Atomic Spectra Module Q28

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Patricia Macalalag 2E
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

Atomic Spectra Module Q28

Postby Patricia Macalalag 2E » Wed Oct 18, 2017 10:48 pm

How would you approach this problem from the modules?

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?

Erik Khong 2E
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: Atomic Spectra Module Q28

Postby Erik Khong 2E » Wed Oct 18, 2017 10:56 pm

So one meter fits 1,650,763.73 wavelengths. This means that it's a very high frequency wave (there's a LOT of squiggles in one meter). To find the wavelength of krypton-86 radiation, you just need to simply divide 1 meter by 1,650,763.73. This equals to 6.058 x 10^-7 m (simplified to 605.8 nm). So one wavelength of this super high frequency wave is in fact very small.

Justin Chang 2K
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

Re: Atomic Spectra Module Q28

Postby Justin Chang 2K » Fri Oct 20, 2017 1:19 am

Yep! I did:

1m/1,650,763.73 wavelengths = x m/1 wavelength in order to figure out the length of only one wavelength, if looking at the mathematical representation helps at all!

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