is the truth? Binding energy or Binding energy with liquid drop model fit?

This stuff confuses me. Does this affect calculations? What energy values are true? They are all different! Check here (binding energy ldm fit): www.nndc.bnl.gov/chart/reColor.jsp?newColor=be-ldm vs here (binding energy): www.nndc.bnl.gov/chart/reColor.jsp?newColor=beda

Also, can somebody explain what practically binding energy per nucleon means? What will happen if I apply that energy to some nucleus? Will it be ejected? But shouldn't energy that some nucleus recieves be shared uniformly among all nucleons? Why is it not the same as neutron or proton separation energy (they are nucleons)?

What what will happen when I apply energy same as binding energy to some nucleus? What will happen when I apply same energy as binding energy per nucleon? Which energy is more practical or useful to know (be or be/a)?

blogfast25 - 19-4-2016 at 07:09

 Quote: Originally posted by radioboy Also, can somebody explain what practically binding energy per nucleon means? What will happen if I apply that energy to some nucleus? Will it be ejected? But shouldn't energy that some nucleus recieves be shared uniformly among all nucleons? Why is it not the same as neutron or proton separation energy (they are nucleons)? What what will happen when I apply energy same as binding energy to some nucleus? What will happen when I apply same energy as binding energy per nucleon? Which energy is more practical or useful to know (be or be/a)?

The nuclear binding energy (BE) is the energy that is released when a nucleus is formed from its constituent nucleons. Simple case: He24 (an alpha particle):

2 p + 2 n ===> He24 + 28.3 MeV

Similarly, to tear a He24 nucleus apart into its constituent nucleons would cost 28.3 MeV (4.53x10-12 J).

The BE per nucleon is simply the BE divided by the number of nucleons, in this case 4.

Joe Skulan - 30-4-2016 at 05:58

"Apply energy" to a nucleus is too vague. How will you apply the energy? Presumably by bombarding the nucleus with some other nucleus or particle. The kinetics of these reactions will differ with different bombarding particles. The results can not be predicted accurately, but if you added just enough energy to overcome the total binding energy the nucleus would not decompose into its component nucleons because the added energy would not all go into breaking nuclear bonds. A heavy nucleus might break apart into smaller, more thermodynamically favored nuclei (like alpha particles) which would carry away some of the excess energy as kinetic energy. More energy would be lost to gamma photons and neutrinos. Breaking (say) a 238U nucleus into 92 protons and 146 neutrons would require vastly more energy than the binding energy of the 238U nucleus.