43. (a) and (b) Using Eq. 40-6 and the result of problem 3 in Chapter 39, we find
hc 1240 eV·nm
"E = Ephoton = = =2.55 eV .
486.1nm
Referring to Fig. 40-16, we see that this must be one of the Balmer series transitions (this fact
could also be found from Fig. 40-17). Therefore, nlow = 2, but what precisely is nhigh?
Ehigh = Elow +"E
13.6eV 13.6eV
- = - +2.55 eV
n2 22
which yields n = 4. Thus, the transition is from the n = 4 to the n =2 state.
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