p40_005 - 5. With mp = 1.67 1027 kg, we obtain E1 = h2 8mL2...

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5. With m p =1 . 67 × 10 27 kg, we obtain E 1 = µ h 2 8 mL 2 n 2 = µ (6 . 63 × 10 34 J · s) 2 8 m p (100 × 10 12 m) 2 (1) 2 =3 . 29 × 10 21 J=0 . 0206 eV . Alternatively, we can use the mc 2 value for a proton from Table 38-3 (938 × 10 6 eV) and the hc = 1240 eV · nm value developed in problem 3 of Chapter 39 by writing Eq. 40-4 as
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