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problem40_17 - m 10 17 2 J 10 9.15 m/s 10 s(3.00 J 10 63 6...

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40.17: Since , 6 0 = E U we can use the results from Section 40.3, , 09 . 5 , 625 . 0 3 1 = = E E E E and 2 15 27 2 34 2 2 2 2 m) 10 kg)(4.0 10 67 . 1 ( 2 s.) J 10 054 . 1 ( 2 - - - × × × = = π mL π E U J. 10 05 . 2 12 - × = E The transition energy is J. 10 9.15 J) 10 05 . 2 )( 625 . 0 09 . 5 ( 12 12 1 3 - - × = × - = - E E The wavelength of the photon absorbed is then
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Unformatted text preview: m. 10 17 . 2 J 10 9.15 m/s) 10 s)(3.00 J 10 63 . 6 ( λ 14 12 8 34 1 3---× = × × ⋅ × =-= E E hc...
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