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Unformatted text preview: n = 1 and n = 2 we get: E 1 =k 2 Z 2 e 4 m Ï€ 2 ~ 2 =(8 . 99 Ã— 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 Ã— 1019 C ) 4 (273)(9 . 11 Ã— 1031 kg ) 2(1 . 05 Ã— 1034 J Â· s ) 2 =4 . 04 Ã— 1013 J =2 . 52 Ã— 10 6 eV E 2 =k 2 Z 2 e 4 m Ï€ 8 ~ 2 =(8 . 99 Ã— 10 9 Jm/C 2 ) 2 (26 2 )(1 . 60 Ã— 1019 C ) 4 (273)(9 . 11 Ã— 1031 kg ) 8(1 . 05 Ã— 1034 J Â· s ) 2 =1 . 01 Ã— 10 6 J =6 . 31 Ã— 10 5 eV Part c) Finally, the energy of the photon emitted upon transition from level 2 to level 1 is the diï¬€erence between the energies of the two levels: E = E 2E 1 = 1 . 89 Ã— 10 6 eV 1...
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 Spring '08
 LIN
 Energy, Kinetic Energy, Potential Energy, Quantum Physics, 2 K, 31 kg, 13 J

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