This is an informative discussion. I think that sometimes we group the coulomb barrier energy with the energy associated with the strong force once the proton in inside the nucleus. Once I tried to see if they were separate and could find no reason that this would be true. I think that both types of forces result in a change to the mass of the nucleus.
Dave -----Original Message----- From: mixent <mix...@bigpond.com> To: vortex-l <vortex-l@eskimo.com> Sent: Sat, Aug 11, 2012 11:58 pm Subject: Re: [Vo]:the Coil In reply to Axil Axil's message of Sat, 11 Aug 2012 23:37:00 -0400: Hi, [snip] >*This is clearly not true. If it were then hot fusion wouldn't work, and >there* >* * > >*would be no Sun. It takes less energy to get in than it takes to get out.* >* * > >*The difference between the two is the net energy of the reaction.* > >I said: The coulomb barrier is symmetric. It takes as much energy to >leave the nucleus as it takes to get in. > >http://en.wikipedia.org/wiki/Alpha_decay Radioactive nuclei are in an excited state (i.e. above ground state) by definition. That's the reason they are radioactive. The ground state of e.g. 4He is way below that of 2 D's hence it's energetically favourable for 2 D's to combine and form 4He. In the radioactive case, it's easier to get out than to get in. In the case of two D's it's easier to get in than to get out. So my statement about it being easier to get in than to get out was too broad, as it only applies to fusion reactions. It's the other way around for fission reactions. However in neither case is the Coulomb barrier symmetric. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
