Question 1.15 sixth edition

H-Atom ()

Moderators: Chem_Mod, Chem_Admin

Heesu_Kim_1F
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am

Question 1.15 sixth edition

Postby Heesu_Kim_1F » Mon Oct 15, 2018 5:10 pm

1.15) 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.
I know that problem 15 is just the opposite of problem 13, which I solved, yet I don't understand how I can find the values of n for both the initial and final energy levels of the electron.
Here's what I have so far. Wavelength was 102.6 nm = 102.6x10^-9 m. I used the c=(wavelength)(velocity) equation to get 2.92x10^15 s^-1 as the velocity. Then I used the E=hv equation to get 1.93x10^-18 J as the energy change. What do I need to do after?
Thank you in advance!

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

Re: Question 1.15 sixth edition

Postby Tam To 1B » Mon Oct 15, 2018 5:41 pm

First, I think you mean that the frequency you got is 2.92E15 Hz (s^-1)
On page 7 of Section 1.3, there is a figure that shows 102.6 nm is part of the Lyman series, which is on n = 1, so you know that the first energy level is n = 1. In order to find what it transitions to, you can use the equation v (frequency) = R[(1/n^2) - (1/n^2)] where the first n in the equation is initial energy level and the second n is final energy level. R = Rydberg's constant 3.29E15 Hz.
Now that you got the frequency and initial energy level n = 1, you can plug it into the equation 2.92E15 Hz = 3.29E15 Hz [ (1/1^2) - (1/n^2)] and solve for n to get final n = 3.

Fanny Lee 2K
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

Re: Question 1.15 sixth edition

Postby Fanny Lee 2K » Mon Oct 15, 2018 5:47 pm

To find the n values of the orbitals, you need to use the Rydberg equation: v=R((1/n(Initial)^2)-(1/n(Final)^2). First you need to find the frequency which you calculated as 2.922 x 10^15 s^-1. Plug in the frequency and n(initial) as 1. solve for n(final)^2 which should equal to 9. If we want to find the n value, we would square root 9 which leaves us with n = 3. Therefore, the transition is from n =1 to n=3.

Lucy Agnew 3E
Posts: 30
Joined: Fri Sep 28, 2018 12:26 am

Re: Question 1.15 sixth edition

Postby Lucy Agnew 3E » Tue Oct 16, 2018 6:58 pm

Very helpful thank you! I got stuck on this problem too!

Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
Been upvoted: 1 time

Re: Question 1.15 sixth edition

Postby Nicolette_Canlian_2L » Tue Oct 16, 2018 9:11 pm

why is the energy level of 102.6 nm 1?

Isabel Nakoud 4D
Posts: 101
Joined: Fri Sep 28, 2018 12:27 am

Re: Question 1.15 sixth edition

Postby Isabel Nakoud 4D » Tue Oct 16, 2018 10:26 pm

Nicolette_Canlian_3G wrote:why is the energy level of 102.6 nm 1?


This is what i don't understand. Even when I see pictures of the spectroscopy with the Lyman series, I notice that 102.6 is one of the values next to a line, but I'm not sure how to identify that it is n=1 .

Isabel Nakoud 4D
Posts: 101
Joined: Fri Sep 28, 2018 12:27 am

Re: Question 1.15 sixth edition

Postby Isabel Nakoud 4D » Tue Oct 16, 2018 10:31 pm

Also, the solution to this problem says that the transition is from n = 1 to n = 3, but the problem says that energy was emitted during this process. Isn't energy only emitted when an electron transitions from a higher energy to a lower energy? So... if it transitioned from n =1 to n=3, it should have absorbed energy. ?

Jeannine 1I
Posts: 73
Joined: Fri Sep 28, 2018 12:27 am

Re: Question 1.15 sixth edition

Postby Jeannine 1I » Tue Oct 16, 2018 10:37 pm

Nicolette_Canlian_3G wrote:why is the energy level of 102.6 nm 1?


My reasoning for why n1=1, is that the 102.6 nm is a part of the Lyman series, and a general rule in the Lyman series is the first energy level (n1) is always equal to 1, so n1=1. I'm not sure if that line where 102.6 nm is is n1, but that was just my reasoning. I hope that helps at least a little bit(:

Leela_Mohan3L
Posts: 44
Joined: Fri Sep 28, 2018 12:26 am

Re: Question 1.15 sixth edition

Postby Leela_Mohan3L » Tue Oct 16, 2018 10:41 pm

The Lyman series refers to the wavelengths of light that are absorbed by hydrogen in the Ultraviolet spectrum. The principal quantum number for this series is n=1, meaning that whenever one of the specific wavelengths in this series is absorbed by hydrogen, an electron will jump from the n=1 energy level to n= ___ energy level (or fall from the n= ___ energy level to the n=1 energy level. It works either way!).

Since the problem tells you that a 102.6 nm is part of the Ultraviolet spectrum of hydrogen, you know that it is referring to the Lyman series. The Lyman series has a principle quantum number of n=1 (This is just something that you should memorize), so your initial energy level would be n=1. Then just solve the problem as the people who commented before me stated :)

I hope this clears up any confusion!

Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
Been upvoted: 1 time

Re: Question 1.15 sixth edition

Postby Nicolette_Canlian_2L » Sun Oct 21, 2018 7:33 pm

Isabel Nakoud 4D wrote:
Nicolette_Canlian_3G wrote:why is the energy level of 102.6 nm 1?


This is what i don't understand. Even when I see pictures of the spectroscopy with the Lyman series, I notice that 102.6 is one of the values next to a line, but I'm not sure how to identify that it is n=1 .


I just realized that we have to use an equation to solve for the energy level. I assumed that the energy level had to be determined just by looking at the Lyman series spectroscopy.

Isabel Nakoud 4D
Posts: 101
Joined: Fri Sep 28, 2018 12:27 am

Re: Question 1.15 sixth edition

Postby Isabel Nakoud 4D » Tue Oct 23, 2018 2:44 pm

Which equation do we use to figure this out?


Return to “Bohr Frequency Condition, H-Atom , Atomic Spectroscopy”

Who is online

Users browsing this forum: No registered users and 1 guest