p43_056 - was captured ±rom the lowest shell (where the...

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56. Assuming the neutrino has negligible mass, then mc 2 =( m Ti m V m e ) c 2 . Now, since Vanadium has 23 electrons (see Appendix F and/or G) and Titanium has 22 electrons, we can add and subtract 22 m e to the above expression and obtain mc 2 =( m Ti +22 m e m V 23 m e ) c 2 =( m Ti m V ) c 2 . We note that our fnal expression ±or ∆ mc 2 involves the atomic masses, 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 = mc 2 as in Sample Problem 43-7? The answer is “no.” The atom is le±t in an excited (high energy) state due to the ±act that an electron
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Unformatted text preview: was captured ±rom the lowest shell (where the absolute value o± the energy, E K , is quite large ±or large Z – see Eq. 41-25). To a very good approximation, the energy o± the K-shell electron in Vanadium is equal to that in Titanium (where there is now a “vacancy” that must be flled by a readjustment o± the whole electron cloud), and we write Q = − ∆ mc 2 − E K so that Eq. 43-27 still holds. Thus, Q = ( m V − m Ti ) c 2 − E K ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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