Ch42-p002 - 2 Our calculation is similar to that shown in Sample Problem 42-1 We set K = 5.30 MeV=U =(1 4πε 0 qα qCu rmin and solve for the

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Unformatted text preview: 2. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4πε 0 ) ( qα qCu / rmin ) and solve for the closest separation, rmin: rmin −19 9 qα qCu kqα qCu ( 2e ) ( 29 ) (1.60 ×10 C ) ( 8.99 ×10 V ⋅ m/C ) = = = 4πε 0 K 4πε 0 K 5.30 ×106 eV = 1.58 × 10−14 m = 15.8 fm. We note that the factor of e in qα = 2e was not set equal to 1.60 × 10– 19 C, but was instead allowed to cancel the “e” in the non-SI energy unit, electron-volt. ...
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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.

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