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AE = -2.18x10-18 J (1/nana- 1/nnn) 1. Using the equation above, calculate the transition energy of 1 photon emitted when a Hy

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D 222 DE = -2.18 X1018 J Feth ; And we need to find the energy during the transition from n= 5 (initial) to n=2(final). Energ4 DE - hc where x = wavelength (= velocity of light = 3.0x108 m/s. ДЕ = (6.626x103 47.5) (3.0x108 ms) (4.58 x 10-19 J) = (4.3

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