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Problem 20 Easy Difficulty

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the $n$ = 3 level. Determine the wavelength and frequency of the photon.

Answer

$$=2.92 \times 10^{15} \mathrm{Hz}$$

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Meghan M.

McMaster University

Video Transcript

okay, In this problem, we have hydrogen, which is excited from its ground state and equals one to n equals three state. We're asked, and after reaching this excited state event, relaxes down and releases a photon, you're asked to find what the wavelength and frequency of this resulting photons would be. So for the equations we need to know, we can use the simple equation for the energy levels of hydrogen. So for any end one, two, three, four imager natural members, we can find resulting energy at that level. We also need to know how the energy of a photonic relates to its frequency and wavelength. Let's jump into it. It's a relatively simple process. So when going from n equals one to the end equals three state the change in energy he three minus the one would simply be You have thirteen point six TV over three squared minus negative, thirteen point six TV, one squared, combining everything. We have thirteen point six TV times nine nine ce minus eight nights or thirteen point six TV divided by nine. Which gives us the results. Or I should say, I just want to feel that too far to you one night. Which, of course, gives us thirteen point six E. V eight nine What? Which gives us approximately twelve point zero nine TV says the energy difference between the two states, it's that's a photon omitted would be this energy twelve point zero nine TV so that we know this energy. We can then take our expression equals H f self ref equals H. You can plug in twelve point zero nine TV. We can use planks constant in its electron volts form four point one five times ten the negative fifteen evey seconds. We will get a frequency to be relatively small. Yeah, actually, relatively large. Is that two point seven eight times ten to the fifteen arts cool, similarly equals H C over Lambda. So solving for Lambda Lambda equals HC Every we played in our numbers, we will have in resulting wave length of one point zero two six to attend the *** seven meters, which is about a tenth of a microt meter. That's it. Thank you very much.

University of Washington
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Meghan M.

McMaster University