Calculate Wavelength of 6th Line in Balmer Series

AI Thread Summary
The discussion revolves around calculating the wavelength of the sixth line in the hydrogen Balmer series using the Rydberg formula. The user initially set up the calculation correctly but misunderstood the definition of the "sixth line," confusing it with the sixth quantum level. Clarification revealed that the sixth line refers to the sixth observed transition, which is from n=8 to n=2. The correct wavelength calculation, once the proper transition is identified, is crucial for accuracy. This highlights the importance of understanding terminology in physics problems.
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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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