Excited hydrogen atom and wavelength

AI Thread Summary
A hydrogen atom excited to the n=5 level transitions to n=2, emitting photons in the process. The energy difference between these levels is calculated as 2.86 eV, which corresponds to the energy of the emitted photon. Using the equation E=(hc)/lambda, the wavelength can be determined from this energy. The discussion also considers whether any emitted light falls within the visible spectrum. The calculations indicate that at least one of the emitted wavelengths is indeed in the visible region.
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i have been working on this question for quite a while. no help on google, or my textbook at all.

A hydrogen atom in its ground state is excited to the n=5 level. It then makes a transition directly to the n=2 level before returning to the ground state.

A) What are the wavelengths of the emitted phontons?

B)Would any of the emmitted light be in the visible region?

I have searched through my textbook and there is no examples to work off of nor much information on this topic really. so if anyone could help that would be greatly appreciated.
 
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This is my guess at how to do this question (I'm referring to my high school notes for this...)

Basically, you can use the good old E=(hc)/lambda.
Wavelength (lambda) will equal hc/E.

Now, if you look at this diagram:
[PLAIN]http://castlelearning.com/review/reference/phys6.gif

you can see that when n=5, E of electron is -0.54eV. At n=2, E=-3.40eV.
When the electron moves from n=5 to n=2, it loses Ef-Ei=-3.40-(-0.54) eV=-2.86eV
That means that the photon is emitted with that same energy (conservation of energy). So, Ep=2.86eV.
From which, you can find the wavelength using equation I stated above.

I hope this helps :)
 
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