What is the total energy needed to see all 15 spectral lines for the electronic transitions involving n = 1 up to and including n = 6, assuming every transition occurs only once? What wavelength is needed to cause this in angstroms?
1 Answer
Well, if you want 15 spectral lines, then you're going to need 15 transitions. That occurs with 6 energy levels involved.
n = 1->6,5,4,3,2
n = 2->6,5,4,3
n = 3->6,5,4
n = 4->6,5
n = 5->6
If we assume that SOMEHOW, all the energy that went in managed to accomplish ALL of these transitions WITHOUT ANY WEIGHTED BLEEDING ONTO ANY GIVEN TRANSITION, then we need to calculate
DeltaE_16 = -"13.61 eV" cdot (1/1^2-1/6^2) = ???
DeltaE_15 = -"13.61 eV" cdot (1/1^2-1/5^2) = ???
DeltaE_14 = -"13.61 eV" cdot (1/1^2-1/4^2) = ???
DeltaE_13 = -"13.61 eV" cdot (1/1^2-1/3^2) = ???
DeltaE_12 = -"13.61 eV" cdot (1/1^2-1/2^2) = ???
DeltaE_26 = -"13.61 eV" cdot (1/2^2-1/6^2) = ???
DeltaE_25 = -"13.61 eV" cdot (1/2^2-1/5^2) = ???
DeltaE_24 = -"13.61 eV" cdot (1/2^2-1/4^2) = ???
DeltaE_23 = -"13.61 eV" cdot (1/2^2-1/3^2) = ???
DeltaE_36 = -"13.61 eV" cdot (1/3^2-1/6^2) = ???
DeltaE_35 = -"13.61 eV" cdot (1/3^2-1/5^2) = ???
DeltaE_34 = -"13.61 eV" cdot (1/3^2-1/4^2) = ???
DeltaE_46 = -"13.61 eV" cdot (1/4^2-1/6^2) = ???
DeltaE_45 = -"13.61 eV" cdot (1/4^2-1/5^2) = ???
DeltaE_56 = -"13.61 eV" cdot (1/5^2-1/6^2) = ???
I'll leave the mathematical slog for you to work out, and when you do that,
DeltaE = DeltaE_16 + . . . + DeltaE_56 (15 terms)
= E_"photon" = hnu = (hc)/lambda
Therefore,
lambda = (hc)/(DeltaE) " m"
And since
color(blue)(lambda = (10^10 hc)/(DeltaE) " angstroms")