## Atomic Spectra Post-Module Assessment #28

$E=hv$

Sofia Lucido 3L
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### Atomic Spectra Post-Module Assessment #28

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 $\lambda =\frac{c}{\nu }$ to find the wavelength and $E=\nu h$ 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
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### Re: Atomic Spectra Post-Module Assessment #28

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.

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### Re: Atomic Spectra Post-Module Assessment #28

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 $c=\lambda\nu$ 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: $\frac{1m}{1,650,763.73 wavelengths}$. 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 $c=\lambda\nu$ to obtain the frequency of radiation from the wavelength.

4) I then used the equation $E=h\nu$ in order to obtain the energy from the frequency obtained in step 3.

Hope this helps!