1.57 [ENDORSED]

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

Wesley_Rugen_1E
Posts: 23
Joined: Fri Jul 22, 2016 3:00 am

1.57

Lines in the Balmer series of the hydrogen spectrum are observed at 653.6, 486.1, 434.0, and 410.2 nm. What is the wavelength of the next line in the series?

I am totally lost on this question. How would I find the wavelength of the next line in this series?

Preston_Dang_1B
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

Re: 1.57

The wavelengths listed correspond to the different wavelengths an electron can absorb at each energy level (i.e. 653.6 nm for n = 1, 486.1 for n = 2, etc.). Since it's asking for the next wavelength in the series, you would need to find the energy difference of the electron at n = 5 since you're given 4 wavelengths. Use $E_{n}=\frac{-hR}{n^2}$ to find the energy difference at n = 5 and then plug what you get for energy into $E=h\frac{c}{\lambda }$ and then solve for wavelength.

Kevin Tam 1J
Posts: 23
Joined: Wed Sep 21, 2016 2:59 pm

Re: 1.57  [ENDORSED]

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

Shushanna S 3F
Posts: 24
Joined: Sat Jul 09, 2016 3:00 am

Re: 1.57

@Kevin Tam 1D

I tried your method, but got 511nm instead of 397nm (the answer in the back of the book). But I agree with your logic...so I'm puzzled too.

@Preston_Dang_1C

I also tried your method just for comparison and got 434nm, which once again, isn't the correct answer of 397nm.

What are we missing here?

georgia_perris4G
Posts: 19
Joined: Wed Sep 21, 2016 2:57 pm

Re: 1.57

I am also lost on this question and when I used the two solutions given, the answers were not right. Is there any other clearer explanation to help solve this problem?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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Re: 1.57

Kevin is correct, but maybe I can help be a little bit clearer!

The Balmer series corresponds to transitions from the n=2 state. Therefore:
656.6 nm corresponds to the transition from n=2 to n=3
486.1 nm corresponds to the transition from n=2 to n=4
434.0 nm: n=2 n=5
410.2 nm: n=2 to n=6

We are trying to find the transition from the n=2 to n=7 case. Remember, after finding the energy difference, the answer we get is in joules. This answer is the energy of light so we have to change energy of the photon into wavelength