Atomic Spectra Post-Module Assessment #28


Moderators: Chem_Mod, Chem_Admin

Sofia Lucido 3L
Posts: 100
Joined: Wed Sep 30, 2020 9:51 pm
Been upvoted: 1 time

Atomic Spectra Post-Module Assessment #28

Postby Sofia Lucido 3L » Wed Oct 07, 2020 10:35 am

The meter was defined in 1963 as 1,650,763.73 wavelengths of radiation emitted by krypton-86 (it has since been redefined). What is the wavelength of this krypton-86 radiation? To what region of the electromagnetic spectrum does this wavelength correspond (i.e. infrared, ultraviolet, x-ray, etc.)? What energy does one photon of this radiation have?

I am not sure how to do this problem. I tried using the equation to find the wavelength and to find the energy per photon.


I used the definition of a meter in the first sentence of the question as the frequency, but I have a feeling that isn't right. Can someone walk me through the steps of how to approach this problem?

Thanks!

Chem_Mod
Posts: 19551
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 882 times

Re: Atomic Spectra Post-Module Assessment #28

Postby Chem_Mod » Wed Oct 07, 2020 10:49 am

Since there is 1,650,763.73 wavelengths in one meter, each wavelength is 1/1,650,763.73 meters. You should get around 606 nm, which is in the visible light region. Since
wavelength= c/v and E=vh, you can use substitution to get the equation E=(hc)/wavelength. h and c are constants, and you solved for the wavelength, so you can get the energy per photon.

Nika Gladkov 1A
Posts: 114
Joined: Wed Sep 30, 2020 9:33 pm
Been upvoted: 6 times

Re: Atomic Spectra Post-Module Assessment #28

Postby Nika Gladkov 1A » Wed Oct 07, 2020 11:03 am

1)Since one meter contains 1,650,763.73 wavelength of krypton, to find the wavelength I divided 1m by 1,650,763.73. The equation is not really applicable in this case because you are not given the frequency; you are instead given a definition in terms of meters. It was helpful for me to imagine it as a proportion: . From this proportion, you can see that, to obtain one wavelength, you would need to divide the top and bottom by 1,650,763.73.

2) I then converted my answer to nm in order to identify what region of the electromagnetic spectrum it belongs to and referenced the wavelength ranges that Dr. Lavelle provided in the module in order to certain what type of radiation it was.

3) Next, I used the equation to obtain the frequency of radiation from the wavelength.

4) I then used the equation in order to obtain the energy from the frequency obtained in step 3.

Hope this helps!


Return to “Einstein Equation”

Who is online

Users browsing this forum: No registered users and 1 guest