## Problem 1A.15

$c=\lambda v$

Melody Haratian 2J
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### Problem 1A.15

I’m having a lot of trouble getting started with problem 1A.15 from the textbook. The problem is:
“ In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.”
Can someone help guide me through the first few steps of this problem? Thank you!

AnnaNovoselov1G
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### Re: Problem 1A.15

The UV spectrum of the hydrogen atom corresponds to the Lyman series, which represents a jump to the principle quantum energy level n=1.

first, you would find the frequency of the 102.6 nm wave using the equation: frequency= c/lambda.

Then, you can solve for the initial energy level (from which the electron went down to n=1) using the equation frequency= R (((1/(nf)^2) - 1/(ni)^2. Use R=3.29 x 10^15 and nf=1 (final energy level). You're trying to find n initial.

Astha Sahoo 3I
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### Re: Problem 1A.15

AnnaNovoselov1G wrote:The UV spectrum of the hydrogen atom corresponds to the Lyman series, which represents a jump to the principle quantum energy level n=1.

first, you would find the frequency of the 102.6 nm wave using the equation: frequency= c/lambda.

Then, you can solve for the initial energy level (from which the electron went down to n=1) using the equation frequency= R (((1/(nf)^2) - 1/(ni)^2. Use R=3.29 x 10^15 and nf=1 (final energy level). You're trying to find n initial.

Thanks so much for the explanation!I think I missed this in the reading, but how would you know that the hydrogen atom's UV spectrum corresponds to the Lyman series?

Katie Lam 2J
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### Re: Problem 1A.15

Hi Astha! In the Atomic Spectrum audio-visual module video, Dr. Lavelle drew out the hydrogen atom spectrum, and the UV spectrum is named the Lyman series. The visible light spectrum is also named the Balmer series. Hope this helps!