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Unformatted text preview: 50. Using Eq. 406 and the result of problem 3 in Chapter 39, we ﬁnd
∆E = Ephoton = hc
1240 eV · nm
=
= 12.09 eV .
λ
102.6 nm Referring to Fig. 4016, 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
−
+ 12.09 eV
12 which yields n = 3. Thus, the transition is from the n = 3 to the n = 1 state. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Photon

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