Question 1.15 on homework

[Sidney Kantono 3C]

"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."

Can anyone explain how to do this question? I noticed that there was an earlier post about this, but the answer to it still did not clarify it for me. Thank you!

[Chem_Mod]

The fact that the emission is ultraviolet tells you that this transition is in the Lyman series, so n₂=1

Convert 102.6 nm into the energy E using the two relations: and

Then plug into the rydberg formula and solve for n₁

[Hank Wang 2J]

Why does the fact that it is in the Lyman series make n2=1?

[Alex Nguyen 3I]

Spectral lines in the same series always end with the same n2 or final energy level. The Lyman series happens to contain all energy transitions that end with n=1. The initial energy level could be any energy level greater than 1 though. For the hydrogen atom, the electron that drops to n=1 will emit UV radiation that can range from any of the lines/wavelengths in the Lyman series.

[Alex Nguyen 3I]

According to the Rydberg Formula, n1 is actually the final energy level I guess you could say. Therefore solving will only work if you designate n1 as being equal to 1. That's just the way the formula is set up. I'm pretty sure n=1 is still the final energy level though.

[Ana Pedreros]

My way to approach this problem was to:
1. figure out the frequency using the wavelength of 102.6 nm and the formula v=c/lambda
2. then use the frequency and plug it into Rydberg formula
3. since the problem states its in the UV spectrum the initial energy should be 1 leaving n2 as the only undefined variable
4. distribute the R constant and solve for n2 and you should get 3

[Nick Lewis 4F]

I had a bit of trouble with this problem at first, but I looked at the problem on pg.7&8 in the textbook as an example. It goes from n1 and n2 to the wavelength of the emitted radiation. My only trouble in going back was how do we know n1=1? Otherwise it was easy to use the equations c = lambda * v and v = R(1/n1^2-1/n2^2)

[Justin Quan 4I]

I had a bit of trouble with this problem at first, but I looked at the problem on pg.7&8 in the textbook as an example. It goes from n1 and n2 to the wavelength of the emitted radiation. My only trouble in going back was how do we know n1=1? Otherwise it was easy to use the equations c = lambda * v and v = R(1/n1^2-1/n2^2)

You know n1=1 because the problem tells you the that it was an ultraviolet wave. Ultraviolet has a n value of 1

[Camille 4I]

How can you tell that the transition is in the Lyman series from the fact that it is in the ultraviolet spectrum of atomic hydrogen?

[Aprice_1J]

When the question says that it is ultraviolet, we know that the line is observed at n=1. If there is a final n value of 3, it is paschen series, if it is 2 it is balmer and if it is 1 it is lynman. Because n final is 1, we can then say that it is lynman series.

[Aman Sankineni 2L]

How can you tell that the transition is in the Lyman series from the fact that it is in the ultraviolet spectrum of atomic hydrogen?

Lyman series are the only series found in the ultraviolet spectrum of atomic hydrogen. By knowing that the transition is in the ultraviolet spectrum allows us to know that it is a Lyman series.

[Camille 4I]

How would you solve this problem using the equation En = -hR/n^2? This is the equation included on the Constants and Equations sheet.