prb4 - 1.The first step in this problem is to convert the...

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1.The first step in this problem is to convert the wavelength, λ , to a frequency, ν , and then calculate the energy of the photon using ε = h ν . In such problems you must take care to use a consistent set of units. We will use S.I. units. Thus, λ = 18900 Å = 18900 x 10 –10 m = 1.89 x 10 –6 m The frequency associated with this wavelength is given by: ν = c/ λ = (3.00 x 10 8 m s –1 ) / (1.89 x 10 –6 m) = 1.59 x 10 14 s –1 The energy ε associated with a photon of this frequency is: ε = h ν = (6.626 x 10 –34 J s) (1.59 x 10 14 s –1 ) = 1.05 x 10 –19 J Note -- the ENERGY of this photon corresponds to a DIFFERENCE between energy levels of the H atom The energy levels for the H atom are given by: ε n = (–21.8 x 10 –19 J) / n 2 ; where n = 1,2,3,4,. ............ Thus, ε 1 = –21.8 x 10 –19 J; ε 2 = –5.45 x 10 –19 J ε 3 = –2.42 x 10 –19 J; ε 4 = –1.36 x 10 –19 J We now seek an ENERGY LEVEL DIFFERENCE which matches the energy of the photon . We see that the electronic transition which gives rise to an EMITTED photon with energy ε = 1.05 x 10 –19 J is: n initial = 4 n final = 3 2. A series of spectral lines in emission is characterized by a common value of
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prb4 - 1.The first step in this problem is to convert the...

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