56. Assuming the neutrino has negligible mass, then ∆mcm ce22=−−mmTiVbg. Now, since Vanadium has 23 electrons (see Appendix F and/or G) and Titanium has 22 electrons, we can add and subtract 22meto the above expression and obtain ∆mmee22223=+−−=−TiVTiVbgWe note that our final expression for ∆mc2involves the atomicmasses, and that this assumes (due to the way they are usually tabulated) the atoms are in the ground states (which is certainly not the case here, as we discuss below). The question now is: do we set Q = – ∆mc2as in Sample Problem 42-7? The answer is “no.” The atom is left in an
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.