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

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Homework Help Overview

The discussion revolves around identifying the lower energy level associated with a blue emission in the hydrogen spectrum, specifically linked to a wavelength of 434 nm. The original poster attempts to apply the Rydberg formula to find the energy level from which the electron transitions.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Rydberg formula and the algebra involved in manipulating the equation. There are questions about the clarity of the original poster's algebraic steps and whether the calculations are being handled correctly. Some suggest working with variables instead of numbers to simplify the process.

Discussion Status

There is ongoing dialogue about the algebraic manipulation of the formula, with some participants offering guidance on how to approach the problem more clearly. The original poster expresses frustration at repeatedly arriving at the same incorrect conclusion, indicating a lack of resolution but a productive exchange of ideas.

Contextual Notes

Participants note that the emission in question belongs to the Balmer series, which involves transitions to the second energy level, suggesting that this context may influence the understanding of the problem.

Mitchtwitchita
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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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