Help on A1.15

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Gabriella Bates 2L
Posts: 94
Joined: Thu Jul 11, 2019 12:15 am

Help on A1.15

Postby Gabriella Bates 2L » Thu Oct 10, 2019 4:00 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 please walk me through this problem using Professor Lavelle's method? I've gone through it two times and I haven't been able to arrive at the correct answer. Thank you!

Kavya Immadi 3D
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Joined: Sat Aug 24, 2019 12:17 am
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Re: Help on A1.15

Postby Kavya Immadi 3D » Thu Oct 10, 2019 4:21 pm

Image


c is 2.998 * 108 m/s
Convert the wavelength to meters (nm-->m)
R is 3.29 * 1015 s-1
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20191010_161438.jpg

Gabriella Bates 2L
Posts: 94
Joined: Thu Jul 11, 2019 12:15 am

Re: Help on A1.15

Postby Gabriella Bates 2L » Thu Oct 10, 2019 5:15 pm

Thank you, but can anyone explain how to do this using Lavelle's method? He doesn't use the Rydberg equation, he uses the change in energy.

Victoria Zheng--2F
Posts: 87
Joined: Fri Aug 09, 2019 12:17 am

Re: Help on A1.15

Postby Victoria Zheng--2F » Fri Oct 11, 2019 10:22 am

To use the change in energy method to solve this question, you would first calculate the frequency of the light using its wavelength, which involves the equation c=λv. This would give you the change in energy of the transition from the initial energy level to the final energy level.
The next step is to set the change in energy equal to the energy of the final level minus the energy of the initial level.
This should give you the equation ΔE=(hR/nf^2)-(hR/ni^2). Then, you can rearrange the equation to be ΔE=hR(1/nf^2-1/ni^2).
The next step is to divide ΔE by hR to give you the value of (1/nf^2-1/ni^2). Then you convert the value to fractions and find nf and ni. I hope this is helpful.

Ashley Tran 2I
Posts: 95
Joined: Thu Jul 11, 2019 12:17 am

Re: Help on A1.15

Postby Ashley Tran 2I » Fri Oct 11, 2019 10:28 am

In the textbook there is a small paragraph that says in the UV region of the spectrum, n1 is equal to 1. From there you can solve for n2.


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