## Knowing the Balmer series HW q1.57

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

Katarina Ho -1B
Posts: 29
Joined: Fri Apr 06, 2018 11:03 am

### Knowing the Balmer series HW q1.57

Hi, In this homework question it gives 4 wavelengths of the balmer series and asks you to name the next wavelength in the series. I was wondering if there was a certain equation to solve or you just need to memorize the balmer series?

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

### Re: Knowing the Balmer series HW q1.57

balmer is all the visible regions. so like from 700nm-400nm

Maria Zamarripa 1L
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### Re: Knowing the Balmer series HW q1.57

I also have a question pertaining to this question, in the answer manual it said n2=7. How did they solve to get that?

Tiffany Chen 1A
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Joined: Fri Apr 06, 2018 11:02 am
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### Re: Knowing the Balmer series HW q1.57

I figured this problem out by first solving for n2 when the wavelength is 656.3 nm (first line in the Balmer series). It's a fact that the Balmer series starts at n=2, so you know that n1=2. Then using hc/lambda = -(hR/n^2)+(hr/n^2) where lambda=656.3, the first n=2, and the second n is 3.

So now you know that 656.3 nm corresponds with n2=3. Then, counting up the series, 486.1 nm corresponds with n2=4, 434.0 nm corresponds with n2=5, and 410.2 nm corresponds with n2=6. Therefore, theoretically, the next line in the series should correspond with n2=7.

Then, plug in 7 for n2, 2 for n1, and solve for lambda.

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