## Atomic Spectra- Question 1.15 [ENDORSED]

Kaitlin_Ryan_ 3K
Posts: 28
Joined: Fri Jul 15, 2016 3:00 am

### Atomic Spectra- Question 1.15

When solving this problem, I ran into a problem when trying to determine which n value should be used for E(final) and which should be solved for E(initial).

When I plugged in n=1 for E(final) and solved for n, I came up with a decimal energy level, which is not possible.

When I plugged in n=1 for E(initial), I got n=3, which is the correct answer.

Why do I need to plug in n=1 for E(initial) and not E(final) to obtain the correct answer to this question?

Aileen Nguyen 3N
Posts: 26
Joined: Wed Sep 21, 2016 2:56 pm
Been upvoted: 1 time

### Re: Atomic Spectra- Question 1.15  [ENDORSED]

I solved this problem using the Rydberg equation instead of the bohr frequency condition.

You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1

Then use the Rydberg to solve for n2 which would give you n2=3

104822659
Posts: 23
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Atomic Spectra- Question 1.15

How did you guys know that n=1 in this problem?

Lauren_Bui_1A
Posts: 10
Joined: Sat Jul 09, 2016 3:00 am

### Re: Atomic Spectra- Question 1.15

In the question it states "In the ultraviolet spectrum of atomic hydrogen..."
The ultraviolet spectrum is the Lyman series, so n final=1.