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

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

Elias Ruben 1O
Posts: 47
Joined: Wed Sep 21, 2016 2:56 pm

### Homework Problems #11 and #15

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.

Elias Ruben 1O
Posts: 47
Joined: Wed Sep 21, 2016 2:56 pm

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

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

Helen_Onuffer_1A
Posts: 15
Joined: Fri Jul 22, 2016 3:00 am

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

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. )

Beata_Vayngortin_3L
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Joined: Fri Jul 22, 2016 3:00 am
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### Re: Homework Problems #11 and #15

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.

704709603
Posts: 92
Joined: Wed Sep 21, 2016 2:59 pm

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

Hi,
How do you solve 15 without the Rydberg equation?
Thank you!