danny8522003 Posted March 28, 2006 Share Posted March 28, 2006 Hey guys, Doing a project on star formation and fusion. We did mass defects in class today, but we were using [math]{\Delta}E={\Delta}mc^2[/math]. Seeing as the mass is the rest mass of an object, can this be applied to protons when they are both at rest and in the nucleus? Thanks, Dan Link to comment Share on other sites More sharing options...
ydoaPs Posted March 28, 2006 Share Posted March 28, 2006 if it is moving in your frame of reference, you use [math]E^2=(mc^2)^2+(pc)^2[/math] Link to comment Share on other sites More sharing options...
danny8522003 Posted March 28, 2006 Author Share Posted March 28, 2006 But it's at rest.. Link to comment Share on other sites More sharing options...
insane_alien Posted March 28, 2006 Share Posted March 28, 2006 so the p in (pc)^2 is equal to zero. which makes (pc)^2 equal zero which leaves you with E^2=(mc^2)^2 which simplifies to E=mc^2. since not all mass is consumed you go back the equation you posted and that gives you the binding energy(the energy released) Link to comment Share on other sites More sharing options...
danny8522003 Posted March 28, 2006 Author Share Posted March 28, 2006 Thank you. Just confused me having the rest mass change while it's stationary both in and out of the nucleus, stupid quantum physics Link to comment Share on other sites More sharing options...
swansont Posted March 28, 2006 Share Posted March 28, 2006 You don't look at the rest mass of the proton or neutron while it's in the nucleus, you are looking at the rest mass of the nucleus itself. It will be less than the sum of the individual parts because it has given up the binding energy. Link to comment Share on other sites More sharing options...
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