## Why are they using the energy levels like this?

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

luca rutigliani 3g
Posts: 4
Joined: Wed Sep 28, 2016 3:01 am

### Why are they using the energy levels like this?

"in the ultraviolet spectrum of hydrogen, a line is observed at 102.6nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line"

I get that you have to use (-hr/n^2)-(-hr/n^2). but shouldn't it be final - initial? so in this case the first n should be 1 and the second one is the unknown? because you get the answer only if you set the "final" n as the unknown and the initial n as n=1. why?? isn't ultraviolet produced when you go from a the higher energy level to the lower energy level??

thanks

Kareem
Posts: 18
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Why are they using the energy levels like this?

The question states that we observe this spectrum during the emission of light, which means that the electron is giving off energy, ergo it is moving from a higher energy level to a lower energy level. If it was going from low to high it would be absorbing energy rather than giving it off. This is why the initial is the unknown and the final is n=1.

luca rutigliani 3g
Posts: 4
Joined: Wed Sep 28, 2016 3:01 am

### Re: Why are they using the energy levels like this?

Exactly! but they are using n=1 as initial and the unknown n as the final, which is the opposite!!

Manpreet Singh 1N
Posts: 41
Joined: Wed Sep 21, 2016 2:59 pm
Been upvoted: 1 time

### Re: Why are they using the energy levels like this?

In this question, it talks about the UV spectrum, which falls under the Lyman series. The Lyman series has a n final of 1. On the other hand, the Balmer series (visible light) has a n final of 2.

Hope this helps :)