p43_001 - . 60 10 19 C) 8 . 03 10 15 m = 28 . 3 10 6 eV ....

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1. In order for the α particle to penetrate the gold nucleus, the separation between the centers of mass of the two particles must be no greater than r = r Cu + r α =6 . 23 fm + 1 . 80 fm = 8 . 03 fm. Thus, the minimum energy K α is given by K α = U = 1 4 πε 0 q α q Au r = kq α q Au r = (8 . 99 × 10 9 V · m / C)(2 e )(79)(1
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Unformatted text preview: . 60 10 19 C) 8 . 03 10 15 m = 28 . 3 10 6 eV . We note that the factor of e in q = 2 e was not set equal to 1 . 60 10 19 C, but was instead carried through to become part of the Fnal units....
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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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