Calculate Wavelength of 6th Line in Balmer Series

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SUMMARY

The wavelength of the sixth line in the hydrogen Balmer series is calculated using the Rydberg formula: 1 / λ = R * (1/n_final² - 1/n_initial²). The Rydberg constant R is 10973731.57 m^-1. The correct transition for the sixth line is from n=7 to n=2, resulting in a wavelength of approximately 410.070 nm. The confusion arose from misinterpreting the term "sixth line" as the sixth quantum level instead of the sixth observed transition.

PREREQUISITES
  • Understanding of the Rydberg formula for hydrogen spectral lines
  • Knowledge of quantum numbers and electron transitions
  • Familiarity with significant figures in scientific calculations
  • Basic concepts of electromagnetic radiation and wavelength
NEXT STEPS
  • Study the Rydberg formula in detail for various elements
  • Learn about quantum transitions in hydrogen and other elements
  • Explore the concept of spectral lines and their significance in spectroscopy
  • Practice calculating wavelengths for different transitions in the Balmer series
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Students studying quantum mechanics, physics educators, and anyone interested in atomic spectroscopy and the behavior of hydrogen's electron 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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