How to find initial number from spectral light

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welcometochillis
Posts: 20
Joined: Fri Mar 17, 2017 3:00 am

How to find initial number from spectral light

Postby welcometochillis » Thu Oct 12, 2017 6:52 pm

Hi everyone, I came upon a question that asked to determine the inital numbers in a series from the given spectral light number (the one that indicated whether it is visible or not, and what color it should be). Anyone know how to confront a problem like this?

nickjadidian 1A
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am

Re: How to find initial number from spectral light

Postby nickjadidian 1A » Fri Oct 13, 2017 1:19 pm

When you say “initial numbers in a series”, this indicates a particular spectral line. Perhaps the most sure-fire way to approach this question would be to memorize an arsenal of these spectral lines, as I'll mention later. Beyond just this, however, you can reason that for each series, there are specific wavelengths you can use to determine the type of light being emitted.

For example, if you are given a particular wavelength that the H is absorbing is λ= 656.2 nm, then you automatically know that type of light it will be. Visible light comprises approximately 400-700nm. From here, you look to the particular series which contains visible light.

H Spectroscopy.gif[/attachment]

To reason through it, you know that Lyman (n=1) contains only UV light (nm UV < 656.2nm ). More specifically, the Lyman series contains lines 91.9 nm to 121.6 nm, which makes sense because it corresponds to UV light which is known to have a shorter wavelength than visible light. So, you move onto the next series. We know that the Balmer series (n=2) contains the visible light, which is has the range 400nm < λ < 700nm. Since this works the data given, you know for sure that you are working with the Balmer series. At least one of the two n’s =2. From this point on, you can refer to the Rydberg equation and solve for the other n, using algebra to rearrange the values as needed.



Where R is equal to 1.097×107 m-1 , as in Rydberg’s original formula. From here, we can derive this formula with the new value of R:

( \frac{1}{{n_{f}}^{2}}-\frac{1}{{n_{i}}^{2}} \right )[/tex][attachment=0]

R (the Rydberg constant) is equal to 2.178×10−18 J

As far as anticipating the actual color of the line, the best tool I can offer is only able to get you close to the correct color. For H, violet is on the low end of the visible spectrum (410 nm), whereas red is on the opposite end (656.2 nm). Based on the location of these colors, you can roughly anticipate the color of the visible light given, but you cannot know the color with certainty without memorization.

Photo Credit: http://chemed.chem.purdue.edu/genchem/t ... /bohr.html
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H Spectroscopy.gif


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