Calculate Wavelength of 6th Line in Balmer Series

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Homework Help Overview

The problem involves calculating the wavelength of the sixth line in the hydrogen Balmer series, which pertains to atomic physics and spectral lines. Participants are tasked with applying the Rydberg formula to determine the wavelength based on specific quantum transitions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the Balmer formula and the calculations performed. There is confusion regarding the identification of the "sixth line," with some participants questioning whether it refers to the sixth observed transition or the sixth quantum level.

Discussion Status

The discussion is ongoing, with participants exploring the correct interpretation of the problem. Some have provided insights into the transitions involved in the Balmer series, indicating a productive direction in clarifying the original poster's misunderstanding.

Contextual Notes

There is a noted ambiguity regarding the definition of the "sixth line," which has led to confusion in the calculations. Participants are reflecting on the importance of precise terminology in the context of quantum transitions.

JJK1503
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Homework Statement



Compute to three significant figures the wavelength of the sixth line in the hydrogen Balmer series.

Homework Equations



1 / lambda = R * ( 1/ n_final^2 - 1 / n_initial^2 )
R = Rydberg constant = 10973731.57 m^-1
lambda = wavelength

The Attempt at a Solution



I set up the Balmer formula I listed above given the info in the problem

1 / lambda = (10973731.57 m^-1) * ( 1/ 2^2 - 1 / 6^2 ) = 1 / 2438607.015555 m^-1

so, lambda = 410.070 nm
or 4.10 x 10^-7 m using 3 sig figs as the question asks for.

Not sure what I am doing wrong. This seems right to me, but it is not.

Any help is appreciated.
 
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JJK1503 said:

Homework Statement



Compute to three significant figures the wavelength of the sixth line in the hydrogen Balmer series.

Homework Equations



1 / lambda = R * ( 1/ n_final^2 - 1 / n_initial^2 )
R = Rydberg constant = 10973731.57 m^-1
lambda = wavelength

The Attempt at a Solution



I set up the Balmer formula I listed above given the info in the problem

1 / lambda = (10973731.57 m^-1) * ( 1/ 2^2 - 1 / 6^2 ) = 1 / 2438607.015555 m^-1

so, lambda = 410.070 nm
or 4.10 x 10^-7 m using 3 sig figs as the question asks for.

Not sure what I am doing wrong. This seems right to me, but it is not.

Any help is appreciated.

The sixth line means the sixth observed line.

Here are the transisitons:
n_i --> n_f
3--> 2 (first line)
4--> 2 (second line)
5--> 2 (third line)
...
Which is the sixth line?
 
Quantum Defect said:
The sixth line means the sixth observed line.

Here are the transisitons:
n_i --> n_f
3--> 2 (first line)
4--> 2 (second line)
5--> 2 (third line)
...
Which is the sixth line?

And the hand meets the forehead...
It states the 6th line of the Balmer series not the sixth quantum level.
Thank you for your help, I had a feeling it was something simple.
 
JJK1503 said:
And the hand meets the forehead...
It states the 6th line of the Balmer series not the sixth quantum level.
Thank you for your help, I had a feeling it was something simple.

We have all done things like this. :wink:
 

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