# 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?

Feb 3, 2018

Well, if you want 15 spectral lines, then you're going to need 15 transitions. That occurs with 6 energy levels involved.

$n = 1 \to 6 , 5 , 4 , 3 , 2$
$n = 2 \to 6 , 5 , 4 , 3$
$n = 3 \to 6 , 5 , 4$
$n = 4 \to 6 , 5$
$n = 5 \to 6$

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 $\Delta E$ for all $15$ transitions...

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,

$\Delta E = \Delta {E}_{16} + . . . + \Delta {E}_{56}$ (15 terms)

$= {E}_{\text{photon}} = h \nu = \frac{h c}{\lambda}$

Therefore,

$\lambda = \frac{h c}{\Delta E} \text{ m}$

And since $\text{1 m" = 10^(10) "angstroms}$,

$\textcolor{b l u e}{\lambda = \frac{{10}^{10} h c}{\Delta E} \text{ angstroms}}$