p40_046 - 46. (a) and (b) Using Eq. 40-6 and the result of...

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Unformatted text preview: 46. (a) and (b) Using Eq. 40-6 and the result of problem 3 in Chapter 39, we find ∆E = Ephoton = hc 1240 eV · nm = = 10.2 eV . λ 121.6 nm Referring to Fig. 40-16, we see that this must be one of the Lyman series transitions. Therefore, nlow = 1, but what precisely is nhigh ? Ehigh 13.6 eV − n2 = = Elow + ∆E 13.6 eV − + 10.2 eV 12 which yields n = 2 (this is confirmed by the calculation found from Sample Problem 40-6). Thus, the transition is from the n = 2 to the n = 1 state. ...
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