## HW Q

$c=\lambda v$

sharonvivianv
Posts: 63
Joined: Fri Apr 06, 2018 11:05 am

### HW Q

In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. 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?

For this question I would think that the final energy level is n=1 since the line is observed at 102.6nm but it is actually the initial energy level. Why is that?

OrAmar-1L
Posts: 35
Joined: Fri Sep 29, 2017 7:04 am

### Re: HW Q

I believe you would need more information, because the equation will leave you with two unknown variables (n1 and n2). With the given wavelength, you can calculate for frequency, but you would still need to know the initial or final energy level to solve for the other. The word "emission" does hint that the energy goes from higher to lower, though.

CORRECTION: Apparently, you know that the lower energy level is n=1 because the ultraviolet spectrum of atomic hydrogen corresponds to the Lymann series (which has n=1). I had to check the solution manual as well/:

arina_m 1A
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

### Re: HW Q

When identifying wavelengths with the Balmer and Lyman series we can always assume that the n1 for the Balmer series will be n1=2 and for the Lyman series it will be n1=1 this identifies the starting energy levels.

From the book:
"The Balmer series consists of the lines with n1=2 (and n2 = 3,4,...)."
"The Lyman series is a set of lines in the ultraviolet region of the spectrum with n1 = 1(and n2 = 2,3,...)."

Hope it helps!

Lenaschelzig1C
Posts: 21
Joined: Fri Apr 06, 2018 11:05 am

### Re: HW Q

Because you know all of the unknown variables except for the final n value, you can solve for it!