## HW 1.15

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Joanna Pham - 2D
Posts: 113
Joined: Fri Apr 06, 2018 11:04 am

### HW 1.15

Could someone please explain how to find the initial and final n-values please? I found the frequency and got 2.9239 x 10^15 Hz. I tried to plug it into the Rydberg formula, but I’m confused how to find the individual n-values...

Kuldeep Gill 1H
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Joined: Fri Apr 06, 2018 11:02 am
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### Re: HW 1.15

I got up to that point as well and plugging it in just does not make sense to me

Kuldeep Gill 1H
Posts: 44
Joined: Fri Apr 06, 2018 11:02 am
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### Re: HW 1.15

Hi, so I figured it out..... I guess we were supposed to know that in the Lyman series ( UV light) that the final energy always goes to 1 so.... after you get 2.923* 10^15 you use the equation V=R*(1/n1^2 - 1/n2^2) and realise that it is final minus initial so set plug in 1 for 1/n1^2 and plug-in 2.923*10^15 for v and 3.29*10^15 for R and solve for n2.. I think it should become a bit clearer now!

Chem_Mod
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### Re: HW 1.15

Could you please provide the complete details of the question?

Johanna Caprietta 1E
Posts: 30
Joined: Fri Apr 06, 2018 11:02 am

### Re: HW 1.15

Where did you find that the final energy goes to 1? Can you explain the reasoning behind this?

kimberlysanchez-1E
Posts: 30
Joined: Tue Nov 14, 2017 3:01 am

### Re: HW 1.15

im having the same problem

Daniel Cho Section 1H
Posts: 29
Joined: Fri Jun 23, 2017 11:40 am

### Re: HW 1.15

For this problem, you kind of have to use to this equation as well as assume that n1= 1 because this is an example of the Lyman series as you look at the wavelength 102.6 nm in terms of the graph on page 7 letter b. Since you know your frequency would be 2.9329X10^15 Hz, you would set up this equation like this: 2.9329X10^15Hz= (3.29X10^15)(Rydberg constant) {(1/(n1)^2)-(1/(n2)^2))] therefore getting (1/n^2)=.112 . You then solve for n in which from then on you get your answer.

I hope this helps. Correct me in places where I might be wrong.

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