What is the Einstein relationship for calculating nuclear binding energy?

jaredogden
Messages
77
Reaction score
0
I was reading over some nuclear physics and I came across the binding energy of a nucleus. The equation is Eb=[ZM(H)+Nmn-M(AZX)]x931.494 MeV/u

I remember seeing this and using this in class when I took it over a year ago but I can't remember exactly how to use it and I see no examples in the book. If anyone can help explain this that would be awesome. Thanks
 
Physics news on Phys.org
This is really just a definition, the observed mass of a nucleus minus the separate masses of all its protons and neutrons. We don't use it directly for anything, but it tells us something about nuclear forces. Often you'll see a graph of Eb per nucleon. The curve is relatively flat at 8 MeV per nucleon, and this suggests that the nuclear forces 'saturate', acting only between a nucleon and its nearest neighbors. (If each nucleon attracted all the others the binding energy would grow proportionally to N2.)
 
Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship:
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top