What is the lower energy level for a blue emission in the hydrogen spectrum?

AI Thread Summary
The discussion centers on determining the lower energy level for a blue emission in the hydrogen spectrum, specifically for a wavelength of 434 nm. The emission occurs when an electron transitions from the fifth energy level to a lower one, but participants struggle with algebraic manipulation in their calculations. It is clarified that visible emissions correspond to the Balmer series, which transitions to the second shell. Participants emphasize the importance of simplifying algebraic steps and using variables instead of numbers to avoid confusion. Ultimately, the correct lower energy level for the blue emission is identified as the second energy level.
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Homework Statement



The hydrogen spectrum contains a blue line with a wavelength of 434 nm. Photons of blue light are emitted when hydrogen's electron drops from the fifth energy level to a lower energy level. What is the lower energy level?

Homework Equations



1/lambda = R[1/m e2 - 1/n e2]


The Attempt at a Solution



1/4.34 x 10 e-7 m = 1.097 x 10 e7 m e-1 [1/m e2 - 1/5 e2]
1/4.34 x 10 e-7 m [1/m e2 - 1/25] = 1/1.097 x 10 e7 m
1/[1/m e2 - 1/25] = 4.34 x 10 e-7m/1.097 x 10 e7
1/m e2 - 1/25 = 1.097 x 10 e7/4.34 x 10 e-7 + 1/25
n e2 = 4.34 x 10 e-7/1.097 x 10 e7 + 25/1
=3.96 x 10 e-14 + 25
=25
n = 5

Grrrrr! I keep getting n = 5 and I don't know what I'm doing wrong. Can anybody please help me?
 
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here, the "m" is the shell number you want, and the "n" is the shell it came from, in this case 5.

Edit: No, that's what you did. Um...

When you write out "1/m e2" do you mean "1/m^2"? No...

Your math is difficult to follow. I'm thinking you are making the algebra harder than it needs to be. Your formula is correct and you should get the right results; the problem here is algebra.

Try solving the problem without cramming in all the numbers first.

ARe you required to work out the math for this? Does the fact that this photon is visible provide enough info?
 
Last edited:
I did mean 1/m e1 = 1/m^2

I guess my algebra is off but I've tried it several times and each time I keep getting 5. Do you have any suggestions?
 
In what you wrote, your second line is all wrong.

Again, work out the algebraic manipulation using the variables, not the numbers, it's less confusing.
 
Aaaaah, I got it now. Thanks Chi!
 
Just wanted to point out, that if a hydrogen emission is visible (with only one red exception) then it is an emission in the "Balmer series," all of which are from transitions to the second shell.

Emissions that are UV, are the "Lyman series" which transition to level 1, and IR emissions are "Paschen series" which transition to level 3. The highest energy paschen emission is barely visible red.
 
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