ch39-p048 - 48. (a) Since E2 = 0.85 eV and E1 = 13.6 eV +...

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48. (a) Since E 2 = – 0.85 eV and E 1 = – 13.6 eV + 10.2 eV = – 3.4 eV, the photon energy is E photon = E 2 E 1 = – 0.85 eV – (– 3.4 eV) = 2.6 eV. (b) From EE nn 21 2 2 1 2 136 11 26 −= F H G I K J = (. ) . eV eV we obtain 112 6 3 16 1 4 1 2 2 2 1 22 2 −=− . . . eV eV Thus, n 2 = 4 and n 1 = 2. So the transition is from the
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