p40_045 - 27 . 2 eV . The kinetic energy is K = E U = ( 13...

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45. Letting a =5 . 292 × 10 11 m be the Bohr radius, the potential energy becomes U = e 2 4 πε 0 a = (8 . 99 × 10 9 N · m 2 / C 2 )(1 . 602 × 10 19 C) 2 5 . 292 × 10 11 m = 4 . 36 × 10 18 J=
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Unformatted text preview: 27 . 2 eV . The kinetic energy is K = E U = ( 13 . 6 eV) ( 27 . 2 eV) = 13 . 6 eV....
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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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