# Chemistry quantum numbers?

I need to find the initial and final energy states from these wavelengths. How do you go about doing that? (n=? to n?)

954.60 nm

1004.94 nm

1093.81 nm

1875.10nm

Thank you!

### 4 Answers

- Dr WLv 72 months ago
those wavelengths correspond to the "paschen" series for hydrogen atoms.

https://en.wikipedia.org/wiki/Hydrogen_spectral_se...

n initial.. ... n final .. . . λ (nm)

... 4.. ... .. .. . .3. .. ... .. . 1875

.. .5. ... .. ... .. 3. .. .. .. . 1282

.. .6.. .. .. .. .. .3. .. .. .. .. 1094

.. .7. ... .. .. .. .3.. .. .. . .. 1005

.. .8. .. .. .. .. ..3. .. ... .. ... 954.6

. ..∞. .. .. ... ...3. .. .. .. .. .. 820.4

those values can be calculated via the rydberg equation

https://en.wikipedia.org/wiki/Rydberg_formula

with Rh = 1.097x10^7 / m

- Roger the MoleLv 72 months ago
First, this question makes no sense because it does not say what kind of atom these electrons are attached to. Since hydrogen is used most commonly in school problems, we could assume hydrogen, but it would be better if the question said so explicitly.

Whether the initial n is greater or less than the final n depends on whether the given wavelength is emitted or absorbed, respectively. The question does not say anything about that either.

Next, I see no way to calculate the answers without using the Rydberg equation and every possible combination of integers taken two at a time to calculate wavelengths, which would consume an absurd amount of time.

The remaining possibility is to look at the various hydrogen spectrum series, and hope to find matching wavelength values. The source below gives many possibilities. The near matches I find there are:

954.60 nm (3 and 8)

1004.94 nm (3 and 7)

1093.81 nm (3 and 6)

1875.10nm (3 and 4)

- davidLv 72 months ago
delta E = hc/L >>> I am using L for lambda (wave length)

delta E = (6.623x10^-34)(3.00x10^8)/(954.60x10^-9)

/ / / / use a calcilator

/ / / hopeflly you have a table somewhere with energies of each emergy level ... n= 1 and n = 2 and ///etc

/ / / subtract then and get the delta E answer above

- ChemTeamLv 72 months ago
First off, a cheat sheet:

https://upload.wikimedia.org/wikipedia/commons/thu...

Look at the 1875.10 value on the image. You'll see that it involves a change from 4 to 3 (for emission) or 3 to 4 (for absorption).

Now, how do you figure out the values without the cheat sheet?

Answer: you have to use the Rydberg equation and try different n_i and n_f values. One point is that the wavelengths are all infrared, so relatively little energy is involved. That immediately eliminates n = 1 since those lines are all ultraviolet. n = 2 is gone since those values are visible to ultraviolet. So, start looking at n = 3.

Make sure to do a couple calculations to make sure you know how to use the Rydberg Equation. Here's an example: