p40_001 - 34 J s) 2 8 m p (100 10 12 m) 2 (1) 2 = 3 . 29 10...

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1. (a) This is computed in part (a) of Sample Problem 40-1. (b) With m p =1 . 67 × 10 27 kg, we obtain E 1 = µ h 2 8 mL 2 n 2 = µ (6 . 63 × 10
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Unformatted text preview: 34 J s) 2 8 m p (100 10 12 m) 2 (1) 2 = 3 . 29 10 21 J = 0 . 0206 eV ....
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