1A15

H-Atom ()

Moderators: Chem_Mod, Chem_Admin

Natallie K 3B
Posts: 61
Joined: Wed Sep 30, 2020 9:51 pm

1A15

Postby Natallie K 3B » Sat Oct 17, 2020 6:04 pm

1A.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 am really confused about how to solve this problem. I do not understand why I keep getting a non real answer. I am unsure how to approach it. I have looked at other solutions on chemistry community, but I don't understand them. Can someone please solve this problem step by step for me?

Faaizah Arshad 1H
Posts: 108
Joined: Wed Sep 30, 2020 10:01 pm
Been upvoted: 2 times

Re: 1A15

Postby Faaizah Arshad 1H » Sat Oct 17, 2020 6:27 pm

Hi Natallie,

I found this challenging the first time I did it as well. Here's a step by step method to solve it:

Step 1) When you read this problem the first time, you should be able to identify that we are discussing atomic spectroscopy in which a hydrogen atom is emitting a certain wavelength of light that appears as a spectral line. Thus, we'll be using Rydberg's constant and the frequency equation for that.

Step 2) Extract all information presented to you.

A. It tells you it's an ultraviolet spectrum, so we know that we are looking at a line in the Lyman series. The Lyman series is any electron transition that involves n = 1. Since the problem tells us this is an emission of energy, we know that the electron dropped from a higher to lower energy level, so n = 1 is the final state.

B. We are told that the line observed is associated with wavelength 102.6 nm. I would change this into frequency in order to use Rydberg's equation.

(2.998 x 10^8 m/s) / (1.026 x 10^-7m) = 2.922 x 10^15 Hz




Step 3) Plug in the information we know, and find the missing n initial value.



When you solve for x, you find that it is 3.

Answer: The initial energy level is n = 3, the final energy level is n = 1

Anil Chaganti 3L
Posts: 67
Joined: Wed Sep 30, 2020 9:52 pm

Re: 1A15

Postby Anil Chaganti 3L » Sat Oct 17, 2020 8:19 pm

Hi, I understand the above work, but how come we can't use the equation: . How come we have to use the frequency equation? Can we not use En=(-hR/n^2)?

Ellison Gonzales 1H
Posts: 63
Joined: Wed Sep 30, 2020 10:00 pm

Re: 1A15

Postby Ellison Gonzales 1H » Sun Oct 18, 2020 9:42 pm

Anil Chaganti 2C wrote:Hi, I understand the above work, but how come we can't use the equation: . How come we have to use the frequency equation? Can we not use En=(-hR/n^2)?


I didn’t use the equation you wrote, but I read from other explanations on chemistry community about this question, that you can use either equation. The only thing is that if you use the equation that you’re suggesting, you have to use it twice (their explanation seemed to be saying that the reason was because there are two 2 energy levels).

Ellison Gonzales 1H
Posts: 63
Joined: Wed Sep 30, 2020 10:00 pm

Re: 1A15

Postby Ellison Gonzales 1H » Sun Oct 18, 2020 9:48 pm

Faaizah Arshad 1E wrote:Hi Natallie,

I found this challenging the first time I did it as well. Here's a step by step method to solve it:

Step 1) When you read this problem the first time, you should be able to identify that we are discussing atomic spectroscopy in which a hydrogen atom is emitting a certain wavelength of light that appears as a spectral line. Thus, we'll be using Rydberg's constant and the frequency equation for that.

Step 2) Extract all information presented to you.

A. It tells you it's an ultraviolet spectrum, so we know that we are looking at a line in the Lyman series. The Lyman series is any electron transition that involves n = 1. Since the problem tells us this is an emission of energy, we know that the electron dropped from a higher to lower energy level, so n = 1 is the final state.

B. We are told that the line observed is associated with wavelength 102.6 nm. I would change this into frequency in order to use Rydberg's equation.

(2.998 x 10^8 m/s) / (1.026 x 10^-7m) = 2.922 x 10^15 Hz




Step 3) Plug in the information we know, and find the missing n initial value.



When you solve for x, you find that it is 3.

Answer: The initial energy level is n = 3, the final energy level is n = 1


Hi! I understand your explanation, thank you very much. However, the thing I am getting hung up on the most may be the simplest thing, and it’s the algebra to isolate n initial. I can’t seem to get the right answer even though I have all the correct numbers plugged into the equation. Thank you


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

Who is online

Users browsing this forum: No registered users and 1 guest