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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 .
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 ?
= Elow + ∆E
+ 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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