# The quantum number of the level at which the

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the quantum number of the level at which the transition terminates . Since the 18900 Å line arises from a transition in which n initial = 4 and n final = 3, the series to which it belongs is characterized by n final = 3 . The series limit for this series of emission lines is the wavelength that would correspond to the transition n initial = to n final = 3 From the energy level expression given in problem1 we see that: ε = (–21.8 x 10 –19 J) (1/ 2 ) = 0

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2 Recall from lecture, that this energy corresponds to an ionized H + ......... e pair both at rest and infinitely separated From problem 1, ε 3 = –2.42 x 10 –19 J Thus, | E| = | ε 3 ε | = | –2.42 x 10 –19 0 | = 2.42 x 10 –19 J Here the vertical lines indicate that we are taking the ABSOLUTE value i.e. just the magnitude. Of course, for emission processes in which the ENERGY OF THE ATOM IS LOWERED, E < 0 λ = {h c} / { E} = {(6.626 x 10 –34 J s) (3.00 x 10 8 m s –1 ) / (2.42 x 10 –19 J)} = 8.21 x 10 –7 m = 821 x 10 –9 m = 821 nm = 8210 Å The energy associated with a photon of wavelength λ = 8210 Å is the energy required TO IONIZE an electron from the n = 3 state of the H atom. 3. The relevant electronic configurations for S and Cl are: S: 1s 2 2s 2 2p 6 3s 2 3p 4 ; Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 For both these atoms the most loosely bound electron is removed from a doubly- occupied 3p orbital on ionization.
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