p43_054 - well. A Fnite well is able to support slightly...

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54. (a) We recall that mc 2 =0 . 511 MeV from Table 38-3, and note that the result of problem 3 in Chapter 39 can be written as hc = 1240 MeV · fm. Using Eq. 38-51 and Eq. 39-13, we obtain λ = h p = hc K 2 +2 Kmc 2 = 1240 MeV · fm p (1 . 0MeV) 2 +2(1 . 0 MeV)(0 . 511 MeV) =9 . 0 × 10 2 fm . (b) r = r 0 A 1 / 3 =(1 . 2 fm)(150) 1 / 3 =6 . 4fm . (c) Since λ À r the electron cannot be conFned in the nuclide. We recall from Chapters 40 and 41, that at least λ/ 2 was needed in any particular direction, to support a standing wave in an “inFnite
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Unformatted text preview: well. A Fnite well is able to support slightly less than / 2 (as one can infer from the ground state wavefunction in ig. 40-8), but in the present case /r is far too big to be supported. (d) A strong case can be made on the basis of the remarks in part (c), above....
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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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