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### Homework Problems #11 and #15

Posted: **Fri Oct 07, 2016 11:37 am**

by **Elias Ruben 1O**

I am confused about the answer in #11. It says that for the Lyman series, the lowest energy level is n=1, and for the Balmer series the lowest is n=2. What does that mean? Can someone please explain it to me?

Also, for #15, I know you have to do the problem in reverse, first by finding change in energy and then plugging that into the Rydberg formula. However, I got -0.888156=(1/n_final^2)-(1/n_initial^2), and I don't know how to solve that since they didn't give us any of the n values. Can someone please help? Thank you.

### Re: Homework Problems #11 and #15

Posted: **Fri Oct 07, 2016 11:39 am**

by **Elias Ruben 1O**

The homework problems from Chapter 1. Sorry I forgot to specify the chapter in the original post.

### Re: Homework Problems #11 and #15 [ENDORSED]

Posted: **Fri Oct 07, 2016 11:47 am**

by **Helen_Onuffer_1A**

When it says that the Lyman series goes to n=1 and the Balmer series goes to n=2, the book means that the final n/energy level for electrons in the two series are 1 and 2, respectively. (So say you're trying to calculate the initial and final energy levels of an electron that emitted a line in the Lyman series spectrum, you know that the final energy/n level of the electron is n = 1. If it was a line in the Balmer series, the final energy level would be n=2. )

### Re: Homework Problems #11 and #15

Posted: **Fri Oct 07, 2016 12:31 pm**

by **Beata_Vayngortin_3L**

For the Lyman series, n=1 is the ground state. For the Balmer series, n=2 is the ground state.

For number 15, the problem is using the lyman series so you know that n_initial=1. Now you can solve for your other n-value through calculator work and plugging in the values since you only have one unknown value now.

### Re: Homework Problems #11 and #15

Posted: **Sat Oct 08, 2016 7:23 pm**

by **704709603**

Hi,

How do you solve 15 without the Rydberg equation?

Thank you!