Prob 1.15 6th Ed.

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Rian Montagh 2K
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

Prob 1.15 6th Ed.

Postby Rian Montagh 2K » Fri Oct 19, 2018 10:58 pm

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 explain this problem with steps and intermediate numbers as well?

spark99
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Joined: Fri Sep 28, 2018 12:18 am
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Re: Prob 1.15 6th Ed.

Postby spark99 » Sat Oct 20, 2018 3:41 pm

First, you have to recognize that ultraviolet light is included in the Lyman series, which means its initial n is n=1. This can be found in your book. Then, you can use the Rydberg equation:
v=R((1/n1^2)-(1/n2^2)
Now that we now that its initial is n=1, you plug in 1 in n1.
v=R((1/1^2)-(1/n2^2)
In the problem, they give you the wavelength, being 102.26nm. You convert this value into meters, and you get 1.026E-7.
You use this wavelength to find the frequency, and you can plug this value into the Rydberg equation. Use the expession
v=speed of light/wavelength.
Your frequency should be 2.9E15.
Now your equation should look like:
2.9E15=R((1/1^2)-(1/n2^2)
Now you can just solve for n2, or final level of the electron.
Answer should be n=1 to n=3.

Sarah Bui 2L
Posts: 61
Joined: Fri Sep 28, 2018 12:29 am

Re: Prob 1.15 6th Ed.

Postby Sarah Bui 2L » Sat Oct 20, 2018 11:27 pm

How would you solve this using the equation Professor Lavelle showed us in class?


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