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.
Homework Problems #11 and #15 [ENDORSED]
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Re: Homework Problems #11 and #15
The homework problems from Chapter 1. Sorry I forgot to specify the chapter in the original post.
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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. )
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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.
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.
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