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It's a standard question on the atomic emission spectrum.

An atomic emission spectrum of hydrogen shows three wavelengths: 1875 nm, 1282 nm, and 1093 nm. Assign this wavelengths to transitions in the hydrogen atom.

I understand how to calculate it, but from both the answers in my book and the answer from slader.com. Both just say

since these wavelengths are larger than those of visible light, then we assume $n_\mathrm{f} = 3.$

However, I don't understand why we could assume $n_\mathrm{f} = 3.$ Is it because we just look at Paschen series (Bohr series, $n′ = 3$) and the Paschen lines all lie in the infrared band? Could anyone explain it?

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  • $\begingroup$ I am sorry for being off topic, but @andselisk how do we create direct hyperlink? $\endgroup$ – Zenix Nov 19 '19 at 20:33
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    $\begingroup$ @Zenix If you are referring to Markdown, then inline markup would be [Text](https://site.domain.com). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Nov 19 '19 at 20:35
  • $\begingroup$ Another method is using the provided button: a "link" symbol, which you can click, and then insert the https:// for your selected text. $\endgroup$ – Molly_K Nov 19 '19 at 20:36
  • $\begingroup$ @Molly_K True, but there is nothing to click on when you post a comment, and this is an "anchoring" method which is fine for a link or two, but when you have a couple of dozens of URLs in your post it quickly becomes tedious to keep a track on what pointer refers to which site. $\endgroup$ – andselisk Nov 19 '19 at 20:39
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As the wavelengths are in the infrared region hence the transition occurs at $n > 2.$

Now, if you may have seen the data, these values are all common values of Paschen series. But even if you haven't, you can do a simple test by finding the shortest wavelength of all the series after Balmer. You will find only the shortest wavelengths of Paschen and Brackett are suitable for having values lying in this range as the shortest wavelength for Pfund is $\pu{2279 nm}.$

Then also as the shortest wavelength for Brackett is $\pu{1458 nm},$ you can be sure that the second and third are definitely lying in the Paschen series and check for the remaining wavelength which would be a fairly easy job now.

Directly assuming that the final energy level is $n = 3$ isn't correct, I think, but some analysis can help us to draw the above-mentioned conclusions.

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    $\begingroup$ For inline formulas, enclose the formula in $...$. For displayed formulas, use $$...$$. Probably you will find a more in-depth overview on Math.SE helpful: MathJax basic tutorial and quick reference. $\endgroup$ – andselisk Nov 20 '19 at 11:14
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    $\begingroup$ @andselisk got it thanks $\endgroup$ – user78585 Nov 20 '19 at 11:17
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    $\begingroup$ No prob. Also, note that here using MathJax for the quantities here wasn't really necessary: 1458 nm looks fine on its own. \pu{…} macro comes in handy in more complex situations like $R = \pu{8.21E-5 m3 atm K-1 mol-1}$ ($R = \pu{8.21E-5 m3 atm K-1 mol-1}$). Keep in mind though that \pu{…} is currently only available here on Chemistry.SE and won't work on other SE sites such as Physics.SE where you have to enforce proper typography with bare hands and use ~, \,, \mathrm{…} etc. :) $\endgroup$ – andselisk Nov 20 '19 at 11:20
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    $\begingroup$ @andeselisk got your point. $\endgroup$ – user78585 Nov 20 '19 at 11:30
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    $\begingroup$ @StackUpPhysics, thanks for your response, I think it would be essential to have the Series spectrum next to the question to make the assumption. (or able to pull that information out of my brain if there's anything to pull...) $\endgroup$ – Molly_K Nov 20 '19 at 20:21

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